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MATHEMATICAL TEXT-BOOKS
BY
GEORGE A. WENTWORTH
Mental Arithmetic
Elementary Arithmetic
Practical Arithmetic
Primary Arithmetic (Wentworth and Reed)
Grammar School Arithmetic
Advanced Arithmetic
First Steps in Algebra
New School Algebra
School Algebra
Elements of Algebra (Revised Edition)
Shorter Course in Algebra
Complete Algebra
Higher Algebra
College Algebra (Revised Edition)
First Steps in Geometry (Wentworth and Hill)
Plane Geometry (Revised)
Solid Geometry (Revised)
Plane and Solid Geometry (Revised)
Syllabus of Geometry
Geometrical Exercises
Analytic Geometry
Plane and Solid Geometry and Plane Trigonometry
(Second Revised Edition i
Plane Trigonometry (Second Revised Edition)
Plane Trigonometry and Tables (Second Revised
Edition)
Plane and Spherical Trigonometry (Second Revised
Edition)
Plane and Spherical Trigonometry and Tables
(Second Revised Edition)
Plane and Spherical Trigonometry, Surveying, and
Navigation (Second Revised Edition)
Surveying and Tables (Second Revised Edition)
Plane Trigonometry, Surveying, and Tables (Second
Revised Editioni
Plane and Spherical Trigonometry, Surveying, and
Tables (Second Revised Edition)
Logarithms, Metric Measures, etc.
ELEMENTS
Analytic Geometry
G. A. WENTWUKTH, A.M.
ACTHOE OF A SKKIKS OK TEXT-BOOKS IN MATHEMATICS
GINN & COMPANY
BOSTON ■ NEW YORK • CHICAGO • LONDON
Copyright, 1886, by
GEORGE A. WENTWORTH
ALL RIGHTS RESERVED
G8.10
Che gtftengum ^regg
GINN & COMPANY • PRO-
FRIETORS • BOSTON • U.S.A.
NOTE TO THE SECOND EDITION.
In this edition such changes have been made as actual experi-
ence in the class-room has shown to be desirable.
A chapter on Higher Plane Curves, and four chapters on Solid
Geometry have been added, making the work sufficiently extensive
for our best schools and colleges.
An effort has been made to have this edition free from errors.
It is not likely, however, that this effort has been entirely success-
ful, and the author will be very grateful to any reader who will
notify him of any needed corrections.
NOTE TO THE EDITION OF 1898.
The old plates have become so worn that it is necessary to have
new plates. A few verbal changes liave been made. No changes,
however, have been made that will prevent the using of old and
new books together.
March, 1898. G. A WENTWORTH.
PREFACE.
o>Ko
rpmS book is intended for beginners. As beginners
-*~ generally find great diiticulty in comprehending the
connection between a locus and its equation, the opening
chapter is devoted maiidy to an attempt, by means of easy
illustrations and exam[)les, to niaki; this connection clear.
Each chapter abounds in exercises ; for it is only by
solving })roblems which require some degree of original
thought that any real mastery of the study can be gained.
The more difficult propositions have been })ut at the ends
of the chapters, under the heading of " Supplementary
Propositions." This arrangement makes it possible for
every teacher to mark out his own course. The simplest
course will be Chapters I. -ill. and Chapters V.-VII., with
Review Exercises and Supplementary Propositions left
out. Between this course and the entire work the teacher
can exercise his choice, and take just so much as time and
circumstances will allow.
The author has gathered his materials from many sources,
but he is particularly indebted to tlie English treatise of
Charles Smith. Special acknowledgment is due to ii. A.
Hill, A.M., of Cambridge, Mass., and to Prof. J. M.
Taylor, Colgate University, Hamilton, New York, for
assistance in the preparation of the book.
Corrections and suggestions will be thankfully received.
G. A. WENTWORTH.
Exeter, N.H., January, 1888.
O O IsT T E N T S.
ol»<c
PART I. PLANE GEOMETEY.
Chapter I. Loci and their Equations.
BECTION
1. Quadrants .....
2. Algebraic Signs ....
3. Axes of Coordinates
4. Rectilinear System of Coordinates
5. Circular Measure of an Angle
6-7. Distance between Two Points
8-9. Division of a Line .
10-16. Constants and Variables
17-24. Locus of an Equation .
25. Definitions
26. Intercepts of a Curve
27. Intersections of Two Curves
28. Curve Passing through the Origin .
29. Equation having no Constant Term
30. Construction of Straight Line and Circle
31-34. Constructions of Loci of Given E<iuations
35. Equation of a Curve ....
Review Exercises
page
1
2
2
3
5
6
8
10
14
22
22
22
23
23
25
25
31
33
Chapter II. The Straight Line.
36. Notation
37-39. Equations of the Straight Line
40. Symmetrical Equation of the Straight Line
41. Normal Equation of the Straight Line .
42-43. General Equation of the First Degree .
36
36
38
39
43
VI
CONTENTS.
SECTION
44.
45.
46.
47.
48-49.
50.
Locus of the First Order .....
Angle Formed by Two Lines ....
Equati(Mis of Parallels and Perpendiculars .
Equation of Line making given Angle with a Line
Distance from a Point to a Line ....
Area of a Triangle
Review Exercises
PAGE
43
45
46
46
50
54
M
Supplementary Propositions.
51-52. Equation of a Line from a Point to tlie Intersection of
Two Lines
53. Condition that Three Lines meet in a Point .
54. Equation of the Bisector of an Angle .
55. Homogeneous Equation of the nlh Degree .
56. Angles between the Two Lines Ax- + Cxy + By^ = 0
57. Condition that a Quadratic represents Two Straight Lines
58. Problems on Loci involving Three Variables.
Chapter III. The Circle.
59-00. Equations of the Circle
61. Condition that a Quadratic represents a Circle
62. Condition that a Point is without, on, or within a
63. Tangents, Normals, Subtangents, Subnormals
64. Equation of a Tangent to the Circle x- + ij- — r^
(55. P^quation of a Normal through the Point (xi, ii\)
66. Equations of the Tangent and Normal to the
(X — a)- + (?/ — 6)2 = r2 .
67. Condition that a Straight Line touches a Circle
Review Exercises
Circle
Circle
Supplementary Propositions.
68. Diameter, Chords of a Diameter ....
69. Equation of a Diameter of the Circle x^ + y- = r^
70. Condition of Two, One, or No Tangents to a Circle
71. Equation of the Chord of Contact ....
72. Pole and Polar. Equation of Polar
73. Pole and Polar of a Circle .....
CONTENTS.
SECTIOK PAOE
74. Relations of Poles and Polars
75. Geometrical Construction of a Polar to a Circle
70. Length of Tangent from a given Point .
77. Radical Axis of Two Circles ....
78. Radical Centre of Two Circles
Chapter IV. Different Systems of Coordinates.
79-81. Rectilinear and Oblique Systems .
82. Polar System ....
83. Polar Equation of the Circle .
84. Transformation of Coordinates
85. New Axes parallel to Old Axes
86. From One Set of Rectangular Axes to Another Set
87. From One Set of Rectangular Axes to Another Set
with Different ( )rigin
88. From Rectangular to ( )blique Axes
89. From Rectangular to Polar Coordinates
90. From Polar to Rectangular Coiirdinates
91. Degree of an Equation not Altered by Transformation
Review Exercises
Chapter V. The Parabola.
92. Simple Proprieties of the Parabola
93. Construction of a Parabola
94. Principal Equation of the Parabola
95. Parabola Symmetrical with Respect to the Axis .
96. Condition that a I'oint is without, on, or within
Parabola .......
97. Latus Rectum a Third Proportional to any Abscissa
and Corresponding Ordinate ....
98. Squares of Ordinates of Two Points are as Abscissas
99. Points in which a Straight Line meets a Parabola.
100. Equations of Tangents and Normals
101. Subtangent and Subnormal
102. Tangent makes E(iual Angles with the Axis and Focal
Radius
Review Exercises
!):5
94
05
05
06
99
101
103
105
105
100
107
107
108
100
109
111
113
113
114
115
115
lie.
lit!
110
no
no
1-20
123
VUl CONTEXTS.
SCPPLEMENTARV PhOPOSITIONS.
SECTION PAGE
103. Condition of Two, One, or No Tangents to a Parabola . 126
104. Equation of tlie Cliord of Contact 127
105. Equation of the Polar with respect to the Parabola . 127
106. Equation of a Diameter of the Parabola . . . 128
107. Tangent through End of a Diameter Parallel to Chords
of Diameter ........ 129
108. Perpendicular from Focus to a Chord, also from Focus
to a Tangent 129
109. Tangents through the Ends of a Chord . . . .130
110. Locus of Foot of Perpendicular from Focus to a Tangent 130
111. Points from which each Point in Tangent is Equidistant 130
112. Tangents at Right Angles intersect in Directrix . . 130
113. Polar of the Focus 131
114. Equation of the Parabola, Axes being Diameter and
Tangent through its End 131
115. Polar Equation of the Parabola 133
Chapter VI. The Ellipse.
116. Simple Properties of the Ellipse 136
117. Construction of an Ellipse 136
118. Transverse and Conjugate Axes 137
119. Equation of the Ellipse 138
120. Characteristics of the Curve learned from its Equation 139
121. Change in the Form of the Ellipse by Changing Semi-
Axes 139
122. Ratio of the Squares of Any Two Ordinates . . . 139
123. Condition that a Point is without, on, or within the
Ellipse 140
124. Form of Equation representing an Ellipse . . . 140
125. Latus Rectum a Third Proportional to Major and Minor
Axes ......... 141
126. Auxiliary Circles 141
127. Ratio of the Ordinates of the Ellipse and Auxiliary Circle 142
128. Construction of the Ellipse by § 127 . . . . 142
129. Area of the Ellipse 143
130. Equations of Tangents and Normals .... 146
131. Subtangents and Subnormals . *. . . . 147
132. Tangents to Ellipses having a Common Major Axis . 148
CONTENTS.
SECTION
133.
134.
135.
136.
137.
138.
139.
140.
141.
142.
143.
144.
145.
146.
147.
148.
149.
150.
151.
152.
153.
154.
155.
156.
157.
158.
159.
160.
161.
162.
163.
The Normal bisects Angle between Focal Radii .
Method of drawing the Tangent and Normal at a Point
on Ellipse
Equation of Tangent in Terms of its Slope
Director Circle of the Ellipse
Review Exercises .
Supplementary Propositions.
Condition of Two, One, or No Tangents to an Ellipse
Equation of Chord of Contact ....
Equation of the Polar with respect to an Ellipse .
Method of drawing a Tangent to an Ellipse .
Equation of a Diameter of an Ellipse .
Conjugate Diameters
Tangents at Ends of Diameter Parallel to Conjugate
Diameter
Relation of Ends of Conjugate Diameters
Sum of Squares of any Pair of Semi-Conjugate Diameters
Difference between Eccenti'ic Angles of Ends of Conju-
gate Diameters
Angle between Two Conjugate Diameters
Conjugate Diameters parallel to Supplemental Chords .
Equation of Ellipse having Conjugate Diameters as Axes
Construction of the Polar of a Focus ....
The Polar Equation with the Left-Hand Focus as Pole
Chapter VII, The Hyperbola.
Simple Properties of the Hyperbola
Construction of an Hyperbola
Centre, Transverse Axis, Vertices
Equation of the Hyperbola
Properties of the Hyperbola
Equilateral Hyperbola .
Conjugate Hyperbolas .
Straight Line through Centre meets Cur
Asymptotes .
Equation of Tangent
Equation of Normal
Subtangent, Subnormal
PAGE
148
149
140
150
152
154
154
155
155
156
156
e in Two Points
157
158
158
158
159
160
161
162
163
168
168
170
171
171
172
172
173
173
175
175
175
X CONTKNTS.
SECTIOK PAGE
l(i4. Cniiditiiin that a Straiglit Line is Tangent . . . 175
1()5. iMiuation of the Director Circle . . . . .175
IGO. Tangent and Normal bisect Angles between Focal Kadii 175
Review Exercises 177
SUPPLEMICNTAUY Puoi'OSITIONS.
107. Condition of Two, One, or No Tangents to an Hyperbola 178
10b. K(iuation of Chord of Contact . . . . .178
16!). Ivjuation of the Polar with respect to the Hyperbola . 179
170. Etiuation of a Diameter of an Hyjierbola . . . 17!)
171. Conjugate Diameters 179
172. Properties of Conjugate Diameters .... 179
178. Length of a Diameter ....... 180
174. Portions of a Line between two Conjugate Hyperbolas
are Ecjual ISO
175. Tangent at End of a Diameter is Parallel to Conjugate
Diameter . . . . . . . .180
176. Given Eml of Diameter, to find Ends of Conjugate
Diameter 181
177. Equation of Hyperbola having Conjugate Diameters as
Axes 182
178. Tangents at Ends of Conjugate Diameters meet in
Asymptotes 182
179. Angle between Two Conjugate Diameters . . . 183
180. P(jrtions of a Line between Hyperbola and Asymptotes
are Eiiual 18.3
181. A Parallel to an Asymptote meets the Curve in only
One Finite Point 184
182. Equation of Hyperbola having the Asynqjtotes as Axes 185
183. The Polar of the Focus 187
184. I'olar Equations of an Hyperbola 188
Chapter VIII. Loci of the Second Order.
185. General Equation of the second Degree
186. Condition that this Equation represents Two Lines
187. Central and Non-Central Curves ....
188. General E(iuation of Central Loci ....
18!). lieduction of this Equation to a Known Form
190. Nature of Locus of Px- + Qy- = R
191
191
192
192
194
195
CONTENTS.
SECTION
191.
192.
193.
194.
195.
196.
197.
198.
199.
200.
201.
202-203.
204.
205.
206.
207.
208.
209.
Locus of Equation when A = 0 and 2 = 0.
Locus of Equation when A is not 0 and S = 0
Summary .......
Examples .......
Definition of a Conic .....
Equation of a Conic .....
Exercises .
Chapter IX. IIic.nER Plane Curves.
Higher Plane Curves .
The Cissoid of Diodes
The Conchoid of Niconicdes
The Lemniscate of Bernnulli
The Witch of Agnesi .
The Cycloid
Spirals ....
The Spiral of Archimedes .
The Hyperbolic Spiral
The Lituus.
The Logarithmic Spiral
The Parabolic Spiral .
PAGE
196
197
200
201
205
205
206
208
208
211
213
215
216
220
221
222
223
223
224
210
211
212-213
214-216
217
218
219
220
PART II. SOLID GEOMETRY.
Chapter I. The Point.
Definitions ......
The Radius Vector of a Point .
Direction Angles and Direction Cosines
Projections upon a Straight Line
Angle Between Two Straight Lines .
Distance Between Two Points .
Polar Coordinates ....
Projections upon a Plane .
Exercises
226
228
228
230
231
232
233
234
235
Chapter II. The Plane.
221-222. Normal Equation of a Plane
236
CONTENTS.
SECTION
223. Sjninietrical Equation of a Plane
224. Angle between Two Planes
225. Distance from a Point to a Plane
226.
227-228.
229.
230.
231.
2.32.
233.
234-235.
236.
237.
238.
239.
240.
241.
242.
243.
244.
245.
Chapter III. The Straight Line
Equations of a Straight Line
Symmetrical Equations of a Straight Line
Angle between Two Straight Lines .
Inclination of a Line to a Plane .
Exercises
Supplementary Propositions.
Traces of a Plane
Equations of the Traces of a Plane .
Condition of Intersection of Two Straight Lines
To pass a Plane through a Point and a Right Line
Chapter IV. Surfaces of Revolution.
A Single Equation in a;, ?/, z, represents a Surface
Ti'aces of a Surface .
Definitions ....
General Equation of a Surface of Revolution
Paraboloid of Revolution .
Ellipsoid of Revolution
Hyperboloid of Revolution .
Central Surfaces
Cone of Revolution .
Conic Sections .
Exercises ....
page
238
239
240
243
245
246
246
247
250
250
250
251
252
254
254
254
255
256
258
259
259
260
263
Supplementary Propositions.
246. General Equation of the Sphere
247. Intersection of Two Spheres
248. Equation of Tangent Plane to Sphere
249-250. Transformation of Coordinates .
251-252. Quadrics
253-257. Central Quadrics ....
258. Non-Central Quadrics
264
265
265
266
267
268
271
ANALYTIC GEOMETRY.
3j«<0«
PART I. — PLANE GEOMETRY.
CHAPTER I.
LOCI AND THEIR EQUATIONS.
Rectilinear System of Coordinates.
1. Let AX' and FF' (Fig. 1) be two fixed lines inter-
secting in the point 0. These lines divide the plane in
which they lie into four portions.
■-D
Y
<B
E
i^
\Pi ^
X'
'F
0
M X
JPa
T
\l^
G
iC
Y'
^-H
Fig. 1.
Let these parts be called Quadrants (as in Trigonometry),
and distinguished by naming the area between OX and
OY the first quadrant; that between OY and OX' the
second quadrant; that between OX' and Oi'^ the third
quadrant ; and that between OY and OX the fourth
quadrant.
ANALYTIC GEOMETKY.
Sui)])Ose the position of a point is described by saying
that its distance from YY', expressed in terms of some
chosen unit of length, is 3, and its distance from XX' is 4,
it being understood that tlie distance from either line is
measured parallel to the other. It is clear that in each
quadrant there is one point, and only one, that will
satisfy these conditions. The position of the point in
each quadrant may be found by drawing parallels to YY'
at the distance 3 from YY', and parallels to XX at the
distance 4 fro.m XX' ; then the intersections JPi, P^, Ps,
Pi satisfy the given conditions.
D
Y
,B
E
[^
\Pi -f
X'
jiV
0
M\ X
iPs
T
.._ \3
a
iC
Y'
\ah
Fig. 1.
2. In order to determine which one of the four points,
Pi, Pg, Ps, Pi, is meant, we adopt the rule that opposite
directions shall be indicated by V7illke signs. As in Trigo-
nometry, distances measured from YY' to the right are
considered positive ; to the left, negative. Distances meas-
ured from XX' upward are positive ; doivnward, negative.
Then the position of Pi will be denoted by + 3, + 4 ; of
P2,by-3,+4; of P3, by -3, -4 ; ofP„by + 3,-4.
3. The fixed lines XX' and. YY' are called the Axes of
Coordinates ; XX is called the Axis of Abscissas, or Axis
of X ; YY', the Axis of Ordinates, or Axis of y. The
intersection 0 is called the Origin.
LOCI AND THEIR EQUATIONS. 3
The two distances (with signs prefixed) that deter-
mine the position of a point are called the Coordinates
of the point ; the distance of the point from YY' is called
its Abscissa ; and the distance from XX', its Ordinate.
Abscissas are usually denoted by x, and ordinates by y.
A point is represented algebraically by simply writing
the values of its coordinates within a parenthesis, that of
the abscissa being always written first.
Thus Pi (Fig. 1) is the point (3, 4) ; P^, the point ( — 3, 4) ;
Ps, the point (—3, —4); P^, the point (3,-4). In general,
the point whose coordinates are x and y is the point (x, y).
4. This method of determining the position of a point
in a plane is called the Rectilinear System of Coordinates.
The coordinates are called rectangular or oblique, accoi'ding
as the axes are rectangular or oblique ; that is, according
as the axes intersect at right or oblique angles. In the
first three chapters we shall use only rectangular coordinates.
Note. The first man to employ this method successfully in
investigating the properties of certain figures was the French philoso-
pher Descartes, whose work on Geometry appeared in the year 1637.
Exercise 1.
1. What are the coordinates of the origin ?
2. In what quadrants are the following points (a and b
being given lengths) :
{-a,-b), (-a,b), (a,b), (a,-b)?
3. To what quadrants is a point limited if its abscissa is
positive ? negative ? ordinate positive ? ordinate negative ?
4. In what line does a point lie if its abscissa = 0 ? if
its ordinate ^ 0 ?
5. A point (x, y) moves parallel to the axis of x ; which
one of its coordinates remains constant in value ?
4 ANALYTIC GEOMETRY.
6. Construct or ^j^onhe points : (2,3), (3, — 3), (—1,-3),
(-4, 4), (3, 0), (-3, 0), (0, 4), (0, -1), (0, 0).
Note. To plot a point is to mark its proper position on paper,
when its coordinates are given. The fii'st thing to do is to draw the
two axes. The rest of the work is obvious after a study of Nos. 1-3.
7. Construct the triangle whose vertices are the points
(2,4), (-2,7), (-6,-8).
8. Construct the quadrilateral whose vertices are the
points (7, 2), (0, -9), (-7, -1), (-6, 4).
9. Construct the quadrilateral whose vertices are ( — 3, 6),
(—3, 0), (3, 0), (3, 6). What kind of quadrilateral is it?
10. Mark the four points (2, 1), (4, 3), (2, 5), and (0, 3),
and connect them by straight lines. What kind of a figure
do these four lines enclose ?
11. The side of a square = a ; the origin of coordinates
is the intersection of the diagonals. What are the coor-
dinates of the vertices (i) if the axes are parallel to the
sides of the square ? (ii) if the axes coincide with the
diagonals ?
(i.)(»V2.o),(o.«V5),(-|V2,«),(o,-^V5).
12. The side of an equilateral triangle = a ; the origin
is taken at one vertex, and the axis of x coincides with
one side. What are the coordinates of the three vertices ?
Ans. (0,0), (a, 0),(^ |^y3).
13. The line joining two points is bisected at the origin.
If the coordinates of one of the points are a and b, what
are the coordinates of the other ?
14. Connect the points (5, 3) and (5, — 3) by a straight
line. What is the direction of this line ?
liOCI AND THE IK EQUATIONS. O
Circular Measure.
5. In Analytic Geometry, angles are often expressed in
degrees, minutes, and seconds ; but sometimes it is very
convenient to employ the Circular Measure of an angle.
In circular measure, an angle is defined by the equation
, arc
angle = — - — ,
radius
in which the word '• arc " denotes the length of the arc
corresponding to the angle when both arc and radius are
expressed in terms of a common linear unit.
This equation gives us a correct measure of angular
magnitude, because (as shown in Geometry) for a given
angle the value of the above ratio of arc and radius is
constant for all values of the radius.
If the radius = 1, the equation becomes
angle =^ arc ; that is.
In circular measure an angle is measured hy the length of
the arc suhte^ided bij it iri a unit circle.
It is shown in Geometry that the circumference of a
unit circle ^ 27r ; as this circumference contains 360°
common measure, the two measures are easily compared
by means of the relation
360 degrees = 27r units, circular measure.
Exercise 2.
1. Find the value in circular measure of the angles 1°,
45°, 90°, 180°, 270°.
"*"^- 180' 4' 2' ''• 2*
2. In circular measure, the unit angle is that angle
whose arc is equal to the radius of the circle. What is
the value of this angle in degrees, etc.?
Ans 57° 17' 46".
ANALYTIC GEOMETRY.
Distance between Two Points.
6, To find the distance between two given points.
Let P and Q (Fig. 2) be the given points, cci and yi the
coordinates of P, ccj and y2 those of Q. Also let d = PQ
■=■ the required distance.
M
Fig. 2.
A^X
Draw PJ/and ^.V || to OT, and PP || to OX.
Then 031 = x„ MP = y,,
ON = x^, NQ = ?/2,
PP := Xa — Xi QR = y^ — y^.
By Geometry,
(^2 = {x^ — x^}^ + (?/2 — yiY ;
whence, rf= V{x2-iCiy-^+ (1/2- X/i)^-
[1]
Since (xi — .rg)^ = (ir2 — a"i)^ it makes no difference which
point is called (x-^, y^) and which (x^, y^).
7. Equation [1] is perfectly general, holding true for
points situated in any quadrant. Thus, if P is in the
second quadrant and Q in the third quadrant (Fig. 3),
^2 — ^1 is obviously equal to the leg RQ ; and since y^ is
negative, y^ — yi is the sum of two negative numbers, and
is equal to the absolute length of the leg RP with the —
sign prefixed.
LOCI AND THEIR EQUATIOXS. 7
Note. The learner should satisfy himself that equation [1] is per-
fectly general, by constructing other special cases in which tlie points
P and Q are in different quadrants. In every case he will find that
the numerical values of the expressions {X2 — Xi) and (1/2 — 2/1) are
the legs of the right triangle, the hypotenuse of which is the re(iuired
distance PQ.
Equation [1] is merely an illustration of the general truth that
theorems and formulas deduced by reasoning loitk points or lines in the
first quadrant {where the coordinates are always positive) must, from
the very nature of the analytic method, hold true when the points or
lines are situated in the other quadrants.
Exercise 3.
Find tlie distance
> 1. From the point (—2, 5) to the point (—8, —3).
2. From the point (1, 3) to the point (6, 15).
3. From the point (—4, 5) to the point (0, 2).
■^ 4. From the origin to the point ( — 6, — 8).
5. From the point (a, b) to the point ( — a, — b).
Find the lengths of the sides of a triangle
6. If the vertices are the points (15, — 4), (—9, 3)
(11, 24).
V 7. If the verticesare the points(2,3),(4,— 5),(— 3, — 6).
8. If the vertices are the points (0, 0), (3, 4), (— 3, 4).
9. If the vertices are the points (0, 0), (—a, 0), (0, —b).
"J 10. The vertices of a quadrilateral are (5, 2), (3, 7),
(—1, 4), (—3, — 2). Find the lengths of the sides and
also of the diagonals.
V 11. One end of a line whose length is 13 is the point
( — 4, 8); the ordinate of the other end is 3. What is its
abscissa?
12. What equation must the coordinates of the point
(x, 7/) satisfy if its distance from the point (7, — 2) is
equal to 11 ?
8
ANALYTIC GEOMETRY.
13. What equation expresses algebraically the fact that
the point (x, y) is equidistant from the points (2, 3) and
(4, 5) ?
14. If the value of a quantity depends on the square of
a length, it is immaterial whether the length is considered
positive or negative. Why ?
Division of a Line.
8. To bisect the line joining two given 2^oints.
Let P and Q (Fig. 4) be the given points (x^, i/i) and
(^2) 3/2)- Let X and y be the coordinates of H, the mid-
point of FQ.
The meaning of the problem is to find the values of x
and y in terms of .Tj, 2/1, and x^, y^.
M b
Fig. 4.
N
Draw P3I, ES, QJVW to OY; also draw PA, EB || to OX.
Then rt. A PEA = rt. A EQB (hypotenuse and one
acute angle equal).
Therefore, PA = EB, and AE = BQ;
also, MS=SN.
By substitution, ic — a-^ = .To — x, and y—y\^yi — y\
whence, ^^S^.^^mpi., ,-2]
LOCI AND THEIR EQUATIONS. M
9. To divide the line joining two given points into two
parts having a given ratio m : n.
Let P and Q (Fig. 5) be the given points (x^ y-^ and
{x^y^. Let R be the required point, such that PR : RQ =
m : n, and let x and y denote the coordinates of R.
Complete the figure by drawing lines as in Fig. 4.
The rt. A PR A and RQB, being mutually equiangular,
are similar ; therefore
PA PR m , AR PR m
-= — , and
RB RQ n' BQ RQ n
Substituting for the lines their values, we have
X — Xi m , 1/ — 7/, m
=— , and ■ ■ —
1^ Vi—V *i
Solving these equations for x and ?/, we obtain
_ ^nx<i + ndC\ . _ tnyi. + ny\ [3]
tn-\- n * tn + n
It m = n, we have the special case of bisecting a line
already considered ; and it is easy to see that the values
of X and y reduce to the forms given in [2],
Exercise 4.
What are the coordinates of the point
^1. Halfway between the points (5, 3) and (7, 9) ?
2. Halfway between the points (— 6, 2) and (4, — 2) ?
- 3. Halfway between the points (5, 0) and (—1, —4) ?
4. The vertices of a triangle are (2, 3), (4, — 5),
(— 3, — 6) ; find the middle points of its sides.
T 5. The middle point of a line is (6, 4), and one end of
. the line is (5, 7). What are the coordinates of the other end ?
^ 6. A line is bisected at the origin ; one end of the line
is the point (— a, h). What are the coordinates at the
other end ?
10 ANALYTIC GEOMETRY.
" 7. Prove that the middle point of the hypotenuse of a
right triangle is equidistant from the three vertices.
8. Prove that the diagonals of a parallelogram mutually
bisect each other.
9. Show that the values of x and y in [2] hold true
when the two given points both lie in the second quadrant.
10. Solve the problem of § 9 when the line PQ is cut
externally instead of internally, in the ratio m : n.
11. What are the coordinates of the point that divides
the line joining (3, — 1) and (10, 6) in the ratio 3:4?
12. The line joining (2, 3) and (4, — 5) is trisected.
Determine the point of trisection nearer (2, 3).
13. A line AB is produced to a point C, such tliat BC =
^ AB. If A and B are the points (5, 6) and (7, 2), what
are the coordinates of C ?
■ 14. A line AB is produced to a point C, such that AB :
BC = 4= :7. li A and B are the points (5, 4) and(6, —9),
what are the coordinates of C ?
15. Three vertices of a parallelogram are (1,2), ( — 5, — 3),
(7, — 6). What is the fourth vertex?
Constants and Variables.
10. In Analytic Geometry a line is regarded as a
geometric viagnitude traced or generated by a moving point,
— just as we trace on paper what serves to represent a line
to the eye by moving the point of a pen or pencil over the
paper.
We shall find that great advantages are to be gained by
defining a line iji this way, but we must be prepared from
the outset to make an important distinction in the use of.
symbols representing lengths. We must distinguish between
symbols which denote definite or fixed lengths and those
which denote variable lengths.
LOCI AND THEIR EQUATIONS.
11
11. Let A (Fig. 6) be the point (3, 4). Then OA =
V 9 + 16 = 5. Now let a jjoint F describe the line OA by
moving from 0 to A, and let the coordinates of Pbe denoted
by X and y; also let z denote the length OP at any position
of P. Then z will increase continuously from 0 to 5.
F
V
/
^
0
JT
1
i X
Fig. 6.
Here the word continuously deserves special attention.
It means that P must pass successively through every
position on the line OA from 0 to ^ ; that, therefore, z
must have in succession every conceivable value between 0
and 5. There will be one position of P for which z is
equal to 2 ; there will be another position of P for which z
is equal to 2.000001; but before reaching this value z must
first pass through all values between 2 and 2.000001.
In the same way x and y, the coordinates of P, both
pass through a continuous change, x increasing continuously
from 0 to 3, and y from 0 to 4.
We may now divide the lengths considered in this
example into two classes :
(1) Lengths supposed to remain constant in value,
namely, the coordinates of A and the distance OA ; (2)
lengths supposed to vary continuously in value, namely,
the coordinates of P {x and y), and the distance OP, or z.
12 ANALYTIC GEOMETRY.
Quantities of the first kind in any problem are called
constant quantities, or, more briefly, Constants.
(Quantities of the second kind are called variable quanti
ties, or, more briefly, Variables.
12. Two variables are often so related that if one oi
them changes in value the other also changes in value.
The second variable is then said to be a function of the
first variable. The second variable is also called the
dependent variable, while the first is called the indej^endent
variable. Usually the relation between two variables is
such that either may be treated as the independent variable,
and the other as the dependent variable.
Thus, in § 11, if we suppose z to change, then both x and
y will change ; the values of x audi/ then will depend upon
the value given to z ; that is, x and y will be functions
of z. But we may also suppose the value of x, the abscissa
of P, to change ; then it is clear that the values of both
y and z must also change. In this case we take x as the
independent variable, and values of y and z will depend
upon the value of x ; that is, y and z will be functions of x.
13. The most concise way to express the relations of
the constants and variables which enter into a problem is
by means of algebraic equations.
The coordinates of P (Fig. 6) throughout its motion are
always x and y ; and the triangle OFM is similar to the
triangle OAB. Hence, for any position of P,
y- = % and z^ = x'^y\
By solving, 2/ "= q ^j ^^^d ~ = o ^.
o o
These equations express the values of y and z, respectively,
in terms of x as the independent variable.
LOCI AND THEIR EQUATIONS. 13
14. In § 11, instead of assuming 3 and 4 as the coordi-
nates of A, we might have employed two letters, as a and
b, with the understanding that these letters should denote
two coordinates that remain constant in value during the
motion of P. If we choose these letters, we obtain,
b -Ja^ + b^
11= -X. z=^ X.
^ a ' a
15. There is a noteworthy difference between the con-
stants 3 and 4 and the constants a and h. The numbers
3 and 4 cannot be supposed to change under any circum-
stances. The numbers a and b are constants in this sense
only, that they do not change in value when we suppose x
OT y ov z to change in value ; in other words, they are not
functions of x or y or z in the particular problem under
discussion. In all other respects they are free to represent
as many different values as we choose to assign to them.
Constants of the first kind (arithmetical numbers) are
called absolute constants. Constants of the second kind
(letters) are called arbitrary or general constants.
16. By general agreement, variables are represented by
the last letters of the alphabet, as x, y, z; while constants
are represented by the first letters, a, b, c \ or by the last
letters with subscripts, as Xj, y^, x^, y^, etc.
Exercise 5.
1. A point P (x, y) revolves about the point Q {x^, yi),
keeping always at the distance a from it. Name the
constants and the variables in this case. What is the
total change in the value of each variable ?
2. A point Q {x, y) moves: first parallel to the axis of //,
then parallel to the axis of x, then equally inclined to the
axes. Point out in each case the constants and the variables.
14
ANALYTIC GEOMETRY.
Locus OF AN Equation.
17. Let us continue to regard x and y as the coordinates
of a point, and proceed to illustrate the meaning of an
algebraic equation containing one or both of these letters.
Take as the first case the equation x — 4=^0, whence
a; = 4. It is clear that this equation is satisfied by the
coordinates of every point so situated that its abscissa is
equal to 4 ; therefore, it is satisfied by the coordinate of
B
Fig. 7.
every point in the line AB (Fig. 7), drawn || to OY, on
the right of OY, and at the distance 4 from OY. And it
is also clear that this line contains all the points whose
coordinates will satisfy the given equation.
The line AB, then, may be regarded as the f/eometric
representation or meaning of the equation x — 4^0 ; and,
conversely, the equation x — 4 = 0 may be considered to be
the algebraic representative of this particular line.
In Analytic Geometry the line AB is called the locus of
the equation x — 4^0; conversely, the equation x — 4 = 0
is known as the equation of the line AB.
The line AB is to be regarded as extending indefinitely
in both directions. If AB is described by a point P,
moving parallel to the axis of y, then at all points x is
LOCI AND THEIR EQUATIONS.
15
constant in value and equal to 4, while y (which does not
appear in the given equation) is a variable, passing througli
an unlimited number of values, both positive and negative.
1 8. The equation x — y = 0,ov x^y, states in algebraic
language that the abscissa of the point is always equal to
the ordinate.
Values of x. Values of y.
0 0.
1 1.
2 ..... 2.
— 1 -1.
etc. etc.
Fig. 8.
If we draw through the origin 0 (Fig. 8) a straight
line AB, bisecting the first and third quadrants, then it is
easy to see that the given equation is satisfied by every
point in this line and by no other points. If we conceive a
point P to move so that its abscissa shall always be equal
to its ordinate, then the point must describe the line AB.
In other words, if the point P is obliged to move so that
its coordinates (which of course are variables) shall always
satisfy the condition expressed by the equation a-— // = 0 ;
then the motion of P is confined to the line AB.
The line AB is the locus of the equation x — y = 0. and
this equation represents the line AB.
19. The equation 2x-\-y — 3 = 0 is satisfied by an un-
limited number of values of x and y. We may find as many
of them as w.e please by assuming values for one of the varia-
bles, and computing the corresponding values of the other.
16
ANALYTIC GEOMETRY.
If we assume for x the values given below, we easily find
for y the corresponding values given in the next column.
Values of x. Values of y.
0
3.
1
1.
2
-1.
3
-3.
4
— 5.
-1
5.
-2
7.
-3
9.
-4
11.
etc.
etc.
Fig. 9.
Plotting these points (as shown in Fig. 9), we obtain a
series of points so placed that their coordinates all satisfy
the given equation. By assuming for x values between
0 and 1, 1 and 2, etc., we might in the same way obtain as
many points as we please between A and B, B and C, etc.
In this case, however, the points all lie in a straight line
(as will be shown later); so that if any tivo points are
found, the straight line drawn through them will include
all the points whose coordinates satisfy the given equa-
tion. Now imagine that a point P, the coordinates of
which are denoted by x and y, is required to move in such
a way that the values of x and y shall always satisfy the
equation 2x-\-y — 3 = 0; then P mtist describe the line
AB, and cannot describe any other line.
The line AB is the locus of the equation 2x-^y — 3 = 0.
20. Thus far we have taken equations of the first
degree. Let us now consider the equation x^ — 2/^ = 0.
By solving for y, we obtain y = ±x. Hence, for every
LOCI AND THEIR EQUATIONS.
17
value of X there are tivo values of y, both equal numerically
to X, but having unlike signs. Thus, for assumed values
of X, we have corresponding values of y given below :
"Values of x.
0 .
1 .
2
3 .
— 1 .
— 2 .
— 3 .
Values of y.
0.
1,-1.
2—2
3,-3.
-1, 1.
-2, 2.
-3, 3.
Fig. 10.
By plotting a few points, and comparing this case with
the example in § 18, it becomes evident that the locus of
the equation consists of two lines, AB, CD (Fig. 10), drawn
through the origin so as to bisect the four quadrants,
21. There is another way of looking at this case. The
equation x'^ — y^ = 0, by factoring, may be written (x — y)
(x-]-y)^= 0. Now the equation is satisfied if eithe?' factor
= 0; hence, it is satisfied if x — y = 0, and also if x-\-y
^0. We know (see § 18) that the locus of the equation
x — y=^0 is the line AB (Fig. 8). And the locus of the
equation x-\-y = 0 (or x = — y) is evidently the line CD,
since every point in it is so placed that the two coordinates
are equal numerically but unlike in sign. Therefore, the
original equation x'^ — jf = 0 is represented by the pair of
lines AB and CD (Fig. 10).
22. Let us next consider the equation cr^+?/ = 25.
Solving for y, we obtain y=± V25 — x\ When x < 5
there are two values of y equal numerically but unlike in
sign. When x = 5, y = 0. When a; > 5 the values of y
are imaginary ; this last result means that there is no point
with an abscissa greater than 5 whose coordinates will
satisfy the given equation.
18
ANALYTIC GEOMETRY.
By assigning values of x differing by unity, we obtain
tlie following sets of values of x and // ; and by plotting
the points, and then drawing through them a continuous
curve, we obtain the curve shown in Fig. 11.
alues of X.
Val
aes of y.
0 . .
±5.
1 . . .
±4.9.
2 . .
±4.6.
3 . .
±4.
4 . .
±3.
5 . .
0.
-1 . .
±4.9.
-2 . .
±4.6.
— 3 . .
±4.
— 4 . .
±3.
-5 . .
0.
In this case, however, the locus may be foiind as follows :
Let P (Fig. 11) be any point so placed that its coordinates,
x=OM, i/ = MP, satisfy the equation x^ -\-y-^^25. Join
OP; then x'-{- f^'oF; therefore, OF = 5. Hence, if P is
any point in the circumference described with 0 as centre
and 5 for radius, its coordinates will satisfy the given
equation ; and if P is not in this circumference, its coor-
dinates will not satisfy the equation. This circumference,
then, is the locus of the equation.
23. The points whose coordinates satisfy the equation
y^ = 4x lie neither in a straight line nor in a circumference.
.Nevertheless, they do all lie in a certain line, which is,
therefore, completely determined by the equation. To con-
struct this line, we first find a number of points that satisfy
the equation (the closer the points to one another, the
better) and then draw, freehand or with the aid of tracing
curves, a continuous curve through the points.
LOCI AND THEIR EQUATIONS.
19
The coordinates of a number of such points are given in
the table below. It is evident that for each positive value
of X there are two values of ?/, equal numerically but un-
like in sign. For a negative value of x, the value of >/ is
imaginary ; this means that there are no points to the left
of the axis of y that will satisfy the given equation.
Values of x. Values of y.
• 0 0.
1 ±2.
2 ±2.83.
3 ±3.46.
4 ±4.
5 ±4.47.
6 ±4.90.
7 ±5.29.
8 ±5.66.
9 ±6.
— 1 imaginary.
In Fig. 12 the several points obtained are plotted, and a
smooth curve is then drawn through them. It passes
through the origin, is placed symmetrically on both sides
of the axis of x, lies wholly on the right of the axis of f/,
and extends towards the right without limit. It is the locus
of the given equation, and is a curve called the Parabola.
24. After a study of the foregoing examples, we may
lay down the following general principles, which form the
foundation of the science of Analytic Geometry:
I. Every algebraic equation involving x and i/ is satis-
fied by an unlimited number of sets of values of x and i/;
in other words, x and // may be treated as variaMes, or
quantities varying continuously, yet always so related that
their values constantly satisfy the equation.
Fig. 12.
20 ANALYTIC GEOMETRY.
II. The letters x and ?/ may also be regarded as repre-
senting the coordinates of a point. This point is not fixed
in position, because x and y are variables ; but it cannot be
placed at random, because x and y can have only such
values as will satisfy the equation; now, since these values
are continuous, the point may be conceived to move con-
tinuously, and will therefore describe a definite line, or
group of lines.
The line, or group of lines, described by a point moving
so that its coordinates always satisfy the equation is called
the Locus of the Equation; conversely, the equation satis-
fied by the coordinates of every point in a certain line is
called the Equation of the Line.
An equation, therefore, containing the variables x and y
is the algebraic representation of a line.
In Analytic Geometry the loci considered are represented
by their equations, and the investigation of their properties
is carried on by means of these equations.
Exercise 6.
Determine and construct the loci of the following equa-
tions (the locus in each case being either a straight line or
a circumference of a circle) :
1. a; — 6 = 0. 9. 9x- — 25 = 0.
2. a; + 5 = 0. 10. ^x-—y'^ — ^.
3. y= — l. 11. a;2 — 16^2 = 0.
4. x = 0. 12. a;2_f_y2^36_
5. ?/ = 0. 13. x''-\-y'' — l=Q.
6. x-\-y = 0. 14. x(?/H-5)=0.
7. x — 2y = 0. 15. (x — 2)(x — 3) = G.
8. 2x + 3?/+10 = 0. 16. (y-4)(3/+l)=0.
LOCI AND TIIEIK EQUATIONS. 21
17. What is the geometric meaning of the equation
5a;2 — 17x — 12 = 0?
Hint. Resolve the equation into two binomial factors.
18. What is the geometric meaning of the equation
19. What two lines form the locus of the equation
xy-\-4:X^0?
20. Is the point (2, — 5) situated in the locus of the equa-
tion 4ic — 3// - 22 = 0 ?
Hint. See if the coordinates of the point satisfy the equation.
21. Is the point (4, — 6) in the locus of the equation
22. Is the point ( — 1, — 1) in the locus of the equation
16a;2 + V + 15cc — 6?/— 18 = 0 ?
23. Does the locus of the equation a---l-y" = 100 pass
through the point ( — 6, 8)?
24. Which of the loci represented by the following
equations pass through the origin ?
(l)3x-[-2 = 0. (5)3x = 2i/.
(2)3x-lUj-{-7 = 0. {e)3x-n>/ = 0.
(3) a;2 - 16/ -10 = 0. (7) a- — 1 Oy- = 0.
(4) aa; + % + c = 0. (8) ax-\-b>/ = 0.
25. The abscissa of a point in the locus of the equation
Sx — 4?/ — 7^0 is 9 ; wliat is the value of the ordinate ?
Ans. 5.
26. Determine that point in the locus of // — 4a- ^0
for which the ordinate =: — 6.
A]is. The point (0, - 6).
27. Determine the point where the line represented by
the equation lx-\- y — 14 = 0 cuts tfie axis of x.
Ans. The point (2, 0).
22 analytic geomexky.
Intersections of Loci.
25. The term Curve, as used in Analytic Geometry,
means any geometric locus, including the straight line as
well as lines commonly called curves.
The Intercepts of a curve on the axes are the distances
from the origin to the points where the curve cuts the axes.
26. To find the intercepts of a curve, having given its
equation.
The intercept of a curve on the axis of x is the abscissa
of the point where the curve cuts the axis of x. The
ordinate of this point = 0. Therefore, to find this inter-
cept, put v/ = 0 in the given equation of the curve, and
then solve the equation for x ; the resulting real values of
X will be the intercepts required.
If the equation is of a higher degree than the first, there
will in general be more than one real value of x; and the
curve will intersect the axis of x in as many points as there
are real values of x.
To an imaginary value of x there corresponds no inter-
cept ; but it is sometimes convenient to speak of such a
value as an imaginary intercept.
Similarly, to find the intercepts on the axis of y, put
a; = 0 in the given equation, and then solve it for y ; the
resulting real values of y will be the intercepts required.
27. To find the points of intersection of two curves, having
given their equations.
Since the points of intersection lie in both curves, their
coordinates must satisfy both equations. Tlierefore, to
find their coordinates, solve the two equations, regarding
the variables x and y as unknown quantities.
If the equations are both of the first degree, there will
LOCI AND THEIR EQUATIONS. 23
be only one pair of values of x and y, and one point of
intersection.
If the equations are, one or both of them, of higher
degree than the first, there may be several pairs of values
of X and y ; in this case there will be as many points of
intersection as there are pairs of real values of x and y.
If imaginary values of either x or y are obtained, there
are no corresponding points of intersection.
28. If a curve 2>asses throxujh the oriyin, its equation, re-
duced to its slmjdest form, cannot have a constant term; that
is, cannot have a term free from both x and y.
Since in this case the point (0, 0) is a point of the curve,
its equation must be satisfied by the values cc = 0, and
y^=0. But it is obvious that these values cannot satisfy
the equation if, after reduction to its simplest form, it still
contains a constant term. Therefore the equation cannot
have a constant term.
29. If an equation has no constant term, its locus must
pass through the origm.
For, the values x = 0, y = 0 must evidently satisfy the
equation, and therefore the point (0, 0) must be a point of
the locus.
Exercise 7.
Find the intercepts of the following curves :
1. 4a- + 3// — 48 = 0. 8. 0- — 3 = 0.
2. 5y — ox — 30^0. 9. a- — 9=0.
3. ic2 + 2/2 = 16. 10. a-'- — //- = 0.
4. 9a- + 4/ = 16. 11. / = 4a-.
5. 9.r2 — 4/ = 16. 12. .T- + // — 4.r — 8// = 32.
6. 9x^ — 4// = 1 6. 13. X- + // — 4.r — 8// = 0.
7. a:'x' -\- by = a'b\ 14. (x — 5)' -\- (y — ())' = 20.
24 ANALYTIC GEOMETKY.
Find the points of intersection of the following curver;
15. Sx — 4:1/ -{-13 = 0, lla;+7^— 104=0.
16. 2a; + 3^ = 7, x—i/ = l.
17. a;-7// + 25 = 0, x'-\-i/^ = 25.
18. 3x-j-ii/ = 2o, x'-\-i/ = 25.
19. x-\-y = 8, x--{-f = 34..
20. 2x=i/, x^ + i/^ — 10x = 0.
21. The equations of the sides of a triangle are 2x-{-
9y + 17=r0, 7a; — ^—38 = 0, a; — 2^-^2 = 0. Find the
coordinates of its three vertices.
22. The equations of tlie sides of a triangle are 5a; + 6^
= 12, 3a; — 4^ = 30, a; + 5y = 10. Find the lengths of its
sides.
23. Find the lengths of the sides of a triangle if the
equations of the sides are a; = 0, //^O, and 4a; + 3^ = 12.
24. What are the vertices of the quadrilateral enclosed
by the straight lines a; — a = 0, a; + a = 0, y — b = 0, y-\-b
= 0? What kind of a quadrilateral is it?
25. Does the straight line 5a; + 4y = 20 cut the circle
26. Find the length of that part of the straight line
3a; — 4y = 0 which is contained within the circle x^-\-if
= 25.
27. Which of the following curves pass through the
origin of coordinates?
(1) 7a; — 2y+4 = 0. (4) ax-\-by = 0.
(2) 7a; — 2// = 0. (5) ax + % + c = 0.
(3) tf — x' = 4:y. (6) x'^ — y-\-a = a-\-xy.
28. Change the equation 4a'H-2// — 7^0 so that its
locus shall pass through the origin.
LOCI AND THEIK EQUATIONS, 25
CONSTKUCTION OF LoCI.
30. If we know that the locus of a given equation is a
straight line, the locus is easily constructed; it is only
necessary to find any two points in it, plot them, and draw
a straight line through them with the aid of a ruler.
Likewise, if we know that the locus is a circumference,
and can find its centre and its radius, the entire locus can
then be described with the aid of a pair of compasses.
It will appear later that the form of the given equation
enables us at once to tell whether its locus is a straight line
or a circumference.
If the locus of an equation is neither a straight line nor
a circumference, then the following method of construction,
which is applicable to the locus of any equation without
regard to the form of the curve, is usually employed.
31. To construct the locus of a given equation.
The steps of the process are as follows:
1. Solve the equation with respect to either x or y.
2. Assign values to the other variable, diifering not
mucli from one another.
3. Find each corresponding value of the first variable.
4. Draw two axes, choose a suitable scale of lengths, and
plot the points whose coordinates have been obtained.
5. Draw a continuous curve through these points.
Discussion. An examination of the equation, as shown
in the examples given below, enables us to obtain a good
general idea of the shape and size of the curve, its position
with respect to the axes, etc. ; in this way it serves as an aid
in constructing the curve, and as a means of detecting nu-
merical errors made in computing the coordinates of the
points. Such an examination is called a discussion of the
equation.
26
ANALYTIC GEOMETRY.
Note 1. This method of constructing a locus is from its nature an
approximate metliod. But the nearer tlie points are to one another,
tlie nearer the curve will approacli the exact position of the locus.
Note 2. In theory, it is immaterial what scale of lengths is used.
In practice, the unit of lengths should be determined by tlie size of the
paper compared with the greatest length to be laid off upon it. Paper
sold under the name of " coordinate paper," ruled in small squares, j^^
of an inch on a side, will be found very convenient in practice.
32. Construct the locus of the equation
9^2 + 4/ -576 = 0.
If we solve for both x and y, we obtain the following values :
'=:±aV64-;
; = ±§Vl44 — ?/2
(1)
(2)
By assigning to x values differing by unity, and finding
corresponding values of ?/, we obtain the results given below.
To each value of x, positive or negative, there correspond two
values of y, equal numerically and unlike in sign. By plotting
the corresponding points, and drawing a continuous curve
through them, we obtain the closed curve shown in Fig. 13.
Values of x
Values of y.
0
. . ± 12.
±1
± 11.91.
±2
±11.62.
±3
±11.13.
±4
± 10.39.
±5
± 9.36.
±6
± 7.93.
±7
± 5.80.
±8
± 0.
±9
± imaginary
LOCI AND THEIR EQUATIONS. 27
Discussion. From equations (1) and (2) we see that
if X =0, ij = ±i 12, and if y = 0, x = ±:8; therefore, the
intercepts of the curve on the axis of x are +8 and — 8,
and those on the axis of y are + 12 and — 12. These inter-
cepts are the lengths OA, OA', and OB, OB', in Eig. 13.
If we assign to a? a numerical value greater than 8, posi-
tive or negative, we find by substitution in equation (1) that
the corresponding value of ij will be imaginary. This shows
that OA and OA' are the maximum abscissas of the curve.
Similarly, equation (2) shows that the curve has no points
with ordinates greater than +12 and — 12.
The greater the numerical value of x, between the limits
0 and 4" 8 or 0 and — 8, the less the corresponding value of
y numerically ; why ?
From equation (1) we see that for each value of x,
between the limits 0 and i 8, there are two real values of
y, equal numerically and unlike in sign. Hence, for each
value of X between 0 and ± 8 there are two points of the
curve placed equally distant from the axis of x. Therefore,
the curve is symmetrical with respect to the axis of x ; iu
other words, if the portion of the curve above the axis of x
is revolved about this axis through 180°, it will coincide with
the portion below the axis. Similarly, it follows from equa-
tion (2) that the curve is also symmetrical with respect to
the axis of y. Therefore, the entire curve is a closed curve,
consisting of four equal quadrantal arcs symmetrically placed
about the origin 0. The name of this curve is the Ellipse.
33. Construct the locus of the equation
4x — /+16 = 0.
Solving for both x and y/, we obtain
y=±2^ir^, (1)
-^-^- (2)
28
ANALYTIC GEOMETRY.
We may either assign values to x, and then compute
values of y by means of (1), or assign values to y, and com-
pute values of x by means of (2) ; the second course is
better, because there is less labor in squaring a number
than in extracting its square root.
By assigning values to y, differing by unity from 0 to
+ 10, and from 0 to — 10, and then proceeding exactly as
in the last example, we obtain the series of values given
below, and the curve shown in J'ig 14.
Values of y.
±0
Fig. 14.
±1
±2
±3
±4
±5
±6
t9
10
Values of x.
■ -4.
— 3.7o.
-3.
-1.75.
0.
2.25.
5.
8.25.
12.
16.25.
21.
Discussiox. An examination of equations (1) and (2)
yields the following results, the reasons for which are left
as an exercise for the learner :
The intercepts on the axes are :
On the axis of x, OA = — 4.
On the axis of y, 0B = + 4, and 6»C= — 4.
If we draw through A the line AD _\_ to OX, the entire
curve lies to the right of AD.
The curve is situated on both sides of OX, and is sym-
metrical with res})ect to OX.
The curve extends towards the right without limit.
LOCI AND THEIR EQUATIONS.
29
The curve constantly recedes from OX as it extends
towards the right.
This curve is called a Parabola ; the point A is called its
Vertex ; the line AX its Axis.
34. Construct the locus of the equation
y = sin X.
If we assume for x the values 0°, 10°, 20°, 30°, etc., the
corresponding values of >/ are the natural sines of these
angles, and are as follows :
Values of
X. Values of y.
Values of x.
Values of y
0°
... 0.
50° . .
. . 0.77.
10°
. . . 0.17.
60° . .
. . 0.87
20°
. . . 0.34.
70° . .
. . 0.94.
30°
. . . 0.50.
80° . .
. . 0.98.
40°
. . . 0.64.
90° . .
. . 1.
If we continue the values of x from 90° to 180°, the
above values of y repeat themselves in the inverse order
(e.g., if a; = 100°, y = 0.98, etc.) ; from 180° to 360° the
values of y are numerically the same, and occur in the
same order as between 0° and 180°, but are negative.
Fig. 15.
In order to express both x and y in terms of a common
linear unit, we ought, in strictness, to use the circular meas-
ure of an angle in which the linear unit represents an angle
30 ANALYTIC GEOMETRY.
of 57.3°, very nearly (see § 5). But it is more convenient,
and serves our present purpose equally well, to assume that
an angle of G0° = the linear unit. This assumption is made
in Fig. 15, where the curve is drawn with one centimeter
as the linear unit.
Discussion. The curve passes through the origin, and
cuts the axis of x at points separated by intervals of
180°. Since an angle may have any magnitude, positive or
negative, the curve extends on both sides of the origin
without limit. • The maximum value of the ordinate is
alternately + 1 and — 1 : the former value corresponds to
the angle 90°, and repeats itself at intervals of 360°; the
latter value corresponds to the angle 270°, and repeats
itself at intervals of 360°. The curve has the form of a
wave, and is called the Sinusoid.
Exercise 8.
Construct the loci of the following equations :
1.
3x — y — 2 == 0.
13.
r-i=o.
2.
y = 2x.
14.
y = x\
3.
x'^^y\
15.
xy = 12.
4.
^2+,/ =100.
16.
X ■= sin y.
5.
^2-^=^ = 25.
17.
y =^ 2 sin x.
6.
4a;2-/r=0.
18.
y =^ sin 2x.
7.
4a;2+V = 144.
19.
2/ = cos X.
8.
y — 16x = 0.
20.
y = tan x.
9.
y/2_|_ 1(5^=0.
21.
y = cot x.
10.
.T^— 2.T — lOy — 5==0.
22.
y = sec x.
11.
y--2y-10a- = 0.
23.
y =^ CSC X.
12.
(.«-3)^+(/y-2y^=:25.
24.
y = sin X + cos x.
loci and theik equatioxs. 31
Equation of a Curve.
35. From what precedes, we may conclude that every
equation involving x and y as variables represents a
definite line (or group of lines) known as the locus of the
equation. Regarded from this point of view, an equation
is the statement in algebraic language of a geometric con-
dition which must always be satisfied by a point (x, y), as
we imagine it to move in the plane of the axes. For ex-
ample, the equation x^=2y states the condition that the
point must so move that its abscissa shall always be equal
to twice its ordinate; the equation x^ -[-1/^=4: states the
condition that the point must so move that the sum of the
squares of its coordinates shall always be equal to 4 ; etc.
Conversely, every geometric condition that a point is
required to satisfy must confine the point to a definite line
as its locus, and must lead to an equation that is always
satisfied by the coordinates of the point.
Hence arises a new problem, and one usually of greater
difficulty than any thus far considered, namely:
Given the geometric condition to be satisfied by a point, to
find the equation of its locus.
The importance of this problem is that in the practical
applications of Analytic Geometry the law of a moving
point is commonly the one thing known, so that the first
step must consist in finding the equation of its locus.
Exercise 9.
1. A point moves so that it is always three times as far
from the axis of x as from the axis of y. What is the
equation of its locus?
2. "What is the equation of the locus of a point that
moves so that its abscissa is always equal to + 6? — 6? 0?
32 ANALYTIC GEOMETRY.
3. What is the e(;[uation of the locus of a point that
moves so that its ordinate is always equal to -f- 4 ? — 1? 0?
4. A point so moves that its distance from the straight
line X == 3 is always numerically equal to 2. What is the
equation of its locus ?
5. A point so moves that its distance from the straight
line y = 5 is always numerically equal to 3. Find the
equation of its locus. Construct the locus.
6. A point moves so that its distance from the straight
lina X + 4 ^ 0 is always numerically equal to 5. Find the
equation of its locus. Construct the locus.
7. What is the equation of the locus of a point equidistant
(1) from the parallels x = 0 and x == — 6 ?
(2) from the parallels y=^l and ?/ = — 3?
8. What is the equation of the locus of a point always
equidistant from the origin and the point (6, 0) ?
Find the equation of the locus of a point
9. Equidistant from the points (4, 0) and (— 2, 0).
10. Equidistant from the points (0, — 5) and (0, 9).
11. Equidistant from the points (3, 4) and (5, — 2).
12. Equidistant from the points (5, 0) and (0, 5).
13. A point moves so that its distance from the origin
is always equal to 10. Find the equation of its locus.
14. A point moves so that its distance from the point
(4,-3) is always equal to 5. Find the equation of its
locus, and construct it. What kind of curve is it? Does
it pass through the origin? Why ?
15. What is the equation of the locus of a point whose
distance from the point (— 4, — 7) is always equal to 8 ?
LOCI AND THEIR EQUATIONS. 33
16. About the origin of coordinates as centre, with aradius
equal to 5, a circle is described. A point outside this circle
so moves that its distance from the circumference of the cir-
cle is always equal to 4. What is the equation of its locus?
17. A high rock A, rising out of the water, is 3 miles
from a perfectly straight shore BC. A vessel so moves that
its distance from the rock is always the same as its distance
from the shore. What is the equation of its locus?
18. A point A is situated at the distance 6 from the
line BC. A moving point P is always equidistant from
A and BC. Find the equation of its locus.
19. A point moves so that its distance from the axis of x is
half its distance from the origin ; find the equation of its locus.
20. A point moves so that the sum of the squares of its
distances from the two fixed points (a, 0) and ( — a, 0) is
the constant 2k^; find the equation of its locus.
21. A point moves so that the difference of the squares
of its distances from (a, 0) and ( — a, 0) is the constant k^;
find the equation of its locus.
Exercise 10. (Review^.)
1. If we plot all possible points for which x = — 5,
how will they be situated ?
2. Construct the point (x, y) if a- = 2 and
(l)7/ = 4x-3, (2)3x-2y-8.
3. The vertices of a rectangle are the points (a, b), ( — a,
b), ( — a,—b), and (a, — b). Find the lengths of its sides,
the lengths of its diagonals, and show tliat the vertices
are equidistant from the origin.
4. What does equation [1]. p. 6, for the distance between
two points, become when one of the points is the origin ?
34 ANALVTIO GEOMETRY.
5. Express by an equation that the distance of the point
(x, if) from the point (4, 6) is equal to 8.
6. Express by an equation tliat the point (x, y) is equi-
distant from the i)oints (2, 3) and (4, 5).
7. Find the point equidistant from the points (2, 3),
(4, 5), and (6, 1). What is the common distance?
8. Prove that the diagonals of a rectangle are equal.
9. Prove that the diagonals of a parallelogram mutually
bisect each other.
10. The coordinates of three vertices of a parallelogram
are known : (5, 3), (7, 10), (13, 9). What are the coordi-
nates of the remaining vertex ?
11. The coordinates of the vertices of a triangle are
(3,5), (7,-9), (2,-4). Find the coordinates of the
middle points of its sides.
12. The centre of gravity of a triangle is situated on the
line joining any vertex to the middle point of the opposite
side, at the point of triseetion nearer tliat side. Find the
centre of gravity of the triangle whose vertices are the
points (2, 3), (4,-5), (-3,-6).
13. Tlie vertices of a triangle are (5, — 3), (7, 9), (—9, 6).
Find the distance from its centre of gravity to tlie origin.
14. If tlie vertices of a quadrilateral are (0, 0), (5, 0),
(9, 11), (0,3), what are the coordinates of the intersection
of the two straight lines that join the middle points of the
opposite sides ?
15. Prove that the two straight lines which join the
middle points of tlie opposite sides of any quadrilateral
mutually bisect each other.
16. A line is divided into three equal parts. One end of
the line is the point (3, 8) ; the adjacent point of division
is (4, 13). What are the coordinates of the other end?
LOCI AND THEIR EQUATIONS. 35
17. The line juiiiiiig the jnjiuts (xi. y^) and (x^, y^) is
divided into four equal parts. Find the coordinates of
the points of division.
18. Explain and illustrate the relation tliat exists
between an equation and its locus.
19. Construct the two lines that form the locus of the
equation x" — 7a" =; 0.
20. Is the point (2, ~5) in the locus of the equation
4a:=^ — V = 36?
21. Tlie ordinate of a certain point in the locus of the
equation .r- + if -\- 20x' — 70 =; 0 is 1. What is the abscissa
of this point?
22. Find the intercepts of the curve
a:^ + y— 5..-7y + 6 = 0.
Find the points common to the curves :
23. x'-\-y^ = 100, and y- — — = ^-
24. a;^ + //^ =^ 5«^, and ic^ = 4«7/.
25. U^x" + ay = tt^W, and a;^ + / = a\
26. Find the lengths of the sides of a triangle, if its
vertices are (6, 0), (0, —8), (—4, —2).
27. A point moves so that it is always six times as far
from one of two fixed perpendicular lines as from the
other. Find the equation of its locus.
28. A point so moves that its distance from the fixed
point A is always double its distance from the fixed line
AB. Find the equation of its locus.
29. A fixed point is at the distance a from a fixed
straight line. A point so moves that its distance from the
fixed point is always twice its distance from the fixed Hue.
Find the equation of its locus.
CHAPTER II.
THE STRAIGHT LINE.
Equations of the Straight Line.
36. Notation. Throughout this chapter, and generally
in equations of straight lines,
a = the intercept on the axis of x.
h = the intercept on the axis of y.
y = the angle between the axis of x and the line,
m = tan y.
p = the perpendicular from the origin to the line.
a = the angle between the axis of x and p.
These six quantities are general constants; a, b, and 7n
may have any values from — ao to + cc ; ^, any value
from 0 to -)- oc ; y, any value from 0° to 180° ; a, any
value from 0° to 360°.
The constant m is often called the Slope of the line ;
its value determines the direction of the line.
In order to determine a straight line, two geometric
conditions must be given.
37. To find the equation of a straight line passing through
two given points (x-^, y^ and (x2, y^.
Let A (Fig. 16) be the point (xj, y-^, B the point
(^2, 2/2) ; and let P be any point of tlie line drawn through
A and B, x and ?/ its coordinates. Draw AC, BD, PM \\
to OY, and AEF \\ to OX.
THE STRAIGHT LIXE.
37
The triangles APF, ABE are similar; therefore,
PF _BE
AF~ AE
A^iow, FF— 1/ — j/i, AF=x — Xi, BE=i/o— >/i, AE=X2 — Xi.
Therefore, ^^-^^^ ^^^^^y W
This is the equation required.
Y
Fig. 16.
Fig. 17.
It is evident that the angle FAP = y. Therefore, each
side of equation [4] is equal to tan y or m. The first side
contains the two variables x and ?/, and the equation tells us
that they must vary in such a way that the fraction ' —
remains constant in value and equals m.
Note. In Fig. 16 the points A, B, and P are assumed in the first
quadrant in order to avoid negative quantities. But the reasoning will
lead to equation [4] whatever be the positions of ,these points. In
Fig. 17 the points are in different quadrants. The triangles APF,
ABE are to be constructed as shown in the figure. They are similar ;
and by taking proper account of the algebraic signs of the quantities,
we arrive at equation [4], as before. The learner should study this
case with care, and should study other cases devised by himself, till
he is convinced that equation [4] is perfectly general.
38
ANALYTIC (JEOMETRV.
38. To find the equation of a straujld line, given one
point (xy, ?/i) in the line and the slope ni.
Let the figure be constructed like Fig. 16, omitting the
point B and the line BED. Then it is evident that
whence,
PF
y—yx
AF ic — aji'
y — y\ = tn(3c — oci).
Y
ir
I
Y
c
y ()
J
I X
Y
[5]
Fig. 18.
M A\ X
Fig. 19.
39. To find the equation of a strai/jht line, gicen the
intercept h and, the angle y.
Let the line cut the axes in the points A and B (Fig. 18).
Let P be any point (x, y) in tlie line. Draw PM^ to OY,
and7?C|| to OX.
Then OB = h, rBC = y, BC = x, PC = y — b;
therefore, m = ' :
X
whence, y = utx + h. [6]
40. To find the equation of a straight line, given its inter-
cepts a and h.
Let the line cut the axes in tlie points A and B (Fig. 19),
and let P be any point (x, y) in the line. Then OA = a,
THE STKAIGHT LINK.
39
OB = h. Drawi^J/Xtu OX. The triangles F MA, BOA
are similar ; therefore,
PM MA OA — OM
J
OA OA
y -
b
a — X ^ X
J- 5
a a
a b
whence, 7; + a ~ -'^ [^]
This is called the Symmetrical Equation of the straight
line.
41. To find the e(ivat\oii of a straight line, given its dis-
tance p from the origin, and tlce angle a.
i
/>'
vS
r/
/
'*'
\r
\^i
0
.1/
\.Y
Fig. 20.
Let AB (Fig. 20) be the line, P any jioint in it. Draw
OS L to AB, meeting AB in ,S; 7^J/ _L to OA'; Mii || to
AB, meeting OS in U\ ami /'^ J_ to All.
Tlien p=OS= Oil + QB, a = XOS = PMQ.
Now, OR — OM cos a = .^' cos a,
and QP = PM sin a — y sin a.
Therefore, OR + (>F =;> = .r cos a + v/ sin a,
or -z^ cos a + // sin a = 7>. [8]
This is called the Normal Equation of the straight line.
The coefficients cos a and sin a determine the dii-ection
of the line, and p its distance from the origin.
40 AXALYTIC GEOMETRY.
Note. Observe that all the equations of the straight line that
have been obtained are of the first degree. Their differences in form
are due to the constants which enter them. The form of each, and
the signification of its constants, should be thoroughly fixed in mind.
Exercise 11.
Find the equation of the straight line passing through
the two points :
1. (2, 3) and (4, 5). 7. (2, 5) and (0, 7).
2. (4, 5) and (7, 11). 8. (3, 4) and (0, 0).
3. (— 1, 2) and (3, — 2). 9. (3, 0) and (0, 0).
4. (—2,-2) and (-3, -3). 10. (3, 4) and (—2, 4).
5. (4, 0) and (2, 3). 11. (2, 5) and (-2,-5).
6. (0, 2) and (— 3, 0). 12. (m, 7i) and (— m, — n).
Find the equation of a straight line, given :
13. (4, 1) andy = 45^ 2*9. b = —4,y=120°.
14. (2, 7) and y = 60°. 30. 6 = -4, y = 135°.
15. (— 3, 11) andy = 45°. 31. b = — 4=, y = 150°.
16. (13,-4) and y=:.150°. 32. b = — 4:, y = 180°.
17. (3, 0) and y = 30°. 33. a = 4:,b = S.
18. (0, 3) andy = 135°. 34. a^ — 6,b = 2.
19. (0, 0) andy = 120°. 35. a = — 3,b = — 3.
20. (2, — 3) and y == 0°. 36. a = 5,b = — o.
21. (2, — 3) and y = 90°. 37. a = — 10,b = 5.
22. b = 2, y = 4:0°. 38.a = l,b = — l.
23. b = o, y = 4o°. 39. a = n, b = — n.
/ 24. 6 = — 4, y = 45°. 40. a = n,b = 'in.
25. & = — 4, y = 30°. 41. 79 = 5, a = 45°.
26. ^- = -4, y = 0°. 42. ^J = 5, a =120°.
27. i = - 4, y = 60°. 43. 2i = 5,a = 240°.
28. ^; = — 4, y = 90°. 44. J^ = 5,a = 300°.
THE STRAIGHT LINE. 41
Write the equations of the sides of a triangle :
45. If its vertices are the points (2, 1), (3, — 2), ( —4,-1).
46. If its vertices are the points (2, 3), (4, — 5), ( — 3, — 6).
47. Form the equations of the medians of the triangle
described in example 46.
48. The vertices of a quadrilateral are (0, 0), (1, 5),
(7, 0), (4, — 9). Form the equations of its sides, and also
of its diagonals. '
Find the equation of a straight line, given :
49. a = 7^, y = 30°. '51. j!> = 6, y = 45°.
50. « = — 3, (xi, ?/i) is (2, 5). 52. p = 6, y = 135°.
Reduce the following equations to the symmetrical
form, and construct each by its intercepts :
53. 3x — 2^ + 11 = 0 and ?/=7x + l.
54. 3x + 5^ — 13 = 0 and4a; — y — 2 = 0.
55. Reduce Ax-{- Bi/=C to the symmetrical form; also
y = mx-{-h. What are the values of a and b in terras of
A, B, C, and m?
Reduce the following equations to the form i/ = mx + b,
and construct each by its slope, and intercept on the axis
of y:
-56. y + 13 =5x and ?/ + 19 = 7x.
57. 3:c + ?/ + 2 = 0 and2y = 3a' + 6.
58. Reduce Ax -\- Bi/= C to the form // = tnx + b ; also
--}-- = l. What are the values of 7?i and b in terms of
a b
A, B, C, and a ?
59. Find the vertices of the triangle whose sides are the
lines 2.r + 9y + 17 = 0. // = 7.r - 38, 2i/-x= 2.
42 ANALYTIC GEOMETRY.
' • 60. Find the oqiuitioii of the str;iiglit line passing tlirongh
the origin and the intersection of tlie lines '3x — 1*//-[-4 = 0
and o.r-|-4y = 5. Also find the distance between these
two points.
61. AVhat is the equation of the line passing through
(•^'i; ^i)) ^^^^ equally inclined to the two axes ?
62. Find the equations of the diagonals of the parallelo-
gram formed by the lines x = a, x = h, y = c, y^^d.
63. Show that the lines ^=2x + 3, ?/ = 'ox + 4, y = 4.^ + 5
all pass tlirough one point.
Hint. Find the intersection of two of tlie lines, and then see if its
coordinates will satisfy the equation of the remaining line.
64. The vertices of a triangle are (0, 0), (xi, 0), (.rg, y^).
Find the equations of its medians, and prove tliat they
meet in one point.
7 65. What must be the value of in if the line y = mx
passes through the point (1. 4)?
r 66. The line y = mx + 3 passes through the intersection
of the lines y =^ x -\- 1 and y = 2x- -|- 2. Determine the
value of /«.
67. Find the value of h if the line v/ = Ox + ^ passes
through the point (2, 3).
68. What condition must be satisfied if the points
(•^1) yi); (^2) Z/2) (^3? ys) lie in one straight line?
Hint. Let equation [4] represent the line through (xi, ?/i) and
(-'^2, 2/2); then (X3, 2/3) must satisfy it.
69. Discuss equation [5] for the following cases : (i)
(xi, yi) is (0, 0), (ii) m = 0, (iii) ni = cc.
70. Discuss equation [6] for the following cases: (i)
h = 0, (ii) m = 0, (iii) 7n = 00, (iv) m = 0, and b^=0.
71. Discuss equation [7] for the following cases: (i)
a^^h, (ii) rt^O, (iii) a = x, (iv) b = co.
the stkaioht link. 43
General Equation of tijk First Dkguee.
42. If any -point P(xi, 7/1) is connected with the origin 0 ;
then -^ = cos XOP, i^~ = sin XOF, and OP = \/x{' + yf .
OF OF i I ji
Hence, if an// two real quantities are each divided by the
square root of the sum, of their squares, the quotients are the
cosine and sine of some angle.
43. The locus of every equation of the first degree in x
and y is a strciight line.
Any simple equation in ,r and // can be reduced to the
form
Aoc+By=C, [9]
in which C is positive or zero.
Dividing both members of [9] by yj A'-\- r>'-^, we obtain
^ ^ ^' /1^
V^'+5' V^P+7>''' V^-'+/^'
Now, by § 42, the coefficients of x and y in (1) are a set of
values of cos a and sin a, and the second member, being posi-
tive, is some value of ^; (§41). Hence (1) is in the normal
form, and its locus is some straight line. Whence the
proposition.
Cor. 1. To reduce any simple equation to the normal
form, put it in the form of [9], and divide both members by
the square root of the sum of the squares of the coefficients
of X and y.
CoK. 2. To construct (1), locate the point '{A, B), con-
nect it with the origin, and on this line lay off OS equal to
the second member of (1) ; the perpendicular to OiS through
Sis the locus of (1). or [9].
44. The locus of an cquntidu of the first degree in .t and
y is called a Locus of the First Order.
44 ANALYTIC GEOMETRY.
Exercise 12.
Reduce the following equations to the normal form, and
thus determine p, or the distance of eacli locus from the
origin :
1. 3x- — 2^+11 = 0. 5. y-\-VS = 5x.
2. 3^3 + 5^ — 13 = 0. 6. y+ 19 = 70:.
3. 4x— y— 2 = 0. 7. e.r + ry + ?i = 0.
4. 2x-{-o//=7. 8. 7i//-\-cx — r^^O.
Eeduce the following equations to one of the forms [6],
[7], [8], and determine by the signs of the constants which
of the four quadrants each lodus crosses :
9. y = ^x — 9. 14. 5a;4-4y — 20 = 0.
10. 3x-+2 = 2y. 15. i/ = 6x-\- 12.
11. 4.y = o.r — 1. -^16. )/-\-2 = x—4:.
12. 4?/ = 3a: + 24. 17. a; + V% + 10 = 0.
13. 5x + 3?/ + 15 = 0. 18. a;— VSy — 10 = 0.
19. Discuss equation [9] for the following cases :
(i) ^ = 0. (iv) A = cc. (vii) A = B,C=0.
(ii) i? = 0. (v) A=C = 0. (viii) A = -B,C = 0.
(iii) C = 0. (vi) A = B.
20. Reduce equation [7] to the form of equation [6],
and find the value of m in terms of a and b.
21. What must be the value of C that the line 4x — 5y = C
may pass through the origin ? through (2, 0) ?
22. Determine the values of A, B, and C, that the line
Ax-\- By=C may pass through (3, 0) and (0, — 12).
Hint. Since the coordinates of the given points must satisfy the
equation, we have the two relations ZA = C and — 12B = C.
23. From [9] deduce [4] by the method used in No. 22.
THE STRAIGHT LINE.
45
24. If equations [4] and [9] represent the same line,
what are the values of A, B, C, in terms of Xi, i/i, x^, y-i"^
25. In equation [4] find the values of vi and b in terms
of Xi, iji, X2, 1/2-
Angles.
45. To find the angle formed by the lines y ^ mx -\- b,
and y = m'x -\- b'.
Let AB and CD (Fig. 21) represent the two lines respec-
tively, meeting in the point P.
Let the angle AFC^^cf) ; then, by Geometry, <^ = y — y'.
Whence, by Trigonometry,
7n — in'
tan <t> = -, ,
1 + mm
This equation determines the value of <^
Y
[10]
Fig. 21.
Fig. 22.
Cor. 1. If the lines are parallel, tan <^= 0 ; hence, 7n= m'.
Conversely, if m^=m', <^ = 0, and the lines are parallel.
CoR. 2. If the lines are perpendicular, tan <^=x ; hence,
1 4- m )ii' = 0, or ni' — . Conversely, if 1 + m m' = 0,
m
<^:=90°, and the lines are perpendicular.
46 ANALYTIC GEOMETRY.
46. To find the equatioti of a strnujlit line jjcissiiif/ throiujk
the ijoint (a'l, y/i) and (i) 'parallel, (ii) ijerjiendicMlat', to the
line y = mx -f- i>-
The slope of the required line is m in case (i), and ■
in case (ii) ; and in both cases the line passes through a
given point (xi, y^).
Therefore (§ 38), the required equation is
(i) !/ — I/i = '>^K^—'-^i),
i3xercise 13.
Find the equation of tlie straight line :
1. Passing through (3, — 7), and || to the line y = 8x — 5.
2. Passing through (5, 3), and || to the line ^y — ^x=l.
3. Passing through (0, 0), and || to the line // — 4x^10.
4. Passing through (5, 8), and || to the axis of x.
5. Passing througli (5, 8), and || to the axis of y.
6. Passing through (3, —13), and _L to the line y=4:X — 7.
7. Passing through (2, 9), and _L to the line 7y-\-23x
-5 = 0.
8. Passing through (0, 0), and _L to the line x-\-2y=^l.
9. Perpendicular to the line 5x — 7y + 1 = 0, and erected
at the point whose abscissa = 1.
47. To find the eqiiation of a straight line passing
through a given point (xi, y^, and malting a given angle ^
with a given line y = mx -\- b.
Let the required equation be
y — !Ji = '":^'{^ — ^i)j
where in' is not yet determined.
THE STRAIGHT LINK. 47
Since the required line may lie eiciier as PQ or I'R
(Fig. 22), we shall liave (§45),
m' — m VI — m'
tan <^ = -T— ^ or
1 + fti^^' 1 + tnm'
, m ± tan <b
Hence, m' = - ^— >
1 ip ni tan ^
and the required equation is
tn ± tail <j) ^ . . ^
2/ - 2/1 = 7^„, tan 4. ^'^ " '^i)' C^l]
and (as Fig. 22 sliows) there are in general two straight
lines satisfying the given conditions.
Exercise 14.
1. Find the angle formed by the lines .r-f-2// + t=0
and x — 3]/ — 4^0.
The two slopes are — -} and ?,. If we put m.— — ^, in' = ?., we
obtain tan </> = — 1, <p = l.*]5°. If we put ni = J, m' = — i, we get
tan <p = I, (f) = 45°. Sliow that both these results are correct.
Find the tangent of the angle formed by the lines :
, ' 2. 3x — 4:1/ = 7 and 2x — // = 3.
3. 2.r + .3// + 4=0 and Ih- + 4// + .^> = 0.
4. y — nx = 1 and 2 (// — 1) ^ "•^•
Find the angle formed by the lines :
5 . .r -f- ?/ = 1 and ?/ = ,T + 4.
6 . .V + 3 = 2x and ?/ + 3,r = 2.
7. 2x + 3// + 7 = 0 and 3.r — 2// + 4=0.
8. G.r = 2// + r> and // — 3.r = 1 0.
9. X + 3 = 0 and y — V3.r + 4 = 0.
48 ANALYTIC GEOMETRY.
10. Discuss equation [11] for the cases when <^ = 0*
and <^ = 90°.
Note. The learner should solve the next four exercises directly
without using equation [11] ; then verify the result by means of [11].
Find the equation of a straight line :
11. Passing through the point (3, 5), and maliing the
angle 45° with the line 2x — 3y + 5 = 0.
12. Passing through the point ( — 2, 1), and making the
angle 45° with the line 2// = 6 — 3x.
13. Passing through that point of the line y^=2x — l for
which x = 2, and making the angle 30° with the same line.
14. Passing through (1, 3), and making the angle 30°
with the line x — 2// + 1 = 0-
• 15. Prove that tlie lines represented by the equations
Ax + Bi/-{-C = 0, A'x + B'i/-\-C' = 0
are parallel if AB'=^A'B ; i:>erpendicular, if AA'^ — BB'.
y 16. Given the equation 3aj + 4y-|- 6 = 0; show that the
general equations representing (i) all parallels and (ii) all
perpendiculars to the given line are
(i) 3.r + 4y + 7i = 0.
(ii) 4x — 3y + /i = 0.
17. Deduce the following equations for lines passing
through (xi, ?/i) and (i) parallel, (ii) perpendicular, to the
line y = vix + b.
(i) y — mx^ ?/i — mxi.
(ii) my + x = my^ + x^.
18. Write the equations of three lines parallel, and
three lines perpendicular, to the line 2a; + 3^ + 1=^0.
THE STRAIGHT LINE. 49
y 19. Among the following lines select parallel lines ; per-
pendicular lines ; lines neither parallel nor perpendicular :
(i) 2a; + 3?/ — 1 = 0. (v)a;-?/ = 2.
(ii) ?>x — 2ii = 20. (vi) 5 (a; + ?/) - 1 1 = 0.
(iii) ^x-\-^y = 0. (vii) cc = 8.
(iv) 12x = 8^ -I- 7. (viii) 7/ + 10 = 0.
p^ 20. Prove that the angle <^, between the lines
Ja; + % + C' = 0 and A'x + BUj + C'= 0,
is determined by the equation
, A'B — AB'
21. From the preceding equation deduce the conditions
of parallel lines and perpendicular lines given in Xo. 15.
Find the equation of the straight line :
V22. Parallel to 2x -f-3^ + 6 =0, and passing through (5, 7).
^23. Parallel to 2x-{- 1/ — 1 = 0, and passing through the
intersection of 3x + 2^ — 59 ^ 0 and 5x — 7v/ + 6 = 0.
y^ 24. Parallel to the line joining ( — 2, 7) and (—4, — 5),
and passing through (5, 3).
25. Parallel to y = mx-\-b, and at a distance d from the
origin.
26. Perpendicular to Ax-\- Bi/-{- C=^0, and cutting an
intercept b on the axis of i/.
X ■?/
27. Perpendicular to - + j- = 1, and passing through (a, b).
28. Making the angle 45° with --\-j = l, and passing
through (a, 0).
29. Show that the triangle whose vertices are the points
(2, 1), (3, — 2), (— 4, — 1) is a right triangle.
50 ANALYTIC GEOMKTRY.
30. The vertices of a triangle are ( — 1, — 1), ( — 3, 5),
(7, 11). Find the equations of its altitudes. ^*rove that
the altitudes meet in one point.
y^ 31. Find the equation of the perpendicular erected at
the middle point of the line joining (5, 2) to the intersec-
tion of a- + 2y — 11 = 0 and 9a; — 2y — 59 = 0.
32. Find the equations of the perpendiculars erected at
the middle points of the sides of the triangle whose vertices
are (5,-7), (1, 11), (— -1, 13). Prove that these perpen-
diculars meet in one point.
-^3. The equations of the sides of a triangle are
x-\-i/-\-l = 0, 3.r + 5^ + ll = 0, cr + 2^ + 4 = 0.
Find (i) the equations of the perpendiculars erected at
the middle points of the sides ; (ii) the coordinates of their
common point of intersection ; (iii) the distance of this
point from the vertices of the triangle.
34. Show that the straight line passing through («, h)
and (c, d) is perpendicular to the straight line passing
through (b, — a) and ((1, — c).
35. What is the equation of the straight line passing
through (iPi, ?/i), and making an angle ^ witli the line
Ax + Bij+C^O?
Distances.
48. Find the distance from the point ( — 4, 1) to the line
3a; - 4y + 1 = 0. . Ans. 3.
The required distance iS the length of the perpendicular
from the given point to the given line. The method that
first suggests itself is to form the equation of this perpen-
dicular, find its intersection with the given line, and com-
pute the distance from this intersection to the given point.
Let this metliod be followed in solving the above problem
and the first iiv(; pro1)Ienis of Exercise 15.
THE STKAIGHT LINE. 51
49. To find tlte distance from the point (rt\, //i) to tJia Hue
X cos a 4".'/ siu a^=j).
Let the line a cos a + // sin a=:y/, (1)
which is evidently parallel to the given line, pass tlirough
the given point (oi\, y/j); then we have
Xi cos a + i/y sin a =y>'.
Therefore, x^ cos a + .'/i sin a — p^p — p-
But p'' — p equals numerically the required distance.
Tliereforc, the distance from the point (a-,, ij-^ to tlie line
X cos a-\- ij sin a =7> is obtained by substituting Xx for x and
2/1 for y in the expression x cos u + // sin a — p.
CoR. 1. The distance as obtained from the formula will
evidently be positice or negatwe according as the point and
origin are on o})posite sides of the line, or on the same side.
CoR. 2. If the equation of the line is
xix -\- Bij = C,
and d denotes the distance from (.r,, ij{) to the line ; then,
evidently,
Axx -t B\)x — € p..,-.
Hence, to find the distance from the point (x^, y^) to the
line Ax-\-B}j— C, write x^ for x, and iji for y in the ex-
pression Ax-\- By— C, and divide the residt by ^A--\- />'-.
For example, let (—1, 3) be the point, and 2x-\-4: = 'Sy
the equation of the line.
Putting this equation in tlie form of [9], we have
-2x-j-3y = 4.
-2 (-1)4-3x3-4 , ,_
Whence, d = ^pf^l^^^ = + /rr Vl3.
Hence, the point and origin are on opposite sides of the line.
If only the length of d is sought, its sign nuiy be neglected.
52 ANALYTIC GEOMETRY.
Exercise 15.
- 1. Find the distance from (1, 13) to the line Sx=^y — 5.
2. Find the distance from (8, 4) to the line y^=1x — 16.
3. Find the distance from the origin to the line
3x- + 4y = 20.
4 . Find the distance from (2, 3) to the line 2x-\-y — A. =0,
5. Find the distance from (3, 3) to the line y=^^x — 9.
6. Prove that the distance from the point (x^, y^) to the
line y = mx -(- & is
v/i — mxi — h
d^^'- , — >
VI +m^
the upper or lower sign being used according as b is posi-
tive or negative. Express this result in the form of a rule
for practice.
7. Find the distances from the line 3a: -f 4?/ + 15 ^ 0 to
the following points: (3, 0), (3,-1), (3,-2), (3 — 3),
(3, - 4), (3, - 5), (3, - 6), (3, - 7), (0, 0), (- 1, 0), (- 2, 0),
(-3,0), (-4,0), (-5,0), (-6,0).
8. Find the distances from (1, 3) to the following lines:
3a;4-4y + 15 = 0. 3a; + 4?/— 5 = 0.
3a; + 4//+ 10 = 0. 3a; + 4^—10 = 0.
3a; + 4y+ 5 = 0. Sx-\- ^y — 15 = 0.
3x^Ay =0. 3x + 4?/ — 20 = 0.
Find the following distances:
9. From the point (2, — 5) to the line y — 3x = 7.
10. From the point (4, 5) to the line 4:y-\-5x = 20.
11. From the point (2, 3) to the line x-\-y^l.
.12. From the point (0, 1) to the line Sx — 3y = l.
13. From the point (—1, 3) to the line Sx -{-4y-\-2 = 0.
THE STRAIGHT LINE. 53
-/ 14. From the origin to the line ?>x-\-2i/ — G^O.
15. From the point (2, — 7) to the line joining (— 4, 1)
and (3, 2).
16. From the line y^=lx to the intersection of the lines
y = 3x — 4 and y=^ox-\-2.
17. From the origin to the line a(x — a) + b()/ — ^) = 0.
18. From the points (a, h) and ( — a, —b) to the line
^+1=1.
a b
19. From the point {a, b) to the line ax -\-bij=^0.
20. From the point (Ji, k) to the line Ax-\- Bf/-\- C = D.
Find the distance between the two parallels :
21. 3x' + 4^ + lo = 0 and 3a: + 4y + 5 = 0.
^22. 3a; + 4y + lo = 0 and 3.>; + 4y — 5 = 0.
23. Ax-\-Bt/=C and Ax-\-B/j=C'.
24. Ax + Bi/=Caud—Ax — Bi/=C'.
25. y = 5x — 7 and y = 5x -\- 3.
26. - + f = 2 and - + 7 = 0-
a b a b Z
27. Show that the locus of a point which is equidistant
from the lines Sx + 4y — 12 =0 and 4.r + 3// — 24 = 0 con-
sists of two straight lines. Find their equations, and draw
a figure representing the four lines.
28. Show that the locus of a point which so moves that
the sum of its distances from two given straight lines is
constant is a straight line.
54
ANALYTIC GEOMETRY.
r
/
\
/
/
\
>
^r
0
L
I
v"
~1
IX
Areas.
50. To find fhe area of a triangle, having given its vertices.
Solution I. Let PQR (Fig. 23) be the given triangle,
and let the vertices of PQR be (x^, y{), (x^, y^, {x^, y^, re-
spectively. Drop the perpendiculars PM, QN, RL; then
area PQR = PQNM+ RLNQ — PMLR.
By Geometry,
PQNM= ^ NM{MP -\- NQ)
^^(a-i — 3-2)(//i + y2).
Similarly,
RL]S'Q = ^{x^-x,)iy, + y.^.
PMLR = ^{x,-x,)(y, + y,).
Substituting these values, we have
Fig 23. °
area PQR = ^ \_{j\ — x^) {y. + ?/i)
+ (.^2 — x^){yz + .^2) — (^1 — ^?)(!h + //i)]
= i [— -'"i.'/i + ^i//2 ~ -"^zlh + -^'"J/i ~ •'•1//3 + ■'•;;//!]•
.-. area = i [ici (2/2 - */a) + x-i (j/3 -!/i)+ .^s (yi-2/2)]. [13]
Solution II. Since the area of a triangle is equal to
one half the product of its base and its altitude, this prob-
lem may be solved as follows :
(i) Find the length of any side as base,
(ii) Find the equation of the base.
(iii) Find the distance of the base from the opposite vertex,
(iv) Multii)ly this distance by one half the base.
Exercise 16.
Find the area of the triangle whose vertices are the points :
1. (0,0), (1,2), (2,1). 3. (2,3),(4,-5),(-3,-G).
2. (3, 4), (-3, -4), (0,4). 4. (8, 3), (-2, 3), (4, -5).
5. (a,0), (-«,0), (0,^.).
THE STKAIGIIT LINE.
6. Compare the ionuula for the area of a triangle with
the result obtained by solving No. 68, p. 42. What, then,
is the geometric meaning of that result?
Find the area of the figure having for vertices the points :
7. (3, 5), (7, 11), (9, 1).
8. (3, -2), (5, 4), (-7, 3).
9. (-1,2), (4, 4), (6, -3).
10. (0,0), (x,,!/,), (x„,!/,).
11. (2,-r,), (2,8), (-2,-5).
12. (10,5), (-2,5), (-5,-3), (7,-3).
13. (0, 0), (5, 0), (9, 11), (0, 3).
14. (a, 1), (0, b), (c, 1).
15. (a, b), {h, a), (c, c).
16. {a,h), {h,a), {c, — <').
17. Find tlie angles and the area of the triangle whose
vertices are (3, 0), (0, 3 V3), (G, 3 V3).
What is the area contained by the lines :
18. a: = 0, y = 0, 5j-1-4// = 20?
19. x-\-l/=l, .r — //=(), v/=0?
20. a; + 2// = 5, 2.r + // = 7, y = x-\-l?
21. a' + y = 0, .r = //, // = 3ft ?
22. // = 3a-, If = 7.r, // ^ r- ?
23. a- = 0, >j = 0, j--4 = 0, // + (;=0?
24. 3x + // 4- 4 = 0, 3.r - 5// + 34 = 0, 3.i- - 2// +1=0?
25. X- 5// + 13 = 0, 5.r + 7// + 1 = 0. 3.r + //-9 = 0?
26. X — I/ — 0, X + // = 0, X— i/ — (f, X -\- // = b ?
50 ANALYTIC GEOMETRY.
Find the area contained by the lines:
27. x = 0, y = 0, y^:^mx-\-b.
28. a; = 0, tj^Q, - + '{ = 1.
29. x = 0, v/ = 0, Ax-{-Bi/-\-C = 0.
30. 7/== 3a; — 9, y — Sx-'rS, 2i/ = x — 6, 2?/ = a; + 14.
31. What is the area of the triangle formed by drawing
straight lines from the point (2, 11) to the points in the
line y= 5x — 6 for which x = 4:, a: = 7 ?
Exercise 17. (Revie\Ar.)
1. Deduce equation [7], p. 39, from equation [6].
2. The equation y = mx -\-b \^ not so general as the
equation Ax-\- By-\- C=0, because it cannot represent a
line parallel to the axis of y. Explain more fully.
Determine for the following lines the values of a, b, y, p, and a :
3. a; = 2. 6. x+\/3y = 2.
4. x = 7/. 7. x — \/3y = 2.
5. 7/ + l = V3(.x + 2). 8. V3x — ^ = 2.
9. Find the equations of the diagonals of the figure formed
by the lines 3a- — y + 9 = 0, 3x = y—l, 5a; + 3// ^18,
5x + 3y = 2. What kind of quadrilateral is it ? Why ?
10. Find the distance between the parallels 9x=^y-\-l
and 9a; = ?/ — 7.
11. The vertices of a quadrilateral are (3, 12), (7, 9),
(2, — 3), (— 2, 0). Find the equations of its sides and its area.
12. The vertices of a quadrilateral are (6, — 4), (4,4),
(—4,2), ( — 8,-6). Prove that the lines joining the
middle points of adjacent sides form a parallelogram.
Find the area of this parallelogram.
THE STRAIGHT LINE. 57
Find the equation of a line passing through (3, 4), and also :
13. Perpendicular to the axis of x.
14. Making the angle 45° with the axis of x.
15. Parallel to the line ox + 6^ -}- 8 == 0.
16. Intercepting on the axis of y the distance — 10. "
17. Passing through the point halfway between (1, — 4)
and (- 5, 4).
18. Perpendicular to the line joining (3, 4) and (—1,0).
Find the equations of the following lines:
19. A line parallel to the line joining (xi, y^) and (a-j, ya)?
and passing through (ccj, y^.
20. The lines passing through (8, 3), (4,3), (—5,-2).
21. A line passing through the intersection of the lines
2ic + 5y + 8 = 0 and 2>x — 4v/ — 7 = 0, and X to the latter
line.
22. A line JL to the line 4x — ?/ =; 0, and passing through
that point of the given line whose abscissa is 2.
23. A line || to the line 3a:-|-4y = 0, and passing
through the intersection of the lines x — 2y — a = 0 and
x-\-'6y — 2a = ^.
24. A line through (4, 3), such that the given point bi-
sects the portion contained between the axes.
25. A line through (a-j, ?/i), such that the given point
bisects the portion contained between the axes.
26. A line through (4, 3), and forming with the axes in
the second quadrant a triangle whose area is 8.
27. A line through (4, 3), and forming with the axes in
the fourth quadrant a triangle whose area is 8.
28. A line through ( — 4,3), sucli that the jwrtion be-
tween the axes is divided by the given point in the ratio 5:3.
58 ANALYTIC GEOMETRY.
29. A line dividing the distance between ( — 3, 7) and
(5, — 4) in the ratio 4:7, and JL to the line joining these
points.
30. The two lines through (3, 5) making the angle 45°
with the line 2x — 3i/ — 7=0.
31. The two lines through (7, — 5) that make the
angle 45° with the line ijx — 2i/-\-o = 0.
32. The line making the angle 45° with the line joining
(7, — 1) and ( — 3,5), and intercepting the distance 5 on
the axis of x.
33. The two lines that pass through the origin and
trisect the portion of the line x-\- i/ = l included between
tlie axes.
34. The two lines || to the line 4^- -1-5^ + 11 = 0, at the
distance 3 from it.
35. The bisectors of the angles contained between the
lines // = 2x-{-4: and — ]/^3x -\-6.
Hint. Every point in the bisector of an angle is equidistant from
tlie sides of the angle.
36. The bisectors of the angles contained between the
lines 2x — 5y == 0 and 4a- + 3y = 12.
37. The two lines that pass through (3, 12), and whose
distance from (7, 2) is equal to V58.
38. The two lines that pass through ( — 2, 5), and are
each equidistant from (3, — 7) and (—4, 1).
Find the angle contained between the lines:
39. // + 3 = 2x and // + 3.r = 2.
40. >/ = 5x — 7 and 5y -{- Ji — 3 = 0.
THE STRAIGHT JASK. 59
Find the distance :
41. From the intersection of the lines 3x-\-2i/-\- 4 = 0,
2x + 5y + 8 = 0 to the line y = 5x + 6.
X II
42. From the point (A, Ic) to the line -4--= 1.
ah
43. From the origin to the line hx -\- hy = c'^.
44. From the point (a, 0) to the line y=:??u-|- -•
m
Find the area included by the following lines:
45. x = //, X -\- y =^ 0, X = r.
46. x-\-i/ = k, 2x = y + /.-, 2// — X -\- k.
47. - + '^ = 1, y = 2x-\-h, x=2ii-\-a.
a h ^ '
48. ?/=4.r + 7 and the lines that join the origin to
those points of the given line whose ordinates are — 1 and 19.
49. The lines joining the middle points of the sides of tlie
triangle formed by the lines a: — 5^+1 1 =0, 11a- -|-C// — 1 =<>.
a; + ?/ + 4 = 0.
50. Find the area of the quadrilateral whose vertices are
((VO), (0,5), (11, 9), (7,0).
51. What point in the line 5.)- — 4// — 28 = 0 is equidis-
tant from the points (1, 5) and (7, — 3) ?
52. Prove that the diagonals of a square are perpen-
dicular to each other.
53. Prove that the line joining the middle points of two
sides of a triangle is parallel to the third side.
54. What is the geometric meaning of the equation
xy = 0?
60 ANALYTIC GEOMETKY.
55. Show that the three points (3a, 0), (0, 'SO), (a, 2U)
are in a straight line.
56. Show that the three lines 5a; + 3//— 7 = 0, 3a; — Aij
— 10 = 0, and x-\-2i/^0 meet in a point.
57. What must be the value of a in order that tlie three
lines 3x + ?/ — 2 = 0, 2x—y — S=0, and aa; + 2y — 3 = 0
may meet in a point ?
What straight lines are represented by the equations :
58. x^-\-{a — b)x — ah=^0?
59. xy-[-hx-\-(.uj-\-ah—^'^
60. x~ij = xif ?
61. 14a;^ — 5a-^ — y- = 0 ?
In the following exercises prove that the locus of the
point is a straight line, and obtain its equation :
62. The locus of the vertex of a triangle having the base
and the area constant.
63. The locus of a point equidistant from the points
(a;i, yi) and {x^, y^.
64. The locus of a point at the distance d from the line
Ax-\-By^-C = ^.
65. The locus of a point so moving that the sum of its
distances from the axes shall be constant and equal to k.
66. The locus of a point so moving that the sura of its
distances from the lines Ax-\-By-\- C=0, A'x-\-B'y-\- C"=0
shall be constant and equal to k.
67. The locus of tlie vertex of a triangle, having given
the base and the difference of the squares of the other sides.
THE STRAIGHT LINE. 61
SUPPLEMENTARY PPtOPOSITIONS.
Lines passing through One Point.
51. If S=0, S' = 0 represent the equations of any tivo
loci with the terms all transposed to the left-hand side, and k
denotes an arbitrary constant, then the locus represented by
the equation S-\- kS' = 0 pxisses throuc/h every point common
to the two given loci.
For if any coordinates satisfy the equation *S'=0, and
also satisfy the equation *S' = 0, they must likewise satisfy
the equation S + kS^ = 0.
For what values of k will the equation ^'-|- A-,S'' = 0 rep-
resent the lines *S' = 0 and >S" = 0, respectively ?
52. Find the equation of the line joining the point (3, 4)
to the intersection of the lines
3a? — 2^+17=0 and a; + 4?/ — 27 = 0.
The method of solving this question that first suggests
itself is to find the intersection of the given lines and
then apply equation [4], p. 37.
Another method, almost equally obvious, is to employ
equation [5], which gives at once
y — 4 = m(x — 3),
and then determine m by substituting for x and y the coor-
dinates of the intersection of the given lines.
The following method, founded on the principle stated
in § 51, is, however, sometimes preferable on account of its
generality and because it saves the labor of solving the
given equations. According to this principle, the required
equation may be immediately written in the form
3x - 2y + 1 7 + A-(.r -f Ay - 27 ) - 0.
62 ANALYTIC OEOMETKY.
And since the line passes through (3, 4j, we must have
9 - 8 + 17 + /c(3 + 16 - 27) = 0,
whence, k = ' •
Therefore, 12^; — 8// + 08 -\-9x-\- 36i/ — 243 = 0,
or ox + 4// — 25 = 0.
This is the equation of the required line
53. . If the equations of three straight lines are
Ax +/.'// -f-C =0,
A'x ^ />"// + 6" =0,
A"x +/>'"// + C"' ^0,
and we can find three constants, I, m, n, so that the relation
l{Ax-\- By-[- C)^m{A'x+ir ij^C)-\-n{A"x-]- B" i/-^ C")^0
is identically trne, that is, true for all values of x and y,
then the three lines 'meet in a 'point.
For if the coordinates of any point satisfy any two of
the equations, then the above relation shows that they will
also satisfy the third equation.
54. To find the equation of the bisector of the angle
between the two lines
X cos a + // sin a =: p,
and X cos a' -\- y sin a' =^jj'.
There are evidently two bisectors : one bisecting the
angle in which the origin lies ; the other bisecting the
supplementary angle.
Now, every point in either bisector is equally distant
from the sides of the angle. Let (x, y) be any point in the
bisector of the angle that includes the origin ; then (§ 49)
X cos a+ //sin a— 7^:= a- cos a' 4-y sin a' — p\ (1)
Since (x, y) is any point in this bisector, (1) is its equation.
THE STRAUJIIT LINE. 63
The equation of the other bisector is
xcosa-{- 1/ sin a — 2^ — — (;r cos a' + y sin a* — //). (2)
To distinguish equations (1) and (2) we note that in tlie
first the constant terms in the two members have like signs ;
while in the second the constant terms have unlike signs.
CoR. 1. If the equations of the lines are in the form
Ax +Bi/ = C, A'x + B'l/ = C,
the equations of the bisectors are evidently
Ax + Btj- C A'jr, + By - C
V^2+^^ V^'2+"j5'i
1[4]
Equation [14] represents the bisector of the angle in
which the origin lies, or of its supplementary angle, accord-
ing as we take the upper or lower sign.
For example, let the equations of the lines be
2x = 4?/ -f- 9, and 5{/^=ox — 7.
Putting these equations in the form of [9], we have
2x — 4// = 9, and 3x — 5// = 7.
Hence, the equations of the bisectors of tlieir included
angles are ^ . ^ ., ^ ^
2x — 4y — 9 ox — oy — /
V20 ~ V34
in which the upper sign gives the equation of the bisector
of the angle in which tlie origin lies.
CoK. 2. If S=0 and *S"=:0 represent two simple equa-
tions in tlie normal form, with the terms all transposed to
the first members, then the equations of the bisectors of
their included angles may be written
^=±^', or .S'^^' = 0.
64 AJTALYTIC GEOMETRY.
Exercise 18.
Find the equation of the line passing through the inter-
section of the lines 3x — 2y + 1^ = 0, x + 4y — 27 = 0, and :
1. Passing also through the origin. '
2. Parallel to the line x + 2i/-\-3 = 0.
3. Perpendicular to the line 6x — 5?/ = 0.
4. Equally inclined to the two axes.
5. Find the equation of the line parallel to the line
x = y, and passing through the intersection of the lines
y = 2x-\-l and i/-\-3x^= 11.
6. Find the equation of the straight line joining (2, 3)
to the intersection of the lines
2ic + 3^+l=0 ■dnd3x — 4i/ = 5.
7. Find the equation of the straight line joining (0, 0)
to the intersection of the lines
5x — 2i/-\-3=0 and 13x-\-2j = l.
8. Find the equation of the straight line joining (1, 11)
to the intersection of the lines
2x-{-5i/ — 8 = 0 2ind3x — 4>/ = 8.
Find the equation of the straight line passing through
the intersection of the lines Ax -\- Bi/ -\- C^O and A'x-{-
B't/-{-C' = 0, and also:
9. Passing through the origin.
10. Drawn parallel to the axis of x.
11. Passing through the point (xi, t/i).
12. Find the equation of the straight line passing through
the intersection of 5ar — 4?/ + 3^0 and 7a;-[- lly— 1^0,
and cutting on the axis of y an intercept equal to 6.
THE STRAIGHT LINE. 65
13. Find the equation of the straight line passing through
the intersection of y^=-lx — 4 and y = — 2a3 + 5, and form-
ing with the axis of "a; the angle 60°.
14. The distance of a straight line from the origin is 5 ;
and it passes through the intersection of the lines 3x — 1y
+ 11 = 0 and 6x + 7^ — 55 = 0. What is its equation ?
15. What is the equation of the straight line passing
through the intersection of bx-{-ay=^ab and y=mx, and
perpendicular to the former line ?
Prove that the following lines are concurrent (or pas?'
through one point) :
16. ?/ = 2a- + l, yz=x-{-3, ?/=: — 5x + 15.
17. Ax—2y — ^ = 0, 3x-7/ + ^ = 0, 5x — 2y — l = 0.
18. 2x — y = 5, 3cc — y = G, 4x — y = 7-
19. What is the value of 77i if the lines
a 0 0 a
meet in one point ?
20. When do the straight lines y = mx -\-b,y=^ vi'x -\~ h',
y^m"x-\-b" pass through one point?
21. Prove that the three altitudes of a triangle meet in
one point.
22. Prove that the perpendiculars erected at the middle
points of the sides of a triangle meet in one point.
23. Prove that the three medians of a triangle meet in
one point. Show also that this point is one of the two
points of trisection for each median.
24. Prove that the bisectors of the three angles of a tri-
angle meet in one point.
06 ANALYTIC GEOMETKY.
25. The vertices of a triangle are (2,1), (3,-2), (—4,
— 1). Find the lengths of its altitudes. Is the origin
within or without the triangle?
26. The equations of the sides of a triangle are
3j^ + /y + 4=0, 3.x- -5^ + 34=^0, 3a; — 2^ + l:=0.
Find the lengtlis of its altitudes.
What are the equations of the lines bisecting the angles
between the lines :
27. 3x — 4// + 7 = 0 and 4.r — 3y + 17 = 0 ?
28. 'Sx + 4// — 9 = 0 and 12x + 5.7 — 3 = 0?
29. i/ = 2.r — 4:and2t/ = x-\-10?
30. x-\-f/ = 2 Sindx—>/ = 0?
31. y = mx + b and y = in'x -\- //?
32. Prove that the bisectors of the two supplementary
angles formed by two intersecting lines are perpendicular
to each other.
Equations representing Straight Lines.
55. A homogeneous eqvatwn of the nth degree rejjresents
n straight lines through the origin.
Let the equation be
j.r" + i;x"-'n + 6'.r" y + + Ay = 0.
Dividing by .///", we have
If ^'i5 >'2> '*3) '>'n denote the roots of this equation, then
the equation, resolved into its factors, becomes
THE STRAIGHT LINE. 67
and therefore is satisfied when any one of these factors is
zero, and in no other cases.
Therefore, the locus of the equation consists of the n
straight lines
x — i\y = 0, x — r.,i/ = 0, , x — r„y = 0;
and these lines evidently all pass through the origin.
56. To find the angle between the two stniiyht lines rep-
resented hxj the equation Ax^ + Cxy -\- Bif =■ 0.
Solving the equation as a quadratic in x, we obtain
2 Ax -I- ((7 ± ^C"-4AB) t/ = 0.
Hence, the slopes of the two lines are
2A . 2A
— C- VC2 - 4:AB ■ — C + VC'2 - 4:AB
Therefore,
, VC'2— 4^^ , A
VI — m ^ , mm = — ;
B B
and (equation [10], p. 45),
m—m' ^C^—AAB
tan (b = ~-, ,= — •
57. To find the condition that the general emtation of the
second degree may represent two straight lines.
We may write the most general form of tlie equation of
the second degree as follows :
Ax' + By- + Cxy + Bx + By + /<^ = 0. (1 )
That this equation may represent two straight lines, its
first member must be the product of two linear factors in
X and y ; that is, the equation can be written in the form
(lx-]-7ny-\-n) (px -\- qy-\-i') = 0. (2)
Equating coefficients in (1) and (2), we obtain
Ij) = A, niq = />, n r = F.
Iq -f- nip = 6', //• + np = Z>, nir -\- nq = E.
68 AXALYTIC GEOMETRY.
The product of C, D, and E is
CDE = 'llmvpqr -\- lp{ii?(f -\- m^i^) + mq(Pi^ + n^p^y
= 2ABF-\- A {E^ — 2BF) + B(I>'- 2AF)
+ F{C^ — 2AB).
Hence, the required condition is
F{C^ — AAB) + AE' + BD- - CDE= 0. (3)
Exercise 19.
1. Describe the position of the two straight lines repre-
sented by the equation Ax^-{-Cxi/-\- Bi/'-^-\-Dx-{-Ei/-\-F=0,
when (i) A=C = I) = (), (ii) B= C = E=0.
2. When will the equation ax>/-\-bx-\-ci/ -j-cl^O repre-
sent two straight lines ?
3. Find the conditions that the straight lines represented
by the equation Ax--\- Cxy-\- Bt/^^0 may be real ; imagi-
nary ; coincident ; perpendicular to each other.
4. Show that the two straight lines x^ — 2xy sec 6+ ^^ = 0
make the angle 6 with each other.
Show that the following equations represent straight
lines, and find their separate equations :
6. x'' — 2xy — ^if-\-2x — 2y-]-l=0.
7. a;2 — 4a!y + 3/ + 6?/— 9 = 0.
8. Show that the equation x^-\-xy — 6^^ -\-lx-\- Sly —
18 = 0 represents two straight lines, and find the angle
between them.
Determine the values of K for which the following equa-
tions will re})resent in each case a pair of straight lines.
Are the lines real or imaginary ?
9. 12x^ — 10xy-\-2y^-{-llx — 5y + K=0.
THE STRAIGHT LIXE. 69
10. 12x^ + Kx>/-{-2if-\-llx — 5i/-\-2 = 0.
11. 12a;2 + 36a;?/ + Ay + 0./;+6y+3 = 0.
12. For what value of if does the equation Kxy-\-bx
+ 3y+2 = 0 represent two straight lines?
Problems on Loci involving Three Variables.
58. A trapezoid is formed by drawing a line parallel to
the base of a given triangle. Find the locus of the intersec-
tion of its diagonals.
If ABC is the given triangle, and we choose for axes the
base AB and the altitude CO, the vertices A, B, C may be
represented in general by (a, 0), (h, 0), (0, c), respectively.
The equations oi AC and BC are, respectively,
- + ^ = land ? + ^ = l.
a c be
Let y^m be the equation of the line parallel to the base,
and let it cut ^C in D, ^C in ^; then the coordinates of
D and E, respectively, are
f ac — am \ ^ (be — biyi \
( J m 1 and ( > m I •
(1)
X — a be — bm — ac ^ '
If P is the intersection of the diagonals, then the coordi-
nates X and y of the point P must satisfy both (1) and (2) ;
by solving these equations, therefore, we obtain for any
particular value of m the coordinates of the point P. But
what we want is the algebraic relation that is satisfied
Hence, the equation of the diagonal BD is
y cm
X — b ac — am — be
and the equation of the diagonal AE is
y cm
70 ANALYTIC GKOMETllV.
by the coordinates of P, whatever the mine of m may be.
To find this, we have only to eliminate m from equations
(1) and (2). By doing this we obtain
2cx + {a + h)y ={a-\- b)c,
X , >/ .
i(« + ^) c
We see from the form of this equation that the required
locus is the line that joins C to the middle point of AB.
Remark. The above solution should be studied till it is understood.
In problems on loci it is often necessary to obtain relations that
involve not only the x and y of a point of the locus which we are seek-
ing, but also some third variable (as m in the above example).
In such cases we must obtain two equations that involve x and y
and this third variable, and then eliminate the third variable ; tlie
resulting equation will be the equation of the locus required.
Exercise 20.
1. Through a fixed point 0 any straight line is drawn,
meeting two given parallel straight lines in P and Q ;
through P and Q straight lines are drawn in fixed direc-
tions, meeting in R. Prove that the locus of i2 is a
straight line, and find its equation.
2. The hypotenuse of a right triangle slides between the
axes of X and y, its ends always touching the axes. Find
the locus of the vertex of the right angle.
3. Given two fixed points, A and B, one on each of the
axes ; if Z7and Fare two variable points, one on each axis,
so taken that 0U-{- 0F= OA-^OB, find tlie locus of the
intersection of A V and B U.
4. Find the locus of the middle points of the rectangles
that may be inscribed in a given triangle.
5. If PP\ QQ' are any two parallels to the sides of a given
rectangle, find the locus of the intersection of P'Q and PQ'.
CHAPTER III.
THE CIRCLE.
Equations of the Circle.
59. Tlie Circle is the locus of a point which moves so tliat
its distance from a fixed point is constant. Tlie fixed point
is the centre, and the constant distance the radius, of the
circle.
Note. The word "circle," us here detined, means the same thing
as "circumference" in Elementary Geometry. This is the usual
meaning of "circle" in the higher branches of Mathematics.
60. To find the equation of a circle, having given its
centre («, b) and its radius r.
E
Fig. 24.
Let C (Fig. 24) be the centre, and P any point (x, y) of
the circumference. Then it is only necessary to express by
an equation the fact that the distance from P to C is con-
stant, and equal to r : the required equation evidently is (§ 6)
(X - aY +(!/- W = /•■-. [15]
72 ANALYTIC GEOMETRY.
If we draw CR || to OX, to meet the ordinate of P,
then we see from the figure that the legs of the rt. A CPR
are CR = x — a, PR = 1/ — b.
If the origin is taken at the centre, then a = b:=0, and
the equation of the circle is
ic2 + 2/2 = r2. [16]
This is the simplest form of the equation of a circle, and
the one most commonly used.
If the origin is taken on the circle at the point A, and
the diameter AB is taken as the axis of x, then the centre
will be the point (r, 0). Writing r in place of a, and 0 in
place of b in [15], and reducing, we obtain
x"+ y^ = 2rx. [17]
Why is this equation without any constant term?
61. The locus of any equation of the second degree in x
and y in which the term in xy is wanting and the coefficients
of x^ and y^ are equal is a circle.
Any such equation can evidently be reduced to the form
x''-^if-\-2Dx-\-2Ey-{-F=0. (1)
Therefore,
{x^ + 2Dx + 2)2) + (^fJ^2Ey + U^) = D" -\- E^ — F,
or (x-\-Dy+{y-\-Ey=(iy-\-E^ — F). (2)
Now, from [15] it follows that the locus of (2) is a circle
whose centre is ( — i>, — E), and whose radius is
\/l)-^ + E^ — F.
Cor. If Z)2 + ^2 > F, the radius is real and the circle
is readily constructed. If D^ -\- E^=^F, the radius is zero,
and the locus is the single point ( — D, — E). If D^-\-E'^
< F, the radius is imaginary, and the equation represents
no real locus.
THE CIRCLE. 73
'62. Ajit/ point (h, k) is without, on, or within the circle
x^-{- y^:= r^, according as h^ -\- k- >, ==, or < r^.
For a point is without, on, or within a circle, according
as its distance from the centre >, =, or < the radius.
Exercise 21.
Find the equation of the circle, taking as origin :
1. The point B (Fig. 24) ; and BA as axis of x.
2. The point D (Fig. 24) ; and DE as axis of y.
3. The point E (Fig. 24) ; and ED as axis of y.
Write the equations of the following circles:
4. Centre (5, — 3), radius 10.
5. Centre (0, — 2), radius 11.
6. Centre (5, 0), radius 5.
7. Centre ( — 5, 0), radius 5.
8. Centre (2, 3), diameter 10.
9. Centre (Ji, k), radius V/i^ + H
10. Determine the centre and radius of the circle
ic^ + Z — 10^+12^+25 = 0.
Here (x - 5)2 + (y + 6)2 = 36. . •. a = 5, 6 = - 6, r = 6.
Determine the centres and radii of the following circles:
11. x''-\-f/ — 2x — 4.y = 0. 17. 6x2 — 2y(7 — 3y)=0.
12. 3a;2+3/— 5x— 7?/+l=0. 18. x-+if = 9k\
13. x''-\-y^ — Sx = 0. 19. (x + yy- + (x — yy-=8k-.
14. cc2 + /+8.r = 0. 20. x--\-y- = a- + b^
15. x^-{-y--8y = 0. 21. x^ + y' = k{x -\- k).
16. x^-\-f-j-Sy = 0. 22. x'-\-y- = hx-\-ky.
74 ANALYTIC GEOMETRY.
23. When are the circles x- -\- if + />./• -\- Eij -[- 6' = 0 and
^•- + //' + J^>'-'^ + ^^'// + C' = 0 concentric ?
24. Wliat is the geometric meaning of the equation
{x-af-\-{ij-l>f = i'^
25. Find the intercepts of the circles
(i) x2 + /-8.«-8y+ 7 = 0,
(ii) a;- + / — 8a; — 8y+16 = 0,
(iii) x' + }j- — 8ic — 8y + 20 = 0.
Putting 2/ = 0 in each case, we Iiave in case (i) x- — 8x + 7 = 0,
whence x = 1 and 7; in case (ii) x'^ — 8x + 16= 0, whence x = 4; in
case (iii) x- — 8x + 20 = 0, whence x = ± V— 4.
Putting X = 0 in each case, we obtain for y values identical with the
above values of x.
The geometric meaning of these results is as follows:
Circle (i) cuts the axis of x in the points (1, 0), (7, 0), and the axis
of y in the points (0, 1), (0, 7).
Circle (ii) touches the axis of x at (4, 0), and the axis of y at (0, 4).
Circle (iii) does not meet the axes at all.
This is the meaning of the imaginary values of x and y in case (iii).
If, however, we wish to make tlie language of Geometry conform
more exactly to that of Algebra, then in this case we should say that
the circle meets the axes in imaginary points. This form of statement,
however, must be understood as simply another way of saying that
the circle does not meet the axes.
Find the centres, radii, and intercepts on the axes of the
following circles:
26. a-2 4-;//2_5;y — 7y4-6 = 0.
27 . x" + / — 12x — 4?/ + 15 = 0.
28. a-- + if — 4 J- — 8// = 0.
29. x' + // - (;./• + 4// + 4 = 0.
30. x' + y- + Tlx — 1 8y + 57 = 0.
THE CIRCLE. 75
31. Under what conditions will the circle y^ -\- if -\- Dx
-\- Eif-\- C = 0 (i) touch the axis of j! ? (ii) touch the axis
of 1/ ? (iii) not meet the axes at all ?
32. Show that the circle x'^-{-ir -\-10x — 10 i/i- 25 = 0
touches the axes and lies entirely in the second quadrant.
Write the equation so that it shall represent the same
circle touching the axes and lying in the third quadrant.
33. In what points does the straight line 3x-\- u = 25
cut the circle x- -\- if = 65 ?
34. Find the points common to the loci x'^-\-if=^^ and
y = 2x — 4.
35. The equation of a chord of the circle a--+ ?/- = 25 is
y^2x-\- 11. Find the length of the chord.
~ X 1/
36. The equation of a chord is - -f- 'r ^= 1 j that of the
circle is x^-\- )f = 't'^. Find the length of the chord.
37. Find the equation of a line passing through the centre
of a;^ + ^^ — 6a; — ^y = — 21 and perpendicular to a; + 2// = 4.
38. Find the equation of that chord of the circle x--\- if-
= 130 that passes through the point for wliich the abscissa
is 9 and the ordinate negative, and that is parallel to the
straight line 4.x — 5// — 7 = 0.
39. What is the equation of the chord of the circle
a;2_|_^-'_-277 tliat passes through (3, — 5) and is bisected
at this point ?
40. Find the locus of the centre of a circle passing
through the points (xi, y^) and (j-.^, //o).
41. What is the locus of the centres of all the circles
that pass through the points (5, 3) and ( — 7, — 6) ?
76 ANALYTIC GEOMETRY.
Find the equation of tlie circle :
42. Passing through the points (4, 0), (0, 4), ((5, 4).
43. Passing through the points (0, 0), (8, 0), (0,-6).
44. Passing through the points (— 6, — 1), (0, 0), (0, — 1).
45. Passing through the points (0, 0), (—8a, 0), (0, 6a).
46. Passingthroughtliepoints(2, — 3),(3, — 4),(— 2, — 1).
•y 47. Passing through the points (1, 2), (1, 3), (2, 5).
48. Passing through (10, 4) and (17,— 3), and radius = 13.
49. Passing through (3, 6), and touching the axes.
50. Touching each axis at the distance 4 from the origin.
51. Touching each axis at the distance a from the origin.
52. Passing through the origin, and cutting the lengths
a, h from the axes.
53. Passing through (5, 6), and having its centre at the
intersection of the lines y = lx — 3, 4^/ — 3a; = 13.
: 54. Passing through (10, 9) and (5, 2 — 3V6), and hav-
ing its centre in the line 3x — 2y — 17 = 0.
55. Passing through the origin, and cutting equal lengths
a from the lines x = y, x-\-y = 0.
56. Circumscribing the triangle whose sides are the lines
y = 0, y = mx-\-b, - + ^=1.
57. Having for diameter the line joining (0, 0) and (x^, y^).
58. Having for diameter the line joining (x^, ?/i)and(iC2, y^-
59. Having for diameter the line joining the points
where y ^= nnx meets ic^ -|- y/^ = Irx.
60. Having for diameter the common chord of the
circles x^ -\-y^^^ r^ and (x — a)^ -\-y^ = r\
THE CIRCLE.
77
Tangents and jSTormals.
63. Let QPQ' (Fig. 25) represent any curve. If the
secant QPR is turned about the point P until the point Q
approaches indefinitely near to P, then the limiting position,
TT, of the secant is called tlie Tangent to the curve at P.
Fig. 25. Fig. 26.
The tangent TT' is said to touch the curve at P, and the
point P is called the Point of Contact.
The straight line PN drawn from P, perpendicular to
the tangent TT', is called the Normal to the curve at P.
Let the curve be referred to the axes OX, OY, and let M
be the foot of the ordinate of the point P. Let also the
tangent and the normal at P meet the axis of x in the
points T, N, respectively. Then TM is called tlio Sub-
tangent for the point P, and MN is called the SubnormaL
64. To find the equation of the tangent to the circle
x^-\-y'^=^ r"^, at the point of contact (xi, y^.
Let P (Fig. 26) be the point (x^, ?/i), and ()any other point
(^2j 2/2) of the circle. Then the equation of the secant PQ is
y — lh^lh — yi
0)
78 ANALYTIC GEOMETRY.
Now, since (xi, ij^) and (jv, y.^ are on this circle, we have
Subtracting, (x^^ — oc^^) + (jh" — y^) = 0.
_ Factoring, (a^ — Xi) (.r., + x,) + (^, — y ,) (//, + ?/i) = 0,
Whence, by transposition and division, we have
y^ — yi _ _ a-2 + a'i
a-2 — a-i v/2 + y\
By substituting in (1), the equation of the secant becomes
y ~ yi _ •'^2 + a^i
Now let Q coincide with P, or a-g^^u y^^^Uu ^^^^ secant
becomes a tangent at P, and the equation becomes
y — yi^ a-1
^ — ^1 Z/i'
or ^ia; + z/iZ/ = •'^1^ + 2/1".
And, since x^ -\- yi^ = r"^, we obtain
Xi(ic+yii/ = r% [18]
which is the equation required.
NoTK. If we had put Xn = xi, 1/2 = 2/1, in (1) before we introduced
the condition that (xi, 2/1) and (X2, 2/2) were on the circle, the slope of
the tangent would have assumed the indeterminate form -•
The above method of obtaining the equation of the tangent to a
circle is applicable to any curve whatever. It is sometimes called the
secant method. Equation [18] is easily remembered from its sym-
metry, and because it may be formed from x^ + y- — r^ by merely
changing x- to XiX, and y"^ to yiy.
THE CIRCLE. 79
65. To find tlie P(p((tf'tmi nf the normal throxujh (xj, ?/i).
The sloi)e of the tangent is — — •
Therefore, that of the normal is — (§ 45, Cor. 2).
Hence, the equation of the normal is (§ 38)
which reduces to the form
yix - XiU = O. [19]
Therefore, the normal passes through the centre.
66. To find the equations of the tangent and normal to the
circle (x — aY -\- (// — Ij)- = r at the point of contact (x^, ij{).
We proceed as in § 64, only now the equations of con-
dition which place (.rj, //j) and (.j-o, ?/,) on the circle are
(.ri-«)'+ 0/1 -*)' = '•".
After subtracting and factoring, \\'e have
{x-x,) {xA-x-2a)+{!,-y,) (^,+^,-2i)=0,
, Vi — III 3*2 + 3"i — 2a
whence, ^ = — — •
X2 — X1 2/2 + 2/1— 2^>
Hence, the equation of a secant tlirough (x^.?/{) and (xn, ?/.,) is
1/ — 2/1 .Tg -{-Xi — 2«
.-»■ — '^-1 y-i + //i — 2^^
Making .ro^.i,, and //2 = //i, and reducing, we obtain
(a?i - a) {x-a) + {iji -h){y-h) = r\ [20]
80 ANALYTIC GEOMETKY.
Equation [20] may be immediately formed from [18] by
affixing — a to the x factors and — b to tlie t/ factors, on the
left-hand side.
By proceeding as in § 65, we obtain for the equation of
the normal
{yi - b) {X - aci) - (oci - a) (y - yi) = O. [21]
67, To find the condition that the straight line y = mx + c
shall touch the circle x"-\-if=^ r^.
I. If the line touches the circle, it is evident that the per-
pendicular from the origin to the line must be equal to the
radius r of the circle. The length of this perpendicular is
— :^^z=: (§ 49). Therefore, the required condition is
Vl -1" nv^
c- = r^(l + «r).
II. By eliminating y from the equations
y = mx -\- c, x^ + ^^ = ^,
we obtain the quadratic in x,
(1 "1- m^a;^ -|- 2mca; = v^ — c^,
the two roots of which are
mc
Vy^(l -f- m^) — c'
1 -\- vi^ ~~ 1 + m^
If these roots are real and unequal, the line will cut the
circle; if they are equal, it will touch the circle; if they are
imaginary, it will not meet the circle at all.
The roots will be equal if Vr^(l + w^) — c^:=0; that is,
if c^= r^(l -|-m^), a result agreeing with that previously
obtained.
If in the equation y^^mx-{-c we substitute for c the
value 7'Vl -\-m^, we obtain the equation of the tangent of a
circle in the useful form
y = mx ± »■ Vl + m^. [22]
THE CIRCLE. 81
This equation, if we regard m as an arbitrary constant,
represents all possible tangents to the circle x'^-\-]f=^r^.
Note 1. Method II is applicable to any curve, and agi-ees with the
definition of a tangent given in § 63.
Note 2. In i^roblems on tangents the learner should consider
whether the cooi'dinates of the point of contact are involved. If they
are, he should use equation [18] ; if they are not, then in general it is
better to use equation [22].
Exercise 22.
1. Explain the meaning of the double sign in equation
[22].
2. Deduce the equations of the tangent and normal to
the circle x'^-\-i/ = t^, assuming that the normal passes
through the centre.
3. Find the equations of the tangent and the normal to
a;2-[-y2_.52 at the point (4, 6). Find, also, the lengths of
the tangent, normal, subtangent, subnormal, and the portion
of the tangent contained between the axes.
4. A straight line touches the circle x^-\-if=r- in the
point (xi, t/i). Find the lengths of the subtangent, the sub-
normal, and the portion of the line contained between the
axes.
5. What is the equation of the tangent to the circle
x^ -\-y^:= 250 at the point whose abscissa is 9 and ordinate
negative ?
6. Find the equations of tangents to x^-{- ij^ = 10 at the
points whose common abscissa = 1.
7. Tangents are drawn through the points of the circle
x^-{-y^ = 25 that have abscissas numerically equal to 3.
Prove that these tangents enclose a rhombus, and find its area.
8. The subtangent for a certain point of a circle is 5^ ;
the subnormal is 3. What is the equation of the circle ?
82 ANALYTIC GEOMETRY.
Find the equation of the straight line :
9. Touching x~-\-;/-=232 at the point whose abscissa=14.
10. Touching (« — 2)- + (// — 3)' = 10 at the point (5, 4).
11. Touching x^ -\- if — 3x — 4?/ ^ 0 at the origin.
12. Touching x'^-^if — 14x — 4y/ — 5=0 at the point
whose abscissa is equal to 10.
What is the equation of a straight line toucliing the
circle x^-\-y'^= i^, and also :
13. Passing through the point of contact (r, 0)?
14. Parallel to the line Ax-\-Bi/~\- (7=0?
15. Perpendicular to the line Ax -)- /:>//+ C= 0 ?
16. Making the angle 45° with the axis of x?
.17. Passing through tlie exterior point (Ji, 0) ?
18. Forming with the axes a triangle of area r^?
19. Find the equations of tlie tangents drawn from the
point (10,5) to the circle .t- -f" //" = 100.
20. Find the equations of tangents to the circle x^-\-if
+ 10a; — Qnj — 2 = 0 and parallel to the line y = 2x — 7.
21. Find the lengths of the subtangent and subnormal in
the cii'cle x^-\-if — l^x^ — 4// = 5 for the point (10,9).
22. What is the equation of the circle (centre at origin)
that is touched by the straight line x cos a-(- .V sina:=/>?
What are the coordinates of the point of contact?
23. When Avill tlie line Ax -\- B//— C = 0 touch the circle
x^ + ?/ = v-^ ? the circle (x — a.y + (y — f')' == r^ ?
24. Find the equation of the straight line toucliing .t^+//
= ax-\-hy and passing through the origin.
THE CIRCLE. 83
Prove that the following circles and straight lines touch,
and tind tlie point of contact in each case :
25. x^ -{- //'■^ -\- ax -\-b//^^0 and ax + iy + «'■' -\-b'^ = 0.
26. x^+i/- — 2ax — 2b y -\-h^ = {) and x = 2a.
27. iK^H- if = ax -\- bij and ax — bij -\- 1!^ = 0.
28. What is the equation of the circle (centre at origin)
that touches the Hue i/ = 3x — 5 ?
29. What must be the value of 7)i in order that the line
t/ = mx -\- 10 may touch the circle x^-\-y^ = 100 ? Show
that we get the same answer for the line y=^mx — 10, and
explain the reason.
y 30. Determine the value of c in order that the line
ox — 4//-l-c = 0 may touch the circle a;^+^^ — 8a- + 12y
— 44- = 0. Explain the double answer.
31. What is the equation of the circle having for centre
the point (5, 3) and touching the line 3x-\-2y — 10 = 0 ?
32. What is the equation of a circle whose radius =: 10,
and which touches the line 4a; -\-oy — 70 = 0 in the point
(10, 10) ?
33. A circle touching the line ■ix-\-oy-\-o=^0 in the
point ( — 3, 3) passes through the point (o, 9). What is
its equation ?
yT 34. Under what condition will the line --f-^^l touch
the circle x^-\-y'^^^ r' ?
35. What is the equation of the circle inscribed in the
triangle whose sides are
x = 0, y=0, - + -^=1?
a 0
84 ANALYTIC GEOMETRY.
36. Two circles touch each other when the distance
between their centres is equal to the sum or the difference
of their radii. Prove that the circles
touch each other, and find the equation of the common
tangent.
37. Two circles touch each other when the length of their
common chord ^=0. Find the length of the common chord of
(x-ay + (l/-by = r', (x-by + (>/-ay = ^^,
and hence prove that the two circles touch each other when
(a — by =^2)^. *
Exercise 23. (Review.)
Find the radii and centres of the following circles :
1. 3x' — 6x-\-3i/'-\-9i/ — 12 = 0.
2. 7x^- + 3>/ — 4u—(l — 2xy^0.
3- y(l/ — 5)=x{3 — x).
4. Vl + a' (x' + /) = 2b(x + uij).
Find the equation of the circle :
5. Centre (0, 0), radius = 9.
6. Centre (7, 0), radius =3.
7. Centre (—2, 5), radius = 10.
8. Centre (3a, 4a), radius = 5a.
9. Centre (b-\-c, b — c), radius = c.
10. Passing through {a, 0), (0, b), (2a, 2b).
11. Passing through (0, 0), (0, 12), (5, 0).
12. Passing through (10, 9), (4, —5), (0, 5).
1 3. Touching each axis at the distance — 7 from the origin.
THE CIRCLE. 85
14. Touching both axes, and radius =r.
15. Centre (a, a), and cutting chord = ^ from each axis.
16. Having the centre (0, 0), and touching ?/ = 2a; -f- 3.
17. Having the centre (1, — 3), and touching 2x — y = 4.
18. With its centre in the line 5x — 7y — 8=0, and
touching the lines 2x — 1/ = 0, x — 2y — 6 = 0.
19. Passing through the origin and the points common to
the circles
a^' + / - 6x — 10?/ — 15 = 0,
x''-\-u''-\-2x-\- 4?/-20 = 0.
20. Having its centre in the line ox — 3)/ — T = 0, and
passing through the points common to the same circles as
in No. 19.
21. Touching the axis of x, and passing through the
points common to the circles
x^-^7f -\- Ax- U/j- 68 = 0,
x^ + f — 6x — 22i/+30 = 0.
22. Find the centre and the radius of the circle which
passes through (9, 6), (10, 5), (3, —2).
23. What is the distance from the centre of the circle
passing through (2, 0), (8, 0), (5, 9) to the straight line
joining (0, — 11) and (— 16, 1) ?
24. What is the distance from the centre of the circle
x^-\-y^ — 'ix-\-8>/=0to the line ix — 3y -f 30= 0 ?
25. What portion of the line ?/ = o./- + 2 is contained
within the circle x^-\-y- — 13a; — 4y — 9^0?
26. Through that point of the circle a;^ + ^=25 for
which the abscissa = 4 and the ordinate is negative, a
straight line parallel to i/^ox — 5 is drawn. Find the
length of the intercepted chord.
86 ANALYTIC GEOMETRY.
27. Through the point (;/•,, y^), within tlie circle x--\-y"
= r% a chord is drawn so as to l)o bisected at this \)oint.
What is its equation ?
28. What rehation must exist among the coefficients of
the equation J (x- -\- ;/) -\- D.r -\- Ey + 6' = 0,
(i) in order that the circle may touch the axis of x ?
(ii) in order that the circle may touch thc^ axis ot y ?
(iii) in order that the circle may touch both axes ?
29. Under what condition will the straight line ?/ = wa;
-{- e touch the circle a?- + y- = 2rx ?
30. What must be the value of k in order that the line
3.x + 4y = k may touch the circle ?/ = IOj; — x'- ?
31. Find the equation of the circle that passes through
the origin and cuts equal lengths a from the lines x=^y,
x-\ry = 0.
32. Find the equations of the four circles whose com-
mon radius =a\/'2, and which cut chords from each axis
equal to 2a.
33. Fin'd the equation of the circle whose diameter is the
common chord of the circles x"^ -{- y''^ =^ r"^, (x. — <i)--\- y-=r-.
Find the equation of the straight line :
34. Passing through (0, 0) and the centre of the circle
x'^-\-y'^=a(x-\-y).
35. Passing through the centres of the circles
x' + y^ = 25 and x" + // + C^x — 8y == 0.
36. Passing through (0, 0) and touching the circle
a^ -\- y- - Crx — 12// + 41=0.
37. Parallel to a; + V3(^ — 12) = 0 and touching x^ + y-
= 100.
THE CIRCLE. 87
38. Passing through the points common to the circles
X- + //-— 2x— 4y— 20 = 0,
X- + / — 14.r — IG// + 100 = 0.
39. Prove that the common chord of the circles in No. 38
is perpendicular to the straight Hue joining their centres.
40. Find the area of the triangle formed by the radii of
the circle cc^-j- y^ = 169 drawn to the points whose abscissas
are — 12 and + 7 and ordinates positive, and the chord pass-
ing through the same two points.
41. Prove that an angle inscribed in a semicircle is a
right angle.
42. Prove that the radius of a circle drawn perpendicular
to a chord bisects the chord.
43. Find the inclination to the axis of x of the line joining
the centres of the circles x^ -|- 2a: + y^ = 0, x^ -\- 2i/ ■{- 1/~=0.
44. Determine the point from which tangents drawn to
the circles ^.2_|_y2_ 2x — ()//+ G = 0,
a;' + f — 22// - 20a: + 52 = 0,
will each be equal to 4 VC.
45. Find the equations of the circles that touch the
straight lines 6.x + 7// + 9 = 0 and 7x -\-6//-\-o^=0, the
latter line in the point (3, — 4).
Obtain and discuss the equations of the following loci:
46. Locus of the centre of a circle having the radius r
and passing through the point (xi,i/i).
47. Locus of the centre of a circle having the radius r'
and touching tlie circle (x — a)- -\- (// — 1))"= r.
48. Locus of all points from which tangents drawn to
the circle (x — ('y-\-(!/ — hy=^r^ have a given length t.
88 ANALYTIC GEOMETRY.
49. Locus of the middle point of a chord drawn through
a fixed point ^ of a given circle.
50. Locus of the point M which divides the chord AC,
drawn through the fixed point J^ of a given circle, in a
given ratio AM : MC = m : n.
51. Locus of a point whose distances from two fixed
points, A, B, are in a constant ratio m : n.
52. Locus of a point, the sum of the squares of whose
distances from two fixed points, A, B, is constant, and
equal to Jc\
53. Locus of a point, the difference of the squares of
whose distances from two fixed points, A, B, is constant
and equal to k^.
54. Locus of the middle point of a line of constant
length d which moves so that its ends always touch two
fixed perpendicular lines.
55. Locus of the vertex of a triangle whose base is fixed
and of constant length, and the angle at the vertex is also
constant.
56. One side, AB, of a triangle is constant in length and
fixed in position; another side, AC, is constant in length
but revolves about the point A. Find the locus of the
middle point of the third side, BC.
57. Find the locus of the intersections of tangents at the
extremities of a chord whose length is constant.
58. A and B are two fixed points, and the point P moves
so that PA = n X PB ; find the locus of P.
THE CIRCLE.
89
SUPPLEMENTARY PROPOSITIONS.
68. A Diameter of a curve is the locus of the middle
points of a system of parallel chords. The chords which
any diameter bisects are called the Chords of that diameter.
69. To find the equation of a diameter of the circle
Fig. 27.
Let the equation of any one of the parallel chords (Fig.
27) he 7/ = mx + c, and let it meet the circle in the points
(xi,i/i) a,nd(x2,1/o).
Then (§§ 37 and 64) m = - ^^^^- (1)
2/1 + 2/2
Let (x,i/) be the middle point of the chord; then 2x^=
Xi-\-x2, 2_y = ?/i -f 2/2 (§ 8), and by substitution we have
or
x
vi = ,
y
1
V = — ^)
(2)
a relation which evidently holds true for the middle points
of all the chords. Therefore (2) is the equation of a diameter.
Cor. From (2) we see that a diameter of a circle is a
straight line passing through the centre and perpendicular
to its chords.
90
ANALYTIC GEOMKTRY.
70. Two distinct, tiro coiiicidcitt, or no tangents can he
i/raivn to a circln tlinnKjIi anij point (A, k), according us this
jjoint is ivithout, on, uf ivlthin the circle.
Let the tancrent
(J = inx -f- /• Vl + ni^
j)uss through the point (/;, /.); then
mli -\-r '\ll -\- ni\
k
Transposing and squaring, we have
{U'—r'')ni'' — 2hkm=i^-
k\
hk ± r \J li' + Ic' — r''
h'
(1)
The values of m given in (1) are the slopes of the tan-
gents through (Ji,k). Now, these values are real and
unequal, real and equal, or imaginary, according as ]i^-\-k^
>, =, or <C r^; that is, according as (A, k) is without, on,
or within the circle. Hence, two distinct, two coincident,
or no tangents can be drawn througli (/<, k), according as
(Ji , k) is without, on, or within the circle.
Fig. 28.
71. To find the equation of the chord joining the points of
contact of the two tangents from any external jjoint (Ji, Jc).
THE CIRCLE. ' 91
Let (rTj, ?/i), (.ro, 7/2) I'f' tlu; points of contact (^ and 7i' ;
then the equations ol' tlie tangents FQ and PR are (§ G4)
Since both tangents pass tlirough ./'(A, h), botli these
equations are satisfied by the coordinates h, k ; therefore,
hxi + hji = r, (1)
hx^ + hj^ = r\ (2)
From equations (1) and (2) we see that tlie coordinates of
both the points (./'i, y^), (x^, 1/2) satisfy the equation
hx-\-kt/ = 7^. (3)
Hence, the locus of (3), wliich is a right line, passes through
both points of contact ; and, therefore, (3) is the eqnation of
the chord QR. The chord QR is called the Chord of Contact.
72. Sii.pjiose a eliord of a cirrlc. to turn round an// fixed
point (h, k) ; to find the lorvs of tlie intei'section of the two
tangents drawn at its extremities.
Let P (Fig. 29 or 30) be the fixed point (A, k), QPR one
position of the revolving chord, and let the tangents at Q
and R intei-sect in Pj, (.rj, 7/,) ; it is required tb find the
locus of Pi as the chord turns about P. Since QR is the
chord of contact of tangents drawn from the point P,
(•*^i) 2/i)) its equation is (§ 71)
3-1^ + 7/1// ='•''. (1)
Since (1) passes through (//, /,•), we have
hx, + k;,, = r\ (2)
But (x^, 7/j) is a7i7/ point in tlio required locus, and by (2)
its coordinates satisfy the eqnation
' />,--{- ki/=r"-; (3)
hence, (3) is the eqnation of tlie recpiirc^d locus.
Since (3) is of the first degree, the locus is a straight line.
92
ANALYTIC GEOMETKT.
The line hx + kij = r- is called the Polar of the point (h, k)
with regard to the circle x^-{-7/= 7^, and the point (h, k) is
called the Pole of the line. The pole (A, k) may be with-
out, on, or within the curve. In Fig. 29 it is within, while
in Fig. 30 it is without the circle.
Fig. 29.
THE CIRCLE. 93
Cor. 1. If the point (h, k) is on the circle, (3) is evi-
dently the equation of the tangent at (A, k) ; hence,
The polar of any point on the circle is identical with the
tangent at that 2^0 int.
Cor. 2. If (Ji, k) is an external point, by § 71, (3) is the
equation of the chord of contact of tangents from (h, k) to
the circle ; hence,
The polar of any external point is the same line as the
chord of contact of tangents draivn from that point.
Thus, in Fig. 30, iOf is the polar of P, or the chord of
contact of tangents drawn from P.
73. The polar and pole of a circle may be defined as
follows : If a chord of a circle is turned round a fixed
point (Ji, k), the locus of the intersection of the two tan-
gents at its extremities is the polar of the pole (h, k) with
regard to that curve.
74. If the polar of a point F passes through P', then the
polar of P' loill pass through P.
Let P be the point (A, k~), P' the point (A', k'), and let the
equation of the circle be x- -{- y'^=^ 1^.
Then the equations of the polars of P and P' are
hx-\-ky=i^ (1)
h'x + k'y = i^. (2)
If P' is on the polar of P, its coordinates must satisfy
equation (1) ; therefore,
hh'-\-kk' = i^.
But this is also the condition that P shall be on the line
represented by (2) ; that is, on the polar of P'. Therefore,
P is on the polar of P'.
This relation of poles and polars is illustrated in the
Figs. 29 and 30.
94
ANALYTIC GEOMKTUY.
75. To find a (jeometvical conslvnction for the polar of a
point with respect to a circle.
Fig. 32.
The equation of the line through any point P (Ji, k) and
the centre of the circle, or the origin, is
hx—hi/ = 0. (1)
Now, the equation of tlie polar of P is
/'.« + %=>^. (2)
But the loci of (l)and (U) are perpendicular (§ 45, Cor. 2).
Hence, if PC is the polar of P, OP is perpendicular to BC,
and
00 =
^h^+Jc"
(§41)
Also, OP^-sJh^'+k-.
Therefore, OPXOQ = A
Hence, to construct the polar of P :
Join OP, and let it cut the circle in A ; take Q in the
line OP, so that
OP : 0A = OA : OQ.
Tlie line through Q per^x'ndicular to OP is tlie ]iolar of P.
To locate the pole of BC, draw OQ perpendicular to BC,
andtakePsoth.t oQ:OA=OA:OP.
TIIK CIRCLE.
95
76. To find the length of tJie tangent drawn from any
point (h, k) to the circle (x — ay-\-(g — Of — >'^' = 0, (1)
Let F (Fig. 33) be the point (h, k), Q the [)oiiit of contact,
C the centre of the circle; then, since FQC is a riglit angle,
Fq'=fc^—qc\
Now, FC' = (h - a)- + (/.■ - by, and 'QC' = i\
Therefore, FQ- = (h — a)"- + (k - hf — i^.
Hence FQ^ is fonnd by simply substituting the coordinates
of F for X and xj in the expression {x — 0)''^+ {ij — hf — v"^.
P
Fig. 33.
Fig. 34.
If for brevity we write <S' instead of (.r — af -\- {ji — hy — r',
then the equation ;iS = 0 will represent the general equation
of the circle after division by the common coefficient of x^
and ?/, and we may state the above result as follows :
If )S^0 is the erp/atioji of a circle, and the coordinates of
any point are suhstitnted for x and y in S, the result will he
equal to the sqnare of the length of the tangent drawn from
the point to the circle.
11. To find the locus of the point from n'hich tangents
drawn to tu'o given circles are equal.
Let the equations of tlic circles O and O' (Fig. 34), be
(.■-«)^+0/-A)'-'-'=0, (1)
and (x — a'y -\-(y — l>'y — r'^=0. (2)
96 ANALYTIC GEOMETRY.
Then, if the tangents drawn from P (x, y) to the circles
(1) and (2) are eq^ual, we have
{x — ay-\-{y-hy- r2 ={x — a'f -{-{y-b'f- r'% (3)
which is the equation of the required locus.
Cor. 1. Performing the indicated operations in (3), and
transposing, we have
2 (« — a') ic + 2 {b — b') y = a' — a'^^b^ — b'^ — ',-^r''', (4)
which shows that the locus is a straight line.
This locus is called the Radical Axis of the two circles.
Hence, if S-i = 0, 82,=^ 0 are the equations of two circles, then
Si = 025 or Si — *S'2 =^^ 0,
7vill be the equation of their radical axis.
Cor. 2. When the circles *S'i = 0 and *S'2=:0 intersect, the
locus of Si = S2 passes through their common points.
Hence, ivhen two circles intersect or are tangent, their
radical axis is their common chord or common tangent.
CoR. 3. The slope of (4) is the negative reciprocal of the
slope of the line joining the centres of (1) and (2).
Hence, the radical axis of two circles is ji&i'petidicular to
the line joining their centres.
78. Let >S'=0, *S'i = 0, *S^2 = 0 be the equations of three
circles, in each of which the coefficient of x^ is unity.
Then the equations of their radical axes, taken in pairs, are
S — Si = U, Si — 02 ■"" ^t s — S2 ^= 0.
The values of x and y that will satisfy any two of these
equations will also satisfy the third. Therefore, the third axis
passes through the point common to the other two. Hence,
The three radical axes of three circles, taken in pairs, vieet
in a point. This point is called the Radical Centre of the
three circles.
THE CIRCLE, 97
Exercise 24.
1. What is the equation of the diameter of the circle
x^ -{- 7j- = 20 that bisects chords parallel to the line
2. What is the equation of the diameter of the circle that
bisects all chords whose inclination to the axis of x is 135° ?
3. Prove that the tangents at the extremities of a
diameter are parallel.
4. Write the equations of the chords of contact in the
circle x^ -\- y^ = 7'^ fov tangents drawn from the following
points: (r, r), {2r, 3?'), {a-\-h, a — b).
5. From the point (13, 2) tangents are drawn to the circle
cc- + y^ = 49 ; what is the equation of the chord of contact ?
6. What line is represented by the equation hx + ky=^ ?'^
when (Ji, k) is on the circle ?
7. Write the equations of the polars of the following
points with respect to the circle a:^ -f ?/- ^ 4 :
(i)(2,3). (ii)(3,-l). (iii)(l,-l).
8. Find the poles of the following lines with respect to
the circle x^-{-y^=^?>5:
(i) 4a; + 6?/ == 7. (ii) 3a; — 2y = o. (iii) ax + h)j = 1 .
9. Find the pole of 3x-\-4:y = 7 with respect to the
circle x--\- y-= 14.
10. Find the pole of Ax-\-By-\-C^O with respect to
the circle x^-\- y- = r.
11. Find the coordinates of the points in which the line
a; = 4 cuts the circle x^ + y = 2o; also find the equations
of the tangents at those points, and show that they inter-
sect in the point i^^-, 0).
98 anai.vtk; gkomktrv.
12. If tlio polavs of two points /', Q meet in B, then R
is the pole of the line FQ.
13. If the polar of (A, k) witli respect to tlie circle
3.2_j_y2_.j.2 touches the circle x^-\->f = 2rx, then k'^-\-'2rh
14. If the polar of (A, k) with respect to the circle
a:^ -\- f/^ = c^ touches the circle 4(^x^-\-y^)^c'^, then the pole
(h, k) will lie on the circle a;^+?/- = 4cl
15. Find the polar of the centre of tlie circle x'^ -\- //- = r^.
Trace the changes in the position of the polar as tlie pole is
supposed to move from the centre to an infinite distance.
16. Wliat is the square of the length of the tangent
drawn from tlie point (Ji, k) to the circle .t" + //' ^ ?•- ?
17. Find the length of the tangent drawn from (2, 5) to
the circle x^- + f — 2x — ,% — 1 = 0.
Find the radical axis of the circles :
18. (.r + 5)^ + (y + 6)~o, (x-7y-\-(>/-ny-=i6.
1 9. .7-2 + ff -f 2x + 3// — 7 = 0, X- + //•-' — 2x — 7/4-1=0.
20. X- + f + hx + />// — c = 0, ax"^ + air + ""-> + />'// = ^^■
21. Find the radical axis and length of tlie common
choiwl of tlie circles
.r," 4- y' -\- (IX -\- Jnj -|- *" = 0, :i^ -\- //- -|- hx -f- (nj 4- c = 0.
22. Find the radical centre of the three circles
a;2 + yH 4x + 7 = 0,
2.^2+ LY- + o.r + 5/y + 9 = 0,
a;^-|- ?/"-(- // = 0.
CHAPTER IV.
DIFFERENT SYSTEMS OF COORDINATES.
Rectilinear System.
79. When we define the position of a point, with refer-
ence to any fixed lines or points, we are said to nse a
System of Coordinates.
In the Rectilinear System, already described, we liave
thus far employed only rectangular axes, or coordinates,
which are to be preferred for most purposes, on account
of their greater simplicity. When the axes of reference
intersect at oblique angles, the axes and coordinates are
called Oblique.
,Y
Let OX, OF (Fig. 35) be two axes making an acute angle,
A'OF=w, with each other. If we draw PN^ to OX, and
PJ/|| to OY, then the coordinates of P are
NP = 0M= X, MP = //.
Since oblique and rectangular coordinates differ only in
the angle included between the axes, any of the previously
deduced formulas that do not depend on any property of
the right angle are apjdicable wlien the axes are oblicjue.
Thus, formulas [2], [3], [4], [7J liold for oblique axes as
100 ANALYTIC GEOMETRY.
well as for rectangular, and therefore are general formulas
for the Rectilinear System.
When the axes are oblique, instead of [1], we evidently
have (Fig. 35)
PQ = \IpW + UQ' — 2PR XBQ cos PBQ,
••• d= '^ (^2 — xiY + (yz — yif -\- ''^ix2 — ^i) (yz — yi)cosa>,
which reduces to [1] when w=:90°.
The Rectilinear System is sometimes called the Cartesian
System, from Descartes, who first used it.
80. To find the equation of the straight line AC, referred
to the oblique axes OX, OF (Fig. 36), having given the inter-
cept OB=^b and the angle XAC=y.
Let P be any point (x, y) of the line. Draw BD || to
OX, meeting PM in D. Then, by Trigonometry,
PD sin y y — h sin y
BD sin (a» — y)' X sin (w — y)
If now we put m = —. — ; — - — r , we obtain as the result an
sin ((!> — y)
equation of the same form as [6], p. 38,
y = mx -\- b.
Here 771 = the ratio of the sines of the angles which the
line AC makes with the axes ; that is, m = sin JT^P-^ sin
PBY, which equals tan X^P^tan y when w = 90°.
DIFFERENT SYSTEMS OF COORDINATES.
101
81. Oblique coordinates are seldom used, because they
generally lead to more complex formulas than rectangular
coordinates. In many cases, however, they may be em-
ployed to advantage. An example of this kind is furnished
by problem No. 23, p. 65 :
To prove that the medians of a triangle meet in one point.
If a, b, c represent the three sides of the triangle, and we
take as axes the sides a and b, then the equations of the
sides and also of the medians may be written with great
ease, as follows :
The sides, !/ = 0, x = 0, - + {^ = 1.
^0, '-+■''■■
a b
The medians,
a 0 a b
1=0,
a b
On comparing the equations of the medians, we see that
if we subtract the second equation from the first, we obtain
the third; therefore, the three medians must pass through
the same point (§ 53).
Polar System of Coordinates.
82. Next to the rectilinear, tlie system of coordinates
most frequently used is the Polar System.
Fig. 37
Let 0 (Fig. 37) be a fixed point, OA a fixed straight line,
P any point. Join OP.
102 ANALYTIC GEOMETRY.
It is evident tliat we know the position of P, provided
we know the distance OP and the angle wliieh OT I'oi'uis
with OA.
Thus, if we denote the distance OP by p, and tlie angle
AOP by 6, the position of P is determined if p and 6 are
known.
The fixed point O is called the Pole, an<l the fixed line
OA the Polar Axis; p and 6 are called the Polar Coordinates
of P; p, its Radius Vector; and^, its Direction or Vectorial
Angle.
Every point in a plane is perfectly determined by a posi-
tive value of p between 0 and oo, and a positive value of 6
between 0° and 3G0° (or 0 and 27r, circular measure). P>ut,
in order to represent by a single equation all the points of
a geometric locus, we often employ negative valnes of p and
6, and adopt the following laws of signs :
(i) 6 is positive when measured from right to left, and
negative when measured in tlie opposite direction.
(ii) p is positive or negative according as it extends in
the direction of the terminal side of 6 or in the opposite
direction. Thus, any given point may be determined in
four different ways.
DIFFERENT SYSTEMS OF COOKDINATES.
103
Fur example, suppose tluit the straight line r()l\ bisects
the first and third quadrants, and that in this line we take
points F, Pi, at the same distance OF = a from O ; then
F is the point(//, i.7r)or( — a,|7r)or(— «, — ^7r)or( a, — In);
Fi is the point (^A, ^Tr)o\-{—a,\Tr) or( a, — ^7r)()r(— «, — ;j7r).
Fig. 39.
Fig. 40.
'Y 83. To Jiml the jJolar equation of a circle.
(i) Let the pole 0 be at the centre (Fig. ,38). Then, if
r denotes the radius, the polar equation is simply p=^r.
(ii) Let the pole 0 be on the circumference (Fig. o*.)),
and let the diameter OF make an angle a with the initial
line OA. Let F be any point (p, 6) of the circle. Join BF.
Then, OF=^OD cos EOF,
or P = 2»' cos (9 - a). [23]
If OB is taken as the initial line, the equation becomes
P = 2r cos 9. [24]
(iii) Let the pole 0 be any point, and the centre the
point (p', 6'). Then in the triangle OCF (Fig. 40),
0F~ — 20FX OCXcos COF-]- OC^— aF- = Q,
or 9- - 2pp' cos (8 - 6') + p'-2 - r'^ = O, [2r)]
the most general form of tlie polar equation of a circle.
104 ANALYTIC GEOMETRY.
Exercise 25.
1 . Find the distances from the point P in Fig. 38 to the
two axes.
2. Prove that the equation of a straight line, referred to
oblique axes in terms of its intercepts, is identical in form
with [7], p. 39.
3. If the straight line P2O-P3 (Fig. 38) bisects the second
and fourth quadrants, what are the polar coordinates of the
points P2 and P3 ? Give more than one set of values in
each case.
4. Construct the following points (on paper, take a ^ 1 in.) :
«, 0 V [a, IT ), [<i,—
9
2„,|), (2„,.), (^„cos|| ), (.,|), (^S»,|
— 3rt, — I, ( 4rt, tan~^ - ), I 4a, tan-^ -
' 3/ V 3/ V 4
4
Note. The expression tan— 1 - in higher Mathematics means the
o
4
angle whose tangent is -■
5. If pi, p2 denote the two values of p in equation [25],
p. 103, prove that pip2 = p'^ — '■^- What theorem of Ele-
mentary Geometry is expressed by this equation if the pole
is outside the circle ? if the pole is inside the circle ?
6. Through a fixed point P in a circle a chord PP is
drawn, and then revolved about P; find the locus of its
middle point.
Note. In such problems as this there is a great advantage in
using polar equations.
7. li p denotes the distance from the pole to a straight
line, a the angle between p and the polar axis, prove that
the polar equation of the line is p cos (0 — a) =p.
DIFFERENT SYSTEMS OF COORDINATES.
105
Transformation of Coordinates.
84. The equation of a curve is oftentimes greatly sim-
plified by referring it to a new set of axes, or to a new
system of coordinates. For example, compare equations
[15] and [16], p. 71. Hence, it is sometimes useful to
be able to deduce from the equation of a curve referred
to one set of axes or to one system of coordinates, its
equation when referred to another set of axes or to an-
other system of coordinates. Either of these operations
is known as the Transformation of Coordinates. It consists
of expressing the old coordinates in terms of the new, and
then replacing in the equation of the curve the old coordi-
nates by their values in terms of the new ; we thus obtain
a constant relation between the new coordinates, that will
represent the curve referred to the new axes or system.
85. To change the, origin to the •point (Ji, k) without
changing the direction of the axes.
Y
Y'
f
w
0
r
0
-
i
M
X
Fig. 41.
Let OX, OF be the old axes, O'X', O'Y' the new; and
let (x, y), (x', y') be the coordinates of the same point P,
referred to the old and new axes respectively.
Then (Fig. 41)
OA = h, AO'^k, OM=x, MP^y, 0'j\r = x\ MP = y'.
x=OA + AM^ 0A-{- O'M' = x' -\- h.
y = MM' + MP =AO'-Y MP = u'-{-k.
106
ANALYTIC GEOMETRY.
These relations are equally true ior rectangular and
oblique coordinates.
Hence, to find what the equation of a curve becomes
wlieu the origin is transferred to a point (Ji, k), the new
axes running parallel to the old, we must substitute for x
and 1/ the values given above.
After the substitution, we may write x and // instead of
x' and y' ; so that })ractically the change is eifected by
siinjjli/ ivrltiny x-\- h in jjlt^f-ce of x, y-\-h i/ij^lcice of y.
If, however, we wish to transform a ^>t»m^ (x, y) from
the new to the old system, we must write x — It in place
of x and y — k in place of y.
86. To chanye the reference of a curve from one set of
rectanyidar axes to another, the orhjin remainintj the same.
Let (a:, y) be a point P referred to the old axes OX, 0 Y;
(x', y') the same point referred to the new axes OA"', OY
(Fig. 42). Then
03f=x, MP = y, ON=x', Nr = y'.
Let the angle XOX' == 6. Draw NQ, NR i. to PM, OX,
respectively; tlien
NPQ= QXO = EON= e.
DIKKKIJENT SYSTEMS OF COOliDIN A'I'KS. 107
Hence, 0M= OR — RM= OR — NQ = ON cos 6 — FN si ii 6.
Or x^=x' COS 6 — y' sin 6.
And PJ/= MQ + QF = KN-\- QF = ON sin Q + 7'.V cos ^.
Or y^=x' sin ^ + y' cos 6.
Therefore, to find what the equation of a curve becomes
when referred to the new axes, we must write
X cos 6 — 1/ sin 6 for x, x sin. 6 -\- // cos 6 for y.
"TT"^ 87. To change the reference of a curve from one set of
rectangular axes to another, both the origin and the direction
of the axes being changed.
First transform the equation to axes through the new
origin, parallel to the old axes. Then turn these axes
through the required angle.
If (Ji, k) is the new origin referred to the old axes, 0 the
angle between the old and new axes of x, we obtain as the
values of x and y for any point P, in terms of the new
coordinates,
a* = A 4" x' cos 6 — y' sin 6,
y^k-{-x' sin 6 -\- y' cos 6.
In making all these transformations, attention must be
paid to the signs of h, k, and 6.
88. To change the reference of a curve from, rectangular
to oblique axes, the origin remaining the same.
Let a, ^ be the angles formed bj'' the 2iositive directions
of the new axes OX', OY' (Fig. 43) with the positive
direction of OX Let tlie old coordinates of a point P
be x, y; and the new coordinates, x\ //'. Then from the
right triangles ORN, PQN we readily obtain tlie formulas
X = x' cos a + y' cos (3,
y = x' sin a-\-y' sin (3.
Investigate the special case when fS=^a-\-[)0°.
108
ANALYTIC GEOMKTltV.
89. To deduce the formulas for finding the polar equation
of a curoe from its rectangular equation.
Let cc, y be the rectangular coordinates of any point P,
and p, 6 its polar coordinates.
Y
'ix-^
\
vr
^^
0
M
R X
Y
/P
/ ^^-^"^^^^
0
M X
Fig. 43.
Fig. 44.
(i) Let the origin of rectangular coordinates be the pole,
and let the polar axis coincide with the axis of x.
Then (Fig. 44)
031= OP cos 3I0P,
PM= OP sin 3£0P.
Or cc = /o cos 6,
y=^p sin 6.
(ii) If the pole is the point (h, k), we have
x = h-{- p cos $,
y = k-\-p sin 0.
(iii) If the pole coincides with the origin, but the polar
axis OA makes the angle a with the axis of x, we obtain
x = p cos (^ + a),
y = p sin (^ + a).
(iv) If the pole is the point (h, k), and the polar axis
makes the angle a with the axis of x,
x = h-]r p cos (6-\-a),
y = k-\-p sin (<9 + a).
DIFFERENT SYSTEMS OF COORDINATES. 109
90. To deduce the fonnulas for finding the rectangular
equation of a curve from its polar equation.
From the results in cases (i) and (ii) of § 89 (the only
cases of importance), we readily obtain
In case (i), p^ = x--\- i/, tan 6 = --
In case (ii), p^ = (x — hy-\-{y — k),^ tan ^ =
X
y—k
X — h
91. The degree of an equation is not altered by passing
from one set of axes to another.
For, however the axes may be changed, the new equation
is always obtained by substituting for x and y expressions
of the form
ax-\-by-\rc and a'x -\- />'// -\- c'.
These expressions are of the first degree, and, therefore, if
they replace x and y in the equation, the degree of the equa-
tion cannot be raised. Neither can it be lowered ; for if it
could be lowered, it would be raised by returning to the
original axes, and therefore to the original equation.
Exercise 26.
1. What does the equation y"^ — 4x + 4?/ + 8 = 0 become
when the origin is changed to the point (1, — 2) ?
Transform the equation of the circle (x — «)" -f (y — by
= j"^ by changing the origin :
2. To the centre of the circle.
3. To the left-hand end of the horizontal diameter.
4. To the npper end of tlie vertical diameter.
5. What does the equation x^-\- y^^r^ become if the axes
are turned through the angle a ?
6. What does the equation x^ — y^ = a^ become if the
axes are turned through — 45° ?
110 ANALYTIC GEOMETRY.
7. The equation of a curve referred to rectangular axes
is X — xy — y^= 0. Transform it to new axes, whose origin
is the point ( — 1, 1), the new axis of y bisecting two of
the angles formed by the old axes.
8. Change the following equations to polar coordinates,
taking the pole at the origin and the polar axis to coincide
with the axis of x :
(i) a;^ -|- y^ = a^. (ii) x^ — )f = a^.
9. Change the equation x^=:4a?/ to polar coordinates,
(i) taking the pole at the origin ; (ii) taking the pole at
the point (a, 0).
10. Change the following equations to rectangular coordi-
nates, the origin coinciding with the pole, and the polar axis
with the axis of x :
(i) p ^= a, (ii) p = a cos 6, (iii) p^ cos 26 == al
Transform the following equations by changing the origin
to the point given as a new origin :
11. x-\-i/-\- 2 = 0; the new origin ( — 2, 0).
12. 2x — 5y — 10 = 0 ; the new origin (5, — 2).
13. ^x"^ -\- -ixij -\- if — 5x — Qij — 0 = 0; new origin (J, — 4).
14. x^ -\-y'^ — 2x — 4y = 20 ; new origin (1, 2).
1 5 . a;^ — Q)xy -\- if — Cj:' + 2 // -|- 1 = 0 ; new origin (0, — 1) .
16. Transform the equation x' — v/2-|-6 = 0 by turning
the axes through 45°.
17. Transform the equation (x-\- y — 2rt)^=: 4a:-^ by turn-
ing the axes through 45°.
18. Transform the equation 9^^ — 16//= 144 to oblique
axes, such that the new axis of x makes witli the old axis
3
of x the negative angle tan~' — - ; and the new axis of y
. . '->
makes with the old axis of x the i)0sitive angle tan ~^-.
4
DIFFKKKNT SYSTEMS OF C06ui>INATFS. Ill
Exercise 27. (Review^.)
1. Find the distance from the point ( — 2b, b) to the
origin, the axes making the angle 60°.
2. The axes making the angle w, find the distance from
the point (1, — 1) to the point (— 1, 1).
3. The axes making the angle w, find the distance from
the point (0, 2) to the point (3, 0).
Determine the distance between the following points
given by polar coordinates :
4. {a, 6) and {b, <j>).
5. {a, 6) and (a, — 6).
6. {a, 6) and (— a, — 0).
7. (2a, 30°) and (a, 60°).
8. Show that the polar coordinates (p, 6), ( — ,o, 7r + ^),
( — p, 6 — tt) all represent the same point.
9. Transform the equation H.r--\-^.ri/-^4//--\-l2x-\-S}/
+ 1 = 0 to the new origin ( — ^, — ^).
10. Transform the equation 6x^+3^" — 24.r + 6 = 0 to
the new origin (2, 0).
11. Transform the equation --|-^ = 1 by changing the
origin to the point ( -, - ) and turning the axes through an
angle cb, such that tan d> =
"^ a
12. Transform the equation 17.r- — 16.r// -f- 1 7;/- = 225 to
axes that bisect the axes of the old system.
Transform the following rectangular equations to polar
equations, the polar axis in each case coinciding with, or
being parallel to, the axis of x, and the pole being at the
point whose coordinates are given :
112 ANALYTIC GEOMETRY.
13. x^-\-i/=Sax ; the pole (0, 0).
14. X- -\- ij-^'6ax; the pole (4a, 0).
15. 3/- — 6y — 5.x + 9 = 0 ; the pole (| , 3).
16. x^—if — Ax-Q>j — U = 0; the pole (2, — 3).
17. ix--{-y-y = k\x''—ij') ; the pole (0, 0).
Transform the following polar equations to rectangular
axes, the origin being at the pole and the axis of x coincid-
ing with the polar axis :
18. p^sm2e = 2a^
19. p = k sin 2$.
20. p(sin 3^ + cos 3^) = 5k sin $ cos d.
21. Through what angle must a set of rectangular axes
be turned in order that the new axis of x may pass through
the point (5, 7) ?
22. The rectangular equation of a straight line is Ax + By
■j- C=0. Through what angle must the axes be turned in
order that
(i) the term containing x may disappear ?
(ii) the term containing y may disappear ?
23. Deduce the following formulas for changing from one
set of oblique axes to another, the origin remaining the same :
x' sin (o> — a) j^ y' sin (w — /3)
sin 0) sin w
x' sin a , y' sin B
y=' — ■■ r' — '■ —
sm 0} sm w
Note. In these formulas w denotes the angle formed by the old
axes, a and ^ those formed by the positive directions of the new axes
with the positive direction of the old axis of x.
24. From the formulas of I^o. 23 deduce those of § 88,
CHAPTEE V.
THE PARABOLA.
The Equation of the Parabola.
92. A Parabola is the locus of a point whose distance
from a fixed point is always equal to its distance from a
fixed straight line.
The fixed point is called the Focus; the fixed straight
line, the Directrix.
The straight line through the focus perpendicular to
the directrix is called the Axis of the parabola.
The intersection of the axis and the directrix is called
the Foot of the axis.
The point in the axis halfway between the focus and
the directrix is, from the definition, a point of tlie curve ;
this point is called the Vertex of the parabola.
The straight line joining any point of the curve to the
focus is called the Focal Radius of the point.
A straight line passing through the focus and limited by
the curve is called a Focal Chord.
The focal chord perpendicular to the axis is called the
Latus Rectum or Parameter.
93. To construct a parabola, having given the focus and
the directrix.
I. Bij Points. Let F (Fig. 45) be the focus, CE the
directrix. Draw the axis FD, and bisect FD in A ; tlien
A is the vertex of the curve. At any point M in the axis
erect a perpendicular. From F as centre, with DM as
114
ANALYTIC (iKOMKTUV.
radius, cut this perpendicular in 1* and (J ; then /* and Q
are two points of the curve, for 7*7^ = X'iJ/= distance of F
or Q from CE. In the same way we can find as many
points of the curve as we please. After a sufficient number
of points has been found, we draw a continuous curve
through them.
Fig. 45. Fig. 46.
II. Bij Motion. Place a ruler so that one of its edges
shall coincide with the directrix DE (Fig. 46). Then place
a triangular ruler BCE with the edge CE against the
edge of the first ruler. Take a string whose length is
equal to BC ; fasten one end at B and the other end at F.
Then slide the ruler BCE along the directrix, keeping the
string tightly pressed against the ruler by tlie point of a
pencil F. The point F will trace a parabola ; for during
the motion we always have FF=FC.
94. To find the rectangular equation oftheparahola, when
its axis is taken as the axis of x and its vertex as the origin.
Let F (Fig. 45) be the focus, CE the directrix, DF.Y the
axis, A the vertex and origin; also let 2^) denote the known
distance FF.
Let F be any point of the curve ; then its coordinates are
A3I=x, MF = y.
THK I'AKAHOLA.
115
Draw PC_Lto CE\ then by the detiuition of the curve
FF^PC = 1)M.
Therefore, FF' = ljJr.
Now FF''^JIF'+l^'=i/--\-(x—jjy,
and lJJl'=^(x-\-py.
Therefore, y^ + (•*' ~ P) ' =^ (•'■f + p)"-
Whence, y^= 4.pjc. [2G]
This is called the principal equation of a parabola.
95. Since ?/ and jj in equation [2G] are positive, x must
always be positive; therefore, the curve lies wholly on the
positive side of the axis of y.
An examination of equation [26] shows that the curve,
(i) passes through the origin, (ii) is symmetrical with respect
to the axis of x, (iii) extends towards the riglit without
limit, (iv) recedes from the axis of x without limit.
96. Any point (h, />-) is outside, on, or inside the jiorahola
jf=^A:px, arrordin;/ <is k^ — -iph is positive, zero, or ner/atire.
Let Q be the point (A, k), and let its ordinate meet the
curve in P.
116 ANALYTIC GEOMETRY.
If k"^ — 4ph = 0, the point (A, k) satisfies equation [26]
and therefore Q coincides with P.
If k^ — 4ph is positive, or k^'^-iph, then, since P3I =
Aph, we have QM''>FJI', or QM>FM; hence, Q is
outside the curve.
If k'^—Aph is negative, we may prove similarly that Q
must be inside the curve.
97. If X =-p, y = ± 2/?. But these two values of y make
up the latus rectum. Hence, the latus rectum = 4:2j.
Cor. From the equation y^ = 4px, it follows that
x:y = y:4p;
that is, the latus rectum is a third proportional to any
abscissa and its corresponding ordinate.
98. If (xi, yi) and (x^, 2/2) s^re any two points on the
parabola, we have
yj^ = 4:pxi, y^^ = ipx2.
Hence, yi^ ■ y2^ = ^i- ^2',
that is, the squares of the ordinates of any two points on the
parabola are to each other as their abscissas.
99. To find the points in which the straight line y=^mx
+ G meets the parabola y"^ = 4px.
Eegarding these equations as simultaneous, and eliminat-
ing X, we have .
Whence, y=^±^_^ jp-mc^ ^2)
-^ 7n 7n \ p ^ ^
From (2) it follows that y = vix-\-c has two distinct, two
coincident, or no points in common with g-^^ipx, according
as ^ — mc >, =, or < 0.
CoR. If pj — mc = 0, or c^=p-^ni, y = mx-\-c will be a
tangent; that is, y = i7ix-\-— (3)
is a tangent to y'^^Apx in terras of its slope.
THE PAKAUOLA. 117
Exercise 28.
1. Show that the distance of any point of the parabola
y^ =: ^px, from the focus is equal to p -\- x.
2. Find the equation of a parabola, taking as axes the
axis of the curve and the directrix.
3. Find the equation of a parabola, taking the axis of
the curve as the axis of x and the focus as the origin.
4. The distance from the focus of a parabola to the
directrix = 5. Write its equation,
(i) if the origin is taken at the vertex,
(ii) if the origin is taken at the focus,
(iii) if the axis and directrix are taken as axes.
5. The distance from the focus to the vertex of a parab-
ola is 4. Write its equations for the three cases enumerated
in No. 4.
6. For what point of the parabola ?/-:=18x is the ordi-
nate equal to three times the abscissa?
7. Find the latus rectum of tlie following parabolas :
y'^=^Qx, f/'^=lox, hif^ax.
Find the points common to the following ])arabolas and
straight lines :
8. if = ^x, 3.C — 7y -f oO = 0.
^ 9. f = ?>x, ic — 4//+12 = 0.
10. if = '^.x, a.- = 9, a=0, a- = — 2.
11. ?/- = 4x, ?/ = 6, y = — 8.
12. What must be the value of ^; in order that the parab-
ola ij'^ = 'ipx may pass through the point (9, — 12) ?
118 ANALYTIC GEOMETRY.
13. For what point of the pavabohx i/'' = o2x is the ordi-
nate equal to 4 times the abscissa ?
14. The equation of a pavaboLa is i/- = Sx. What is the
equation of (i) its axis, (ii) its directrix, (iii) its hitus
rectum, (iv) a focal chord tli rough the point whose ab-
scissa =8, (t) a chord })assing through the vertex and the
negative end of the latus rectum ?
15. The equation of a parabola is ?/^ = 16a;. Find the
equation of (i) a chord through the points whose abscissas
are 4 and 9, and ordinates positive ; (ii) the circle passing
through the vertex and the ends of the latus rectum,
16. If the distance of a point from the focus of the
parabola if = 4:j)x is equal to the latus rectum, what is the
abscissa of the point ?
17. In the parabola y^ = A2jx an equilateral triangle is
inscribed so that one vertex is at the origin. What is the
length of one of its sides ?
18. A double ordinate of a parabola = 8/). Prove that
straight lines drawn from its ends to the vertex are jperpen-
dicular to each other.
Explain how to construct a parabola, having given :
19. The directrix and the vertex.
20. The focus and the vertex.
21. The axis, vertex, and latus rectum.
^^22. The axis, vertex, and a point of the curve.
23. The axis, focus, and latus rectum.
24. Determine, as regards size and position, the relations
of tlie following parabolas :
(i) y^z=4jjx, (ii) f = — 4:px, (iii) x- = 4:j)i/, (iv) x^ = — 4rp//.
the pakabola, 119
Tangents and Nokmals.
1 00. To find the equation of the tangent and of the normal
to tJte parabola y^='^2^x at any j^oint (a*!, ?/i).
Let (xi, yi), (x.2, ?/o) be any two points on the parabola;
then the equation of the secant through them is
Ju «/ 1 i/^2 ~~^ OC-t
Since (.Ti, ?/i) and (.r^, //o) are on tlie cvirve y- = ^jjx, we have
2/2 — !h _ k^ .
a-2 — a-i V/2 + Vi
By substituting in (1), the et^uation of the secant becomes
rzjh^_J^. (2)
x — x^ y^ + yi
Now, if (xo, jfo) is made to coincide witli (x^, //i), (2) be-
comes the equation of tlie tangent ix\>(xi,y^. Putting //a:=yi,
clearing of fractions, and remembering that y^^-^px^, Ave
obtain as the equation of the tangent at (xi,y^,
yr!/=22){.r + x,). [27]
The normal passes through (x^, y^, and is perpendicular
to the tangent; hence, its equation is, by [27] and §415,
101. If in [27] we put // = 0, we obtain
.<• = — .Ti, or TA = AM(Y\g. 48).
Therefore, tlie snbtangent Is bisected at the vertex.
If in [28] we put y=^0, we obtain
x = x^ + 2p, or X — xi [= J/xV] = 2/; (Fig. 48).
Hence, the stibnoi-inal is constant and e</ual to the scini-
latus rectum.
120
ANALYTIC GEOMETRY.
Cor. These properties furnish simple methods for draw-
ing tangents to the parabola. Thus, to draw a tangent to
the parabola at P (Fig. 48), draw the ordinate PM, lay off
AT=AM, and draw PT, which will be the tangent at P
by § 101. Or lay off MN= FD, and draw PN; then PT
perpendicular to PN at P will be the tangent at P.
Fig. 48.
102. In the triangle FPT(Yig. 48) we have
FT= FA + A T=p + X,
FP = PC= DM= DA -I- AM=^p + x.
Therefore, FT=FP.
Hence, the angle
FPT= PTF= TPC, or
The tangent to a parahoht at a mj point makes equal aiigles
7vith the axis of the curve and the focal radius to the point of
contact.
THE PARABOLA. 121
Iizercise 29.
1. The normal to a parabola at any point bisects the
angle between the focal radius and the line drawn through
the point parallel to the axis.
Note. The use of parabolic reflectors depends on this property.
A ray of light issuing from the focus and falling on the reflector is
reflected in a line parallel to the axis of the reflector.
2. Explain how to draw a tangent and a normal to a
given parabola at a given point.
3. Prove that FC (Fig. 48) is perpendicular to FT.
4. Prove that the tangent y=^mx-\-~ touches the
parabola if = Apx at the point [ ^' —
5. Prove that the equation of a normal to the parabola
y^^=^px in terms of its slope is rj = mx — mp (2 + m?).
6. What are the equations of a tangent and a normal
to the parabola y'^=^ox, that pass through the point whose
abscissa is 20 and ordinate positive ?
7. What are the equations of the tangents and the nor-
mals to the parabola ^^=: 12a*, drawn through thg ends of the
latus rectum? Find the area of the iigure they enclose.
8. Through the point on the parabola y^^= Idx whose
abscissa is 7 and ordinate positive a tangent and a normal
are drawn. Find the lengths of the tangent, the normal,
the subtangent, and the subnormal.
9. A tangent to the parabola if = 20x makes with the
axis of X an angle of 45°. Determine the point of contact.
10. Show that the focus i^(Fig. 48) is equidistant from
the points F, T, N. What easy way of drawing a tangent
and a normal is sugiiested by this theorem ?
122 ANALYTIC GEOMETRY.
11. If F is the focus of a parabola, and Q, R denote the
points in which a tangent cuts the directrix ?ind the latus
rectum produced, prove that FQ=FM.
12. Prove that the tangents drawn through the ends of
the hitus rectum are perpendicular to each other.
13. Find the distances of the vertex and the focus from
the tangent v/= mx -\
14. Find the point of intersection of the tangents to the
parabola y^ = ^px at the points (x^, 3/1), (x^, 2/2)-
15. A tangent to the parabola if^=^px cuts equal inter-
cepts on the axes. Wliat is its equation ? What is the
point of contact? What is the value of each intercept?
16. Through what point in the axis of x must tangents to
the parabola if = 4,jjx be drawn in order that they may form
with the tangent, through the vertex, an equilateral triangle ?
17. For what point of the parabola f = 4:jjx is tlie nor-
mal equal to twice the subtangent?
18. For what point of the parabola if=^^px is the nor-
mal equal to the difference between tlie subtangent and the
subnormal ?
19. Find tlie equation of the tangent to the parabola
y^^^x parallel to the straight line ox — 2_y + 7 = 0. Also
find the point of contact.
20. Find the equation of the straight line that touches
the parabola y"= 12.« and makes an angle of 45° with the
line y = ^x — 4. Also find the point of contact.
21. Find the equation of a straight line that touches
the parabola y^ = 16.r and passes through the point ( — 4, 8).
22. If a normal to a parabola for the point P meets the
curve again in the point Q, find tlie length of PQ.
THE PARABOLA. 123
23. Trove by the secuiit method that the e(|uation of a tan-
gent to the parabola if^=^\px — 4^r, at the point (iCi, v/i) is
VxU ^ 2^^ (■^ + -^i) — ^f-
24. Find the equations of the tangents and normals to
the parabola if — '^x — G^ — 63^0, for the points whose
common abscissa =; — 1.
25. What are the general equations of tangents to the
following parabolas :
(i) y- = — ^px ? (ii ) .7-- ^ ^pii ? (iii) x- = — Ajjt/ ?
Exercise 30. (Revie^Ar.)
Note. If not otherwise specified, tlie axis of the parabola and tlie
tangent at the vertex are to be assumed as the axes of coiirdinates.
What is the equation of a parabola :
1. If the axis and directrix are taken as axes, and the
focus is the point (12, 0) ?
2. If the axis and tangent at the vertex are the two
axes, and (25, 20) is a point on the curve ?
3. If the same axes are taken, and the focus is tlie
point (—4^, 0)?
4. If the axis is parallel to the axis of x, the vertex is
the point (5, — 3), and the latus rectum = 5^?
5. If the axis is tlie line ?/ = — 7, the abscissa of the
vertex = 3, and one point is (4, — 5) ?
6. If the curve passes through the points (0, 0), (3, 2),
(3,-2)?
7. If the curve passes through tlie I'.oints (0, 0), (3, 2),
(-3,2)?
124 ANALYTIC GEOMETRY.
8. What is the latus rectum of the parahohi 2//^ = ou' ?
What is the equation of its directrix, and of the focal
chords passing through the points whose abscissa := 6 ?
9. Describe the change of form which the parabola if ==
4px undergoes as we suppose p to diminish without limit.
10. Find the intercepts of the parabola
i/-{-ix — 6ij — 16 = 0.
11. One vertex of an equilateral triangle coincides with
the focus, and the others lie on the parabola ^/^ = 4^0:. Find
the length of one side.
12. The latus rectum of a parabola = 8; find
(i) Equation of a tangent through its positive end.
(ii) Distance from the focus to this tangent,
(iii) Equation of the normal at this point.
13. What is the equation of the chord passing through
the two points of the parabola y^ = 8x for which x = 2,
y > 0, and x = 18, y < 0 ?
14. Find the equation of the chord of the parabola
fz=4,px that is bisected at a given point (x^, y^).
15. In what points does the line x-]ry ^=12 meet the
parabola ^/^ + 2x — 12?/ + 16 = 0 ?
16. In what points does the line 3?/^ 2a; + 8 meet the
parabola y'^ — 4cX~Sy -\-24: = 0 ?
17. Find the equations of tangents from the origin to
the parabola (y — by=^ A'p(x — a).
18. Describe the position of the parabola y^ + 2a:; + 4 = 0
with respect to the axes, and determine its latus rectum,
vertex, focus, and directrix.
19. What is the distance from the origin to a normal
drawn through the end of the latus rectum of the parabola
y'^=^4^a(x — a) ?
THE PARABOLA. 125
Find the equation of the parabola :
20. If the equation of a tangent is 4y = 3x — 12.
21. If a focal radius = 10, and its equation is 3^ = 4a; — 8.
22. If for a point of the curve the focal radius = r, and
the length of the tangent = ^.
23. If for a point of the curve the focal radius = r, and
the length of the normal = n.
24. If for a point of tlie curve tlie length of the tangent
= t, and the length of the normal = n.
25. If for a point of the curve the focal radius = r, and
the subtangent =: s.
26. Two parabolas have the same vertex, and the same
latus rectum 4p, but their axes are J_ to each other. What
is the length of their common chord ?
27. Through the three points of the parabola 7/^ = 12a',
whose ordinates are 2, 3, 6, tangents are drawn. Show
that the circle circumscribed about the triangle formed by
the tangents passes through the focus.
28. A tangent to the parabola 7f=^px makes the angle
30° with the axis of x. At what point does it cut the axis?
29. For what point of tlie parabola if^Apx is the length
of the tangent equal to 4 times the abscissa of the point of
contact ?
30. The product of the tangent and normal is equal to
twice the square of the ordinate of the point of contact.
Find the point of contact and the inclination of the tangent
to the axis of x.
31. Two tangents to a parabola are perpendicular to
each other. Find the product of their subtangents.
126 ANALYTIC GEOMETKY.
32. Prove that the circle described on a focal radius as
diameter touches the tangent drawn through the vertex.
33. Trove that the circle described on a focal chord as
diameter tpuches tiie directrix.
Find the locus of the middle points :
34. Of all the ordinatos of a parabola.
35. Of all the focal radii.
36. Of all the focal chords.
37. Of all chords passing through the vertex.
38. Of all chords that meet at the foot of the axis.
Two tangents to the parabola if=^^px make the angles
6, 6' with the axis of x ; find the locus of their intersection :
39. It cot 6 -\- cot 6' = k. 41. If tan ^ tan ^' = 7c.
40. If cot e — cot e' = k. 42. If sin 6 sin e'^k.
43. Find the locus of the centre of a circle that passes
through a given point and touches a given straight line.
SUPPLEMENTARY PPtOPOSITIONS.
1 03. Two distmct, two coincident, or no real tunyents can he
drawn to a. parabola from any point (Ji, k), accordiny as the
jjoint is without, on, or icit/iin the curve.
Let the tangent y = mx + — pass through the point (h, k)
P
then, k = mh -\- — ,
m
or h7n'- — km -|- ^v = 0.
Whence, m = ^ — •
THK I'AKAI'.OT.A.
127
These values of m arc real ami unequal, real and equal,
or imaginary, according as /.-"•^ — 4yy //>,=, or <0; that is,
according as (Ji, k) is without, on, or within the parabola ;
hence the proposition (§ 90).
104. To find the equation of the chord of contact of two
tangents drawn from any external point {h,k) to the parabola
Let (xi. iji) and {x^, yi) be the points of contact ; then
the equations of the tangents are
yiij = 'lp{x-\-x^.
Since (/i, k) is in both these lines, we have
ky^ = ^p{x,^h), (1)
ky, = 2p{x^^h). (2)
From equations (1) and (2) we see that both tlie points
(xi, 2/1) and (j-2, y^ lie in the straight line whose equation is
ky = 2p{x-^h). (3)
Hence, (3) is the equation required.
105. To find the equation ofth e pola r of the pole (h, k) u'ith
regard to the parabola y'^ = 4yw;.
Let P be the fixed point (A, k),
PQE one position of the revolv-
ing chord, and let the tangents at
Q and R intersect in Piixi, y^) ; it
is required to find the locus of
Pi, as the chord turns about P.
Since PH is the chord of con-
tact of tangents drawn from the
point Pi(xi, yi), its equation is
(§ 104)
yiy=^j'(->- + ^'i)-
Fig. 49.
(1)
128
ANALYTIC GEOMETRY.
Since (1) passes through (Ji, Ic) we have
y,k = 2p{h + x,). (2)
But (xi, ?/i) is any point on the required locus, and by
(2) its coordinates satisfy the equation
ky = 2p(x + h). (3)
Hence, (3) is the required equation, and the polar is a
straight line.
Cor. When the pole (Ji, k) is on the curve, the polar is
evidently a tangent at (A, k) ; when the pole (A, k) is with-
out the curve, the polar is the chord of contact of tangents
from (A, k). Thus the tangent and chord of contact are par-
ticular cases of the polar.
The Proposition of § 74 may be proved for poles and
polars with respect to a parabola.
106. To find the locus of the middle points of parallel
chords in the jiarahola if' = 4/>x.
NT.,-^^^
^
>-7
>
\ /
;
'A
Fig. 50.
Let any one of the chords PQ (Fig. 50) be y^=mx-\-c,
and let it meet the curve in the points (x^, i/i), {x^, y^)-
THE PARABOLA. 129
Then (§ 100), vi= -^' • (1)
Let M (x, y) be the middle point of PQ ; tlien 2^ =
2/1 + 2/2- By substitution in (1) we obtain
7H — — or y = -^' (2)
y ^ m ^ ^
a relation that holds true for all the parallel chords, because
m is the same for all the chords. The required locus, there-
fore, is represented by (2), and is a straight line- parallel to
the axis of x, and called a diameter of the parabola. Hence,
Every diameter of a parabola is a straight line parallel to
its axis.
Every straight Ihie parallel to the axis is a diameter ; for
2»
m, and, therefore -^, may have any value whatever.
m
107. Let the diameter through Jf meet the curve at 8,
and conceive the straight line P^ to move parallel to itself
till P and Q coincide at 8 ; then the straight line becomes
the tangent at S ; therefore.
The tangent draivn through the extremity of a diameter
is parallel to the chords of that diameter.
108. From the focus i^draw FC ± to PQ, and let EC
meet the directrix in the point C. If 6 denotes the angle
which the chord PQ makes with the axis of x, it easily
follows that DCF^d ; then we have
CD = FD cot 0 = ^-^
m
hence, by (2) § 106,
The perpendicular from the focus to a chord meets the
diameter of the chord in the directrix.
Moreover, since DS (Fig. 50) is parallel to QP, the per-
pendicular from the focus to a tangent and the diameter
through the point of contact meet in the directrix.
130 ANALYTIC GEOMETRY.
109. Let the tangents drawn throngh /' and Q meet in
the point T. Regarding their equations,
y^ij = 2p {x + .T2),
as simultaneous, we obtain for tlie value of the ordinate of T
y — -J-LJ h> _ _L. Hence,
y 2 — 1/1 I'l
Tangents draum throufjh the ends of a chord meet m the
diameter of the chord.
110. To find the locus of the foot of a per2yendicular from
tlie focus to a tangent.
Let the equation of a tangent be
ni
Then the equation of the perpendicular will be
X p
y = 1 —
ni VI
Since these two lines have the same intercept on the axis
of y, they meet in that axis ; hence, the tangent tlirough tlie
vertex is the required locus.
111. Since FP = PC (Fig. 48) and angle EPC = EPF,
therefore the tangent P T is perpendicular to FC at its middle
point, and every point in it is equally distant from i^and C.
112. Tangents at right angles intersect in the directrix.
Let the equation of one tangent be
y = inx -\ (1)
Then the equation of the other is
y = — --7np. (2)
Subtracting (2) from (1) we obtain for tlieir common point
THE PARABOLA.
131
(x-\-p)(vi-\- -)=0.
But m -\ cannot be zero ; lience, x -\-p = 0, or x = — p,
which is the equation of the directrix.
113. The polar of the focus (p, 0) is
0 = 2p (x -{-p), or X = —p.
Hence, the polar of the focus is the directrix, or tangents
at the ends of a focal chord intersect in the directrix.
CoR. From this result and § 112 it follows that tanyents
throufjh the ends of a focal chord intersect at right angles.
114. To find the equation of a parabola referred to any
diameter and the tangent thnnigh its extremitij as axes.
Transform the equation ]/ = 4:px to the diameter SX'
(Fig. 51) and the tangent through S as new axes. Let m
be the slope of the tangent, 9 the angle which the tangent
makes with the diameter ; then m = tan 6.
First transform to new parallel axes through S.
Now, by § 106, BS=2p^m; hence, from y" = Apx we
obtain AB=^p-^vr. Therefore, the now equation is
..■+4
III'
or
my- -\- 4py = 4j>iii.v.
(1)
132
ANALYTIC GKOMETRT.
Now retain the axis of x^ and turn the axis of y till it coin-
cides with the tangent at S\ then for any point P we have
The old X = SR. The new x = SN.
The old ?/ =: RP. The new y = NP.
Now it is easily seen from Fig. 52 that
>S'^ = SN + NP cos e, RP = NP sin 6.
Therefore, equation (1) is transformed to the new system
by writing x -\r y cos 6 in place of x, and y sin 6 in place of
y. Making this substitution, remembering that m = tan 6,
and reducing, we obtain
an equation of the same form as y'^=^^px.
Join S to the focus F, and denote FS by p' ; then
^ ^ ?>i'' m^ sin^ ^
Therefore, equation (2) may be more simply written
y- = ^p'x, (3)
where p' is the distance of the origin from the. focus. It is
easy to see that this equation includes the case when the
axes are the axis of the curve and the tangent at the vertex.
The quantity 4/»' is called the Parameter of the diameter
passing through >S^. When the diameter is tlie axis of the
curve, Ap' is called the Principal Parameter.
THE PARABOLA.
133
Cor. Let the equation of a parabola referred to any
diameter, and the tangent at the end of that diameter as
axes, be y^=-\p'x. Since the investigations in §§ 99, 100
hold good whether the axes are at right angles or not, it
r/
follows immediately that the straight line y=^mx-\- — will
touch the parabola for all values of m, and that the equation
of the tangent at any point (xi, y^ is yxy=^'2,xj'{x-\-x^.
115. To find the 2)olar equation of the jmrabola, the focus
being the j^ole.
Let P be any point (p, 6) of the
curve ; then
p = FF = XP = DM= 2p + FM
■=2p-\- p cos Q.
. - 2y>
' ' ^ 1 — cose
Discussion of [29]
[29]
Fig. 53.
Since cos Q cannot exceed + 1, p is positive for all values
of ^.
If ^^0, cos ^==1, and p=^v:.
This shows that the axis of the parabola does not cut
the curve to the right of the focus.
If ^r=|,r, cos ^^ 0
If ^ = TT, cos ^ = — 1
If ^=3^, cos ^= 0
If ^ = 27r, cos ^= 1
p = 2p = semi-latus rectum.
P= p = FA.
p = 2p = FR'.
p = X.
As 6 increases from zero to tt, p decreases from x to p.
As 6 increases from n to 2ir, p increases from p to x.
134 ANALYTIC GEOMETRY.
Exercise 31.
1. Given a parabola, to draw its axis (§106).
2. Prove that the perpendicular dropped from any point
of the directrix to the polar of the point passes through
the focus.
3. To find by construction the pole of a focal chord.
4. Prove that through any point tJiree normals can be
drawn to a parabola.
5. Tangents are drawn through the ends of a chord.
Prove that the part of the corresponding diameter con-
tained between the chord and the intersection of the
tangents is bisected by the curve.
6. Focal radii are drawn to two points of a parabola,
and tangents are then drawn through these poiuts. Prove
that the angle between the tangents is equal to half the
angle between the focal lines.
7. Show that if the vertex is tnken as pole, the polar
equation of a parabola is
A}) cos 6
^~ s\n-B
8. Explain how tangents to a ]>arabola may be drawn
from an exterior point (§ 102).
9. Having given a ])arabola, how would you fiud its
axis, directrix, focus, and latus rectum?
10. From the point ( — 2, 5) tangents are drawn to the
parabola //- = 6,/-. What is the equation of the chord of
contact?
11. The general equation of a system of parallel chords
in the parabola lif = 25a- is 4.r — 7// + ^•' = 0. What is the
equation of the corresponding diameter?
THE PARABOLA. 135
12. lu the parabola if = Vox, \vh;t,t is the equation of the
ordinates of the diameter y -f~ H = 0 ?
13. In the parabola / = (jj', wliat chord is Ijisccted at
the point (4, 3) ?
14. Given the parabola // = 4/?:r ; find the equation of
the chord that passes through tlie vertex and is bisected
by the diameter y = a. How can this chord be constructed ?
15. The latus rectum of a parabola = 16. Y/hat is the
equation of the curve if a diameter at the distance 12 from
the focus, and the tangent through its extremity, are taken
as axes?
16. Show that the equation of that chord of the i)arab-
ola if^^-ipx which is bisected at the point (//, k) is
l-{l,-k)=^2p{x-h).
17. Prove that the parameter of any diameter is equal
to the focal chord of that diameter.
18. Prove that the locus of ij- — 8// — G.'- + -8=:0 is a
parabola wliose axis is parallel to the axis of x ; and deter-
mine the latus rectum, the vertex, the focus, the axis, and
the directrix.
19. Prove that in general the locus of ?/-+-''•'■ + ^'// +
C=0 is a parabola Avhose axis is parallel to the axis of ./• ;
and determine its latus rectum, vertex, and axis.
20. Prove that in general the \oqx\^ oi x^ -\- Ax -\- B i/ -\-
C = Q is a parabola whose axis is parallel to the axis of //;
and determine its latus rectum, vertex, and axis.
21. Find the locus of the centres of circles that touch a
given circle and also a given straight line.
22. The area and base of- a triangle being given, find the
locus of the intersection of perpendiculars dropped from
the ends of the base to the opposite sides.
CHAPTER VI.
THE ELLIPSE.
Simple Properties of the Ellipse.
116. The Ellipse is the locus of a point, the sum of whose
distances from two fixed points is constant.
The fixed points are called Foci; and the distance from
any point of the curve to a focus is called a Focal Radius.
The constant sum is denoted by 2a, and the distance
between the foci by 2c.
The fraction - is called the Eccentricity, and is repre-
sented by the letter e. Therefore, c = ae.
In the ellipse a > c ; that is, e <C1.
If a=-c, the locus is simply the limited straight line
joining the foci.
If a < c, from the definition it is clear that there is no
locus.
117. To construct an ellipse, having giiien the foci and the
constant snm 2a.
I. B>/ Motion. Fix pins in the paper at the foci. Tie a
string to them, making the length of the string exactly
equal to 2a. Then press a pencil against the string so as
to make it tense, and move the pencil, keeping the string
constantly stretched. The point of the pencil will trace
the required ellipse ; for in every position the sum of the
distances from the point of the pencil to the foci is equal
to the lensrth of the strinsr.
THE ELLIPSE.
137
II. Bij Points. Let F, F' be the loci; then FF' = 2c.
Bisect FF' at 0, and from 0 lay off OA=OA' = a.
Then A' A = 2a, F'A' = FA,
A'F-\-A'F' = AF-{-FA =2a,
AF'+AF =AF'-\-F'A' = 2a.
Therefore, A and A' are points of the curve.
Between i^and F' mark any point X; then describe two
arcs, one with F as centre and AX as radius, the other with
F' as centre and AX as radius ; the intersections P, Q of
these arcs are points of the curve. By merely interchang-
ing the radii, two more points, E, S, may be found.
After a sufficient number of points has been obtained,
draw a continuous curve through them.
118. The line AA' is the Transverse or Major Axis,
A, A' the Vertices, and 0 the Centre of the curve.
The line BB\ perpendicular to the major axis at 0, is
the Conjugate or Minor Axis; its length is denoted by 2b.
Show that B and B' are equidistant from the foci, tliat
BF=a, that BO = h, and that a- = Ir -\- r ^ Ir -[- a-e\
138 ANALYTIC GKOMETRY.
119. To find the equation of the ell'qjse, having given, the
foci and the constant sum 2a.
B
Take the line AA' (Fig. 55), passing through the foci, as
the axis of x, and the point 0, halfway between the foci,
as origin. Let F be any point (x, y) of the curve, and let
r, r' denote the focal radii of P. Then from the definition
of the curve, and from the right triangles F'PM, FPM,
r"=>/+{c + xy,
(1)
?•-= if-\- (c — xy.
(2)
By addition,
r''J^v^=^2(x' + y^ + c').
(3)
By subtraction.
r'- — r = 4:cx.
(4)
But
r'-^r = 2a.
(5)
By division,
2ex
a
(6)
By subtraction,
ex
r^=a = \a — ex\.
a '- -^
(7)
By addition,
r' = a + -- = [a + ex~\.
(8)
Substitute in (3) remembering that h'=^a~ — c' (§ 118).
Then U'x''-\-a-if ^a'U',
or ^! + ^ = 1. [30]
THE ELLIPSE. 139
Cor. If the transverse axis is on the axis of y, and tlie
conjugate on the axis of x, the equation of the elli})se is
^+-A=i. (10)
120. To trace the form of ike ellipse from its equxtion.
The intercepts on the axis of x are + a and — a\ on tlie
axis oi y, -\-b and — h.
Only the squares of tlie variables x and y appear in the
equation; hence, if it is satisfied by a point {x,y), it will
also be satisfied by the points (x, — y), ( — x, y), (— x, — //).
Therefore, we infer that
(i) The curve is symmetrical vnfh respect to the axis of x.
(ii) Tlie curve is symmetrical with respect to the axis of y.
(iii) The curve is symmetrical with resperf to the centre O,
which bisects every chord passing/ through it. This explains
why 0 is called the centre.
\ 2 / \ 2
Since the sum of I - | and [ t\ is 1, neither of these
\aj \bj
squares can exceed 1 ; therefore, the maximum value of .r is + a,
and the minimum value — o, while the corresponding values
of y are -f- b and — b. Therefore, the curve is wholly con-
tained within the rectangle whose sides are x^±a, y = ±b.
121, To trace the chanyes in the form of the ellijjse wlicu
the semi-axes are supposed to change.
Let a be regarded as a constant, and b as a variable.
(i) Sujipose b to increase. Then c decreases (since c'^ =
a^ — i'^), e decreases, the foci approach the centre, and the
ellipse approaches the circle.
(ii) Let b = a. Then c = 0,e = 0, the foci coincide with
the centre, the ellipse becomes a circle of radius (/, and
equation [30] becomes the equation of the circle,
140 ANALYTIC GEOMETRY.
(iii) If we suppose b to decrease to 0 (a remaining con-
stant), c will increase to a, e will increase to 1, while the
curve will approach, and linally coincide with, the major
axis, its equation at the same time becoming i/ = 0.
122. Let {Xi,i/i) and (^'2,2/2) be any two points on the
ellipse i^x^ + a-^^ = a-(5''^; then we have
Dividing and factoring, we have
2/1^ : 2/2^^ : : (a — Xj) {a + x^) : (a — x^) {a + a^g).
This is, the squares of any two ordinates of the ellipse are
to each other as the products of the segments into which they
divide the major axis.
123. It follows from § 119 that a point (Ji, k) is on the
ellipse represented by the equation [30], provided
-. + 7-2-1=0.
a^ V
It may be shown by reasoning similar to that employed
in § 96 that the point (A, li) is outside or inside the curve,
h\ k"" , .
according as — 2 ryi — 1 is positive or negative.
1 24. If A, B, C all have the same sign, every equation
of the form ^^2+^^2_C' (1)
may be reduced to the form
x" . y . X , y ^
a^ W W a?
Hence, every equation of the form of (1) represents an
ellipse whose semi-axes are a/— and -y „* The transverse
axis lies on the axis of x or the axis of y, according as A
is less than or greater than B.
THE ELLIPSE.
141
125. The chord passing through either focus perpendicu-
lar to the majoraxis is called the Latus Rectum or Parameter.
To find its length, put a; = c in the. equation of the ellipse.
b"" , „ „. b* b''
Then,
r=^(s^-<^) = ^
Therefore, the latus rectum
21/ _ r4i/
a |_2a
Forming a proportion from this equation, we have
2a : 2b : : 2b : latus rectum ;
that is, the lathis rectum is a third li^'ojiortional to the major
and minor axes.
126. The circle having for diameter the major axis of
the ellipse is called the Auxiliary Circle ; its equation is
cc^ -f- y" = al
The circle having for diameter the minor axis is called
the Minor Auxiliary Circle ; its equation is
a- + u = ti-.
If P (Fig. 56) is any point of an ellipse, and the ordinate
MP produced meets the auxiliary circle in Q, the point Q
is said to correspond to tlie point J^.
The angle QOM is called the Eccentric Angle of the point
P, and is denoted by the letter (^.
142
ANALYTIC GKOMETRY.
127. Lft ij, ij represent the orcliuates of points in an
ellipse and the auxiliary circle respectively, corresponding
to the same abscissa x. Tlieu from the equations of the two
curves we have b .—^ j , , i—^ :j-
y = rb - V « — X , y —± 'si a — X-.
Whence, y:y'^b:a,
or, the ordinates of the eUijise and the auxiliary circle, corre-
sponding to a common abscissa, are to each other in the con-
stant ratio of the semi-minor a7id semi-major axes of the ellipse.
128. The principle of § 127 furnishes the following easy
method of constructing an ellipse by points when its axes
are given:
Construct both the major and minor auxiliary circles
(Fig. 57); draw any radius cutting the circles in B and Q;
through Q draw a line parallel to BO, and through R draw
a line parallel to OA; the intersection P of these lines is a
point on the ellipse. For we have
MP:MQ = OB:OQ,
or MP \%J — b\a.
From this proportion and that in § 127, we have MP = y;
hence, P is a point on the ellipse. In like manner any num-
ber of points may be found.
THE ELLIPSE.
143
(1)
Cor. From Fig. 57, we have
ic= 0M= OQ cos <f)^a cos <f>,
y=MP= 0N= OR sill <i, = b sin </,.
Equations (1), which express the coordinates of any point
of the ellipse in terms of its eccentric angle, may be used as
the equations of the ellij^se. To obtain from them the com-
mon equation, we have
CC 7/
- = cos d>, and 7 = sin <i.
ah
Therefore, —-\-~^ cos^ <i> -}- sin- <i = 1.
a^ b-
129. To find the area of an eUijise.
Divide the semi-major axis OA' (Fig. 58) into any number
of equal parts, through any two adjacent points of division
M, N erect ordinates, and let the ordinate through M meet
the ellipse in P and the auxiliary circle in Q. Through P, Q
draw parallels to the axis of a-, meeting the other ordinate
in B, S, respectively. Then (§ 127)
area of rectangle MPRN_MP _h
area of rectangle MQSN MQ a
144 ANALYTIC GEOMETRY.
A similar proportion liolds true for every corresponding
pair of rectangles.
Therefore, by the Theory of Proportion,
sum of rectangles in ellipse b
sum of rectangles in circle a
This relation holds true however great the number of rec-
tangles. The greater their number, the nearer does the sum
of their areas approach the area of the elliptic quadrant in
one case, and the circular quadrant in the other. In other
words, these two quadrants are the limits of the sums of
the two series of rectangles. Therefore, by the fundamental
theorem of limits,
area of elliptic quadrant h
area of circular quadrant a
Multiplying both terms of this ratio by 4,
area of the ellipse h
area of the circle a
But the area of the circle = ira^ ; therefore,
area of the ellipse = -aah. [31]
Exercise 32.
What are a, b, c, and e in the ellipse whose equation is;
^ . ^^
25^16
2. x^+2if = 2?
4. Ax''-\-Bi/- = l?
5. Find the latus rectum of the ellipse 3a;^ + 7y^ = 18.
6. Find the eccentricity of an ellipse if its latus rectum
is equal to one-half its minor axis.
THE ELLIPSK. 145
What is the equation of an ellipse if :
7. The axes are 12 and 8 ?
8. Major axis = 2G, distance between foci = 24 ?
9. Sum of axes = 54, distance between foci = 18 ?
10. Latus rectum, = ^5*, eccentricity = 5 ?
11. Minor axis = 10, distance from focus to vertex = 1?
12. The curve passes through (1, 4) and ( — 6, 1) ?
13. Major axis = 20, minor axis = distance between foci ?
14. Sum of the focal radii of a point in the curve = 3
times the distance between the foci ?
15. Prove that the semi-minor axis is a mean propor-
tional between the segments of the major axis made by
one of the foci.
16. What is the ratio of the two axes if the centre and
foci divide the major axis into four equal parts?
17. For what point of an ellipse is the abscissa equal to
the ordinate ?
Find the intersections of the loci :
18. 3x^ + 6/ = 11 and y = x + 1.
19. 2x^ + 3y' = 14 and if = 4.r.
20. a;2 + 7/ = 16 and a-^ + // = 10.
21. The ordinates of the circle x^-\- i/- = 7^ are bisected ;
find the locus of the points of bisection.
22. A straight line AB so moves that the points A and
B always touch two fixed perpendicular straight lines.
Show that any point F in AB describes an ellipse, and
find its equation.
146 ANALYTIC GEOMETKV.
23. What is the locus of J.r- -j- />'//' = C when C is zero ?
When is this locus imaginary ?
24. Prove that the al)scissas of the ellipse Ir^x"^ -\-a'^if
= 11%"^ are to the corresponding abscissas of the minor
auxiliary circle, x^-\-if^=U^, SiS a : b.
25. Construct an ellipse by the method of § 128.
26. Construct an ellipse, having given c and b.
27. Construct tlie axes of an ellipse, having given the
foci and one point of the curve.
28. Construct the minor axis and foci, having given the
major axis (in magnitude and position) and one point of
the ellipse.
29. A square is inscribed in the ellipse
a^^ b'
Find the equations of the sides and the area of the square.
Tangents and Normals.
130. To find the equations of a tangent and of <i normal
to an ellipse, harin;/ f/iren the point of contact (.Tj, i/i).
Taking the equation of the ellipse,
b-.r^ -j- a'^//- = a'^b'^,
and the equation of the straight line through (.Ti, iji) and
(•^'2, y^),
V — Vx _ Vi—yx
— >
.r — .7'i .Tg — .r 1
and proceeding, as in § G4, we obtain as the equation of a
secant through (a^i, y^ and (xj, v/o)
y — V\ ^ _ l>\^x + a-o)
a-— a-i «^(/yi + Z/2)
THE ELLIPSE. 147
Now make x^^x^, ij„=zy^-^ then the chord becomes a tan-
gent, and
x—xj^ (r{!/i + 7/2)
y ~ Vi ^^^^'i
becomes '— = — >
X — Xi c^ y\
which reduces to ^+'Jj!/_ = 1, ri^n
From the equation above it appears tliat the value of the
slope of the tangent, in terms of the coordinates of the
point of contact, is
b'^Xi
a%
The normal is perpendicular to the tangent, and passes
through (xi, ?/i) ; therefore, its equation is easily found (by
the method of § 46) to be
i/-!/i = |!^;(^-^i). [33]
131. To find the suhtangent and subnoDnal.
Making y = 0 in [32] and [33], and tlien solving the
equations for x, we obtain :
Intercept of tangent on axis of a- = —
('"^
Intercept of normal on axis of x =: — .r, = e^x^.
Whence, the values of the subtangent and the subnormal
(defined as in § 63) are easily found to be as follows :
Siibtang-eiit =
[34]
Subiiorinal =
a-
[:ib^
148
ANALYTIC GEOMETRY.
132. If tangents to ellipses having a common major axis
are drawn at points having a common abscissa, they will
meet on the axis of x.
For in all these ellipses the values of a -and. x are con-
stant, and therefore (by § 131) the tangents all cut the
same intercept from the axis of x.
Y
Q
Fig. 59.
133. The normal at any point of an ellipse bisects the
angle formed by the focal radii.
The values of the focal radii for the point P (Fig. 59)
were found, in § 119 to be
PF=a — ex^, PF'=a-\- exy.
If the normal through P meets the axis of x in N, ON
= e'^Xi (§ 131) ; and, therefore,
NF =c — e^Xi = ae — e^.r^ = e(a — ex).
NF' ^c-\- e^Xi = ae -\- e-x^ = e(a + ex).
Therefore, NF : NF' = PF : PF',
or the normal divides the side FF' of the A PFF' into
two parts proportional to the other two sides. Therefore
(by Geometry), Z FPN= Z FPN.
The tangent PT, being perpendicular to the normal,
must bisect the angle FPG, formed by one focal radius
with the otlier produced.
THE ELLIPSE.
149
134. To draw a taiujent and a "lornud tliroiujh a (jlven
point of an ellipse.
I. Let P (Fig. 60) be the given point. Describe the aux-
iliary circle, draw the ordinate MP, produce it to meet the
circle in Q, draw QT tangent to the circle and meeting the
axis of X in T, and join PT; then PT is a tangent to tlie
ellipse (§ 132). Draw PNl, to PT; PNis the normal at P.
C
II. Draw the focal radii, and bisect the angles between
them. The bisectors are the tangent and the normal at the
point P (§ 133),
135. To find the equation of a tangent to an ellipse in
terms of its slope.
This problem may be solved by finding under what con-
dition the straight line
y=^mx-]-c (1)
will touch the ellipse b-x- + <rir = a^b^. (2)
Eliminating y from (1) and (2), and tlien solving for .r.
we find two values of x :
— ma^c ± ab\/ni^a'^ -{■ Ir — c^
""^ m'a'-\-b'
These values will be equal if
m^a^ -\-b'' — c- = 0, or c = ± VwVT^'-
150 ANALYTIC fJEOMETRY.
If the two values of x are equal, the two values of y must
also be equal, from equation (1).
Therefore, the two points in which the ellipse is cut by
the line will coincide if c = ± sj vrii- -\- UK
Hence, the straight line
y = ^nx ± ^tti-fi- + ly^ C^*^]
will touch the ellipse for all values of m. .
Since either sign may be given to the radical, it follows
that two tangents liaving the same slo})e may be drawn to
an ellipse.
136. To find the locus of the intersertion of two tangents
to an ellipse tvhich are jjerjiendictdar to each other.
Let the equations of the tangents be
y^mx -\- '\Jm-(r -\-h^, (1)
y = m'x + Vm'V + i^. (2)
The condition to be satisfied is
m m = — 1 , or ?»/ =
m
If we substitute for vi' in equation (2) its value in terms
of m, the equations of the tangents may be written
y — mx = ^Imhv^ + ^^ (3)
my -\-x^ \la--\- m-/j\ (4)
The coordinates, x and y, of the intersection of the tan-
gents satisfy both (3) and (4); but before Ave can find the
constant relation between them we must first eliminate the
variable 7n.
This is most easily done by adding the squares of the
two equations ; the result is
(1 + ur)x' + (1 + m'-)f =(1 + w') (a^- + h%
or x^-\-y^^^a" -\-U-.
The required locus is therefore a circle. This circle is
called the Director Circle of the ellipse.
THE ELLIPSE. 151
Exercise 33.
1. What are the equations of the tangents and normals
to the ellipse 2x^-\-3i/''^^35 at the points whose abscissa ==2?
2. What are the equations of the tangents and normals
to the ellipse 'ix^-\- 9 y^^SG at the points whose abscissa= — §?
3. Find the eciuations of the tangent and the normal to
the ellipse a-^H-4^'==20 at the point of contact (2, 2).
Also find the svibtangent and the subnormal.
4. Show that the line )/ = x-\- V^' touclies the ellipse
2x^ -{- o//'- = 1.
5. Required the condition whicli must be satisfied in
X II
order that the straight line [- - = 1 may touch the ellipse
^+^ = 1.
6. In an ellipse the subtangent for the point (3, J^) is
— y-, the eccentricity = 4, What is the equation of the
ellipse ?
7. What is the equation of a tangent to the ellipse
9x--|-64//- = 57G parallel to the line 2i/^x?
8. Find the equation of a tangent to the ellipse 3x^-\-
bif^ = 15 parallel to the line 4a; — oy — 1 = 0.
9. In what points do the tangents that are equally
inclined to the axes touch the ellipse b'\r- -{- n^ij^ = a-lr?
10. Tlirough what point of the e\\i\)i^e /rx'- -\- tr//- = (rlr
must a tangent and a normal be drawn in order tliat they
may form, witli the axis of ./; as base, an isosceles triangle?
11. Tlirough a point of the ellipse //-.»■- -|- <(-//- = a-7/-, anil
the corresponding point of the auxiliary circle j'--\-if = a-,
normals are drawn. What is tlie ratio of the subnornuxls?
152 ANALYTIC GEOMETRY.
12. For what points of the ellipse ^V + a}if- = a^l? is the
subtangent equal numerically to the abscissa of the point
of contact ?
13. Find the equations of tangents drawn from the point
(3, 4) to the ellipse 16a;2 + 25/ = 400.
14. What are the equations of the tangents drawn through
the ends of thelatera recta of the ellipse 4a:;^ + 9?/" = 36a^?
15. What is the distance from the centre of an ellipse to
a tangent making the angle <^ with the major axis?
16. What is the area of the triangle formed by the tan-
gent in the last problem and the axes of coordinates ?
17. From the point where the auxiliary circle cuts the
minor axis produced tangents are drawn to the ellipse.
Find the points of contact.
18. Prove that the tangents drawn through the ends of
a chord through the centre are parallel.
19. Find the locus of the foot of a perpendicular dropped
from the focus to a tangent.
Exercise 34. (Review.)
1 . Given the ellipse mx" + 100/ = 3600. Find the equa-
tions and the lengths of focal radii drawn to the point (8, -'/)•
2. Is the point (2, 1) within or without the ellipse
2x^ + 3^2 = 12?
Find the eccentricity of an ellipse :
3. If the equation is 2x' + 3/ = 12.
4. If the angle FBF = 90" (see Fig. 54).
5. Show that 4.t2 - 8x + 9/y' - 36?/ + 4 = 0, or 4(.x — 1)^
4- 9(?/ — 2)2 = 36, is an ellipse whose centre is (1, 2), and
semi-axes 3 and 2.
THE ELLIPSE. 153
Find the equations of tangents to an ellipse :
6. If they make equal intercepts on the axes.
7. If they are parallel to BF (Fig. 54).
8. Which are parallel to the line - -|- y = 1 (a and b
being the semi-axes).
9. Find the equation of a tangent in terms of the eccen-
tric angle ^ of the point of contact.
Find the distance from the centre of an ellipse to :
10. A tangent through the point of contact (a-j, ?/i).
11. A tangent making the angle ^ with the axis of x.
12. In what ratio is the abscissa of a point divided by
the normal at that point?
13. At the point (x^, y^) of an ellipse a normal is drawn.
What is the product of the segments into which it divides
the major axis ?
14. Find the length of FN (Fig. 59).
15. Determine the value of the eccentric angle at the
end of the latus rectum.
Prove that the semi-minor axis b of an ellipse is a mean
proportional between :
16. The distances from the foci to a tangent.
17. A normal and the distance from the centre to the
corresponding tangent.
Determine and describe the loci of the following points :
18. The middle point of the portion of a tangent con-
tained between the tangents drawn through the vertices.
19. The middle point of a perpendicular dropped from
a point of a circle (x — ")*■^"■'/^^ ''" ^^ ^^^® '"^-^^^ ^^ U-
154 ANALYTIC GEOMETKY.
20. The middle jtoiiit of a chord of the ellipse b-y--\- u'l/^
^a'-lr drawn through the positive end oi tlie minor axis.
21. The vertex of a triangle whose base 2c and sum of
the other sides 2s are given.
22. The vertex of a triangle, having given the base 2c
and the product k of the tangents of the angles at the base.
23. The symmetrical point of the right-hand focus of an
ellipse with respect to a tangent.
SUPPLEMENTARY PPvOPOSITIONS.
137. Two distinct, two coincident, or no tangents can he
drawn to an ellipse tlirowjli, auij point {It, k), according as
the point is without, on, or ivithin the curve.
Let the tangent ?/ = ?/iic+ V?>iV + i^ pass through the
point (A, k); then
k = mh -\- Vm'^a- + ^^
or (Ji" — a^) vi" — 2hkin + k^ — V = 0.
m-
hk± \IVK'^a^k^ — a^lP'
h:' — a^ ' <^^)
Hence, there will be two distinct, two coincident, or no
tangents through {h, k), according as h^h^-\- aVc^ — ci-h^^,
=, or<0; that is, according as (h, k) is without, on, or
within the ellipse.
138. To fijid the equation of the chord of contact of the two
tangents drawn from an cjcternal point (Jt, k) to the ellipse,
a^~^ V
Let the student prove, by a course of reasoning similar
to that employed in §§71 and 104, that the required
equation is
hx kg .
THE ELLII'SK.
ir>r,
139. To find the equatioii of the iiohir of the pole (h, /:),
with regard to the ellipse.
Let the student prove, by a course of reasoning similar
to that employed in §§ 72 and 105, that the required
equation is
hx ky
a^ 0-
CoR. The tangent and cliord of contact are particular
cases of the polar. The proposition of § 74 holds true for
poles and polars with regard to the ellipse.
140. To draw a tangent to an ellipse from a given jjoint
P outside the curve.
Fig. CI.
Suppose the problem solved, and let the tangent touch
the ellipse at Q (Fig. (11). If F'Q is produced to G, making
QG=QF, then A FQG is isosceles; now ZFQP = Z
GQP (§ 1.33) ; therefore, PQ is per])ondicnlar to FG at its
middle point ; therefore, P is equidistant from /♦" and G.
This reduces the problem to determining the point G.
Since F'G=^2a, (7 lies on the circle with F' as centre
and 2a as radius. And G also lies on the circle with 7'
as centre and PF as radius. Hence the constrnction is
obvious.
156 ANALYTIC GEOMETRY.
141. To Jind the locus of thi; middla points of anij system
of jmrallel chords in the ellipse.
Let any one of the parallel chords y = mx + c meet the
ellipse
b'x'^ -\- (Cif = a-y^
in the points (xi, yi) and {x«. y^) ; then, by § 130,
/v-(a-i+a-2)
If (a*, y) is the middle point, 2a; = iCi + X2, 2y = yi-{- y^,
and (1) becomes
Px
or y^ — —— (2)
This relation holds true for the middle points of all the
chords ; therefore, it is the equation of the locus required.
From (2) we see that any straight line passing through
the centre of an ellipse is a diameter.
142. Let m! denote the slope of the diameter of the
chords whose slope is m; then from (2) of § 141
m = i' or mm' = r. \ ot \
ma^ a- ^ -■
Thus [37] is the equation of condition that the diameter
yz=m'x bisects all chords parallel to the diameter y = mx',
but [37] is evidently also the equation of condition that
y =■ mx bisects all chords parallel to ?/ = vi'x ; hence,
If one diameter bisects all chords parallel to another, the
second diameter bisects all chords parallel to the first.
Two such diameters are called Conjugate Diameters.
CoR. From [37] the slopes of two conjugate diameters
must have opposite signs ; hence, two conjugate diameters
of an ellijyse lie on opposite sides of the miiior axis.
THE ELLIPSE.
157
143. Let a straight line cutting the ellipse in P and Q
move parallel to itself till P and Q coincide with the end
of the diameter bisecting PQ; then the straight line be-
comes the tangent at the end of the diameter. Therefore,
The tangents at the extremities of any diameter are parallel
to the chords of that diameter , and also to its conjugate diameter.
144. Let POP' and HOP' (Fig. 62) be two conjugate
diameters meeting the ellipse in the points P (xi, y^ and
R (^2, 3/2)- The slope of the tangent through P is ^;
a yi
hence, the equation of the diameter ^0^', which is parallel
to this tangent (§ 143), is
h\r^ Xix . yiv „
y = ^x, or — - + •— - = 0.
ayi a^ ¥
(1)
Fig. G2.
Now R {x2, 2/2) is on (1), and also on the ellipse ; hence
we have 3. 3. , , , ,
-J— 2 J_ •ZLZj! = 0 (0\
and
1.
Solving (2) and (3) for x^ and y^, we obtain
a b
o a
(3)
158 ANALYTIC GKOMICTRY.
The up]ter signs give the coordinates of li, and the lower
those of E' in terms of a\ and //i.
Equation (2) is the condition that must be satisfied by
the coordinates of the extremities of every pair of conju-
gate diameters.
145. Denoting the semi-conjugate diameters OP and OB
(Fig. G2) by a' and //, respectively, we have
a'2 — xf + //i" = .rj- + (((- — xA
= ^^ + "^^V^ -= H' + ^'^.\ (1)
a
and h"' = jv + 1/2' = J, Z/r + ^^vTi'^ (§ 144)
= a' - j-;' + K x,^ = a' — eW. (2)
Adding (1) and (2), we have
a'2 + ^/2 = a- + i'^.
That is, the sum of the stianres of any pair of semi^cov jugate
diameters is equal to the sum of the squares of the semi-axes.
Equations (1) and (2) ex])ress the lengths of tlie semi-
conjugate diameters a' and // in terms of a, b, and x^ (the
abscissa of the extremity of a').
146. Let the ordinates of the extremities P, R (Fig. 62)
of two conjugate diameters meet the auxiliary circle in Q, S,
respectively, join QO and SO, and denote Z. QOX by <^,
Z^ SOX by <^'. Then the values of the coordinates of P
and R are (§ 127),
Xi = a cos <f>, Xo = a cos (f>',
l/i^b sin ^, yi'=b -sin <^'.
Whence, by substitution in etpiation (2) of § 144, we obtain
cos <^ cos ^'-|-sin <^ sin <^'=z{).
Therefore, cos (^' — ^) = 0, or <^' — <^ = ^tt.
THK ELLirSE.
159
That is, tlie difference of the eccentric avrjles correspond i>ir/
to the ends of two conjugate diameters is equal to a right angle.
CoK. The angle FOR (Fig. 62) is obtuse, siiu^e QOS=l_'Tr.
147. To find the angle formed by two conjugate semi-diam-
eters, whose lengths a', b' are gioen.
Let the semi-diameters make the angles a, (3, respectively,
with the axis of .r, and let 0 denote the required angle.
Then if (a-j, ?/i) and (x^, i/o) are the extremities of a' and b\
respectively,
sin a
7/
''''f^=b'=7b'-
cos a = —5 COS (3
or., aj/i
«■ ■ 'b'^~Jb''
sin 6^ sin (/S — a)
= sin )8 cos a — cos /8 sin a
b^x-^-\-a^Ui^ aHr
aba'b'
aba'b'
ah
a'l^
(1)
CoR. 1. Clearing (1) of fractions, we liave
a'V sin 6=^ ah,
which shows that the area of the parallelogram HEKE is
equal to the rectangle L21QN. (O ClfJiS=a'b' sin 6.)
R
N
^ jy w
Fig. G3.
That is, the j>arallelogram formed by tangents at the
extremities of any pair of conjugate diameters is equal to the
rectangle on the axes.
IGO ANALYTIC GEOMETRY.
CoK. 2. If CT (Fig. G3) is perpendicular to tlie tangent
KK, then,
CT^ CD sin CDR = a' sin $ = j,-
148. The lines joining any point of an ellipse to the
ends of any diameter are called Supplemental Chords.
Let FQ, F'Q be two supplemental chords (Fig. 64).
Through the centre 0 draw OR parallel to P'Q, and meeting
PQ'm R; also OR' parallel to FQ, and meeting F'Q in R'.
Q
Fig. 64.
Since 0 is the middle point of FF', and OR is drawn
parallel to F'Q, and OR' is drawn parallel to FQ, R and R'
are the middle points of QF, QF', respectively. Therefore,
OR will bisect all chords parallel to QF, and OR' will
bisect all chords parallel to QF'. Hence, OR, OR' are
conjugate diameters.
Therefore, the diameters jxirallel to a pair of supplemental
chords are conjugate diameters.
CoR. 1. This principle affords the following easy method
of drawing a pair of conjugate diameters which shall in-
clude a given angle.
On the transverse axis AA' describe a segment of a circle
which shall include the sriven angle. Let the arc of this
THE ELLirSE. 161
segment cut the ellipse in Q and *S'; then the diameters
parallel to QA and QA', or SA and SA', are conjugate and
include the required angle.
CoR. 2. If B is the upper vertex of the conjugate axis,
the conjugate diameters parallel to BA and BA' will evi-
dently be equal, and will lie on the diagonals of the rec-
tangle on the axes of the ellipse.
149. To find the equation of an ellipse referred to a pair
of conjugate diameters as axes.
Since each of two conjugate diameters of the ellipse
bisects the chords parallel to the other, the curve is
(obliquely) symmetrical with respect to each of the new
axes; hence, as the required equation is of the second degree,
it contains only the squares of x and y, and is of the form
Ax^ + Bi/=C. (1)
The intercepts of the curve on the new axes are the semi-
conjugate diameters. Denoting them by a' and b', we have
Substituting these values in (1), we obtain
which is the required equation in terms of the semi-conjugate
diameters.
This equation has the same form as the equation referred
to the axes of the curve ; whence it follows that formulas
derived from equation [30], by processes that do not pre-
suppose the axes of coordinates to be rectangular, hold true
when we employ as axes two conjugate diameters.
For example, the equation of a tangent at the point(a:i,2/i),
referred to the semi-conjugate diameters a' and b', is
a'2 "•" b'''
1G2
ANALYTIC OKOMKTRY.
150. To consfritrt the, polar of a fucus
Since the po
focus (ae, 0) is
Since the polar of (h, k) is -7 + ^ = 1, tlie polar of the
aex = a; or x
Hence, ae : a ^= a -. x.
Therefore, if OD (Fig. 65) is taken so that
OF: 0A=^ OA : OD,
and DC is drawn perpendicular to OD, DC will be the
polar of the focus F.
The polar of a focus is called a Directrix of the ellipse.
Hence, DC is the directrix corresponding to the focus F.
In like manner we may constnict F'C, or the directrix
corresponding to the focus F'.
Cor. Let Q (x, ?/) bo any ])oint on the ellipse; then,
QS=OD-OM^'^-x = '^ = ^.
Hence, e = FQ-^QS.
That is, tJie distances of ant/ poivt on the ellipse from a
focus, and the corresj)ondi)if/ directrix, bear the constant ratio e.
THE ELLIPSE.
163
Whence, the ellipse is often defined as :
Tlte locus of a point ivliicli moves so that its distances from
a fixed 2)0 int and a fixed stvahjlit line hear a constant ratio
less than unity.
151. To find tlie polar equation of the ellipse, the left-
hand focus being taken as the jjole.
Fig. 66.
Let F be any point (p, 0) of tlie ellipse ; then, from equa-
tion (8) of § 119, we have
p = a^ ex. (1)
Now X = 0M= F'M — F'0 = pcoQe — ae.
Substituting tins value of x in (1), we have
p = « -|- ep cos 9 — ae^.
Whence, p
1 — ecos6
[39]
Cob. Since e < 1, and cos 6 cannot exceed unity, p is
always positive.
), p^ -^ = ./ + ae = F'A.
If
If
1 = 0.
; Itt, p=:a(l — e'-) = F'Ji =■ senii-latus rectum.
164 ANALYTIC GEOMETRY.
If e = ^, p = ''-^^^ = a-ae = F'A'.
^ l-\-e
If 0 = ^-,r, p = «(1 — e^) = semi-latus rectum.
If 6 = 2ir,p = a + ae = F'A.
While 6 increases from zero to tt, p decreases from a-\-ae
to a — ae; and while 6 increases from tt to 27r, p increases
from a — ae to a-\- ae.
If F is taken as the pole, the polar equation is
a(l — e})
" 1 + e cos 6
Exercise 35.
1. Find the area of the ellipse a:;^ + 4y^ = 16.
2. Find the distances of the directrices from the centre
in No. 1.
3. What is the equation of the polar of the point (5, 7)
with respect to the ellipse 4a;^ + %^ = 36 ?
4. Prove that a focal chord is perpendicular to the line
that joins its pole to the focus. In what line does the
pole lie ?
5. Find the pole of the line Ax-\- By -\- C = 0 with
respect to the ellipse b'^x^ -\- a? if = a%^.
6. Each of the two tangents that can be drawn to an
ellipse from any point on its directrix subtends a right
angle at the focus.
7. The two tangents that can be drawn to an ellipse
from any external point subtend equal angles at the focus.
8. Find the slope mi of a diameter if the square of the
diameter is (i) an arithmetic, (ii) a geometric, (iii) an har-
monic mean between the squares of the axes.
9. Given the length 21 of a diameter, its inclination 6 to
the axis, and the eccentricity; find the major and minor axes.
THE ELLIPSE. 165
10. Tangents at the extremities of any chord intersect
on the diameter which bisects that chord.
11. Tangents are drawn from (3, 2) to the ellipse
x^ + 4//- = 4. Find the equation of the chox-d of contact, and
of the line that joins (3, 2) to the middle point of the chord.
12. Find the area of the rectangle whose sides are the two
segments into which a focal chord is divided by the focus.
13. What is the equation of a chord in the ellipse
13^- + 11^^143 that passes through (1, 2) and is bisected
by the diameter ^x — 2?/ = 0 ?
14. In the ellipse 9x^ + 36 ?/^ = 324 find the equation of a
chord passing through (4, 2) and bisected at this point.
15. Write the equations of diameters conjugate to the
following lines :
X — y^O, a" -|- y = 0, ax = hy, ay = bx.
16. Show that the lines 2x — y^=0, x-\-Sy=^0 are con-
jugate diameters in the ellipse 2x^ -\- 3y^ ^= 4.
17. Find the equation of a diameter parallel to tlie
normal at the point (xi, y^), the semi-axes being a and b. .
18. The rectangle of the focal perpendiculars upon any
tangent is constant and equal to the square of the semi-minor
axis.
19. The diagonals of the parallelogram in Fig. 63, § 147,
are also conjugate diameters.
20. The angle between two semi-conjugate diameters is
a maximum when they are equal.
21. The eccentric angles corresponding to equal semi-
conjugate diameters are 45° and 135°.
22. The polar of a point in a diameter is parallel to tlie
conjugate diameter.
166 ANALYTIC GEOMKTKV.
23. Find the ecjuations of equal conjugate diameters.
24. Tlie length of a semi-diameter is I; find the equation
of the conjugate diameter.
25. The angl(! between two equal conjugate diameters
is 120° ; find the (;c(u'ntricity of the ellipse.
26. Given a diameter, to construct the conjugate diameter.
27. To draw a tangent to a given ellipse parallel to a
given straight line.
28. Given an ellipse, to find by construction the centre,
foci, and axes.
29. Find the rectangular equation of the ellipse, taking
the origin at the right-hand vertex.
30. Find the polar equation of an ellipse, taking as i)ole
tlie right-hand focus.
31. Find the polar equation of the ellipse, taking the
centre as pole.
32. If the centre of an ellipse is the point (4, 7), and
the major and minor axes are 14 and 8, find its equation,
the axes being supposed parallel to the axes of C()()rdinates.
• 33. The equation of an ellipse, the origin being at tlie
left-hand vertex, is 25.r--l- 81^ ;=450.r ; find the axes.
34. If the minor axis = 12, and the latus rectum = 5,
what is the equation of the ellipse, the origin being taken
at the left-hand vertex ?
35. Find the eccentric angle <^ corresponding to the
diameter whose length is 2c.
36. At the intersection of the ellipse Ji^x^ -\- c/^ = aHi'^ and
the circle x--\-/f^(ih tangents are drawn to both curves.
Find the ansrle between them.
THE ELLIPSE. 167
37. How would you draw a normal to an ellipse from
auy point in the minor axis ?
38. Find the equation of a chord that is bisected at
the point (Ji, k).
39. Prove that the length of a line drawn from the
centre to a tangent, and parallel to either focal radius of
the point of contact, is equal to the semi-major axis.
40. A circle described on a focal radius will touch the
auxiliary circle.
41. Find the locus of the intersection of tangents drawn
through the ends of conjugate diameters of an ellipse.
42. Find the locus of the middle jtoint of the chord
joining the ends of two conjugate diameters.
43. Find the locus of the vertex of a triangle Avhose
base is the line joining the foci, and whose other sides are
parallel to two conjugated iameters.
44. Show that ^iX^ + y^ + 8a? — 2y + 1 = 0 represents an
ellipse ; find its centre and axes.
45. If A and B have like signs, sliow that tlie locus of
Ax- + 11 y- -\- iKr -f- Jiy + i''= 0 is in general an ellipse whose
axes are parallel to the coordiuate axes ; and determine its
semi-axes.
46. Find the locus of the centre of a circle that passes
through the point (0, 3) and touches internally the circle
x^ + 1/' ^ 25.
CHAPTER VII.
THE HYPERBOLA.
Simple Properties of the Hyperbola.
152. The Hyperbola is the locus of a point the difference
of whose distances from two fixed points is constant.
The fixed points are called the Foci, and a line joining
any point of the curve to a focus is called a Focal Radius.
The constant difference is denoted by 2a, and the dis-
tance between the foci by 2c.
The fraction - is called the Eccentricity, and is denoted
a
by the letter e. Therefore, c-=ae.
Since the difference of two sides of a triangle is always
less than the third side, we must have in the hyperbola
2a <! 2c, or a <C c, or e > 1.
153. To construct an hyperhola, having r/iven the foci, and
the constant difference 2a.
I. B]/ Motion (Fig. 67). Fasten one end of a ruler to
one focus F' so that it can turn freely about F'. To the
other end fasten a string. Make the length of the string
less than that of the ruler by 2a, and fasten the free end
to the focus F. Press the string against the ruler by a
pencil point P, and turn the ruler about F\
The point P will describe one branch of an hyperbola.
The other branch may be described in the same way by
interchanging the fixed ends of the ruler and the string.
THE HYPEKBOLA.
169
II. By Points (Fig. 68). Let F, F' be the foci ; then
FF' = 2c.
Bisect FF' at 0, and from 0 lay off OA = OA' = a.
Then AA' = 2a, FA = F'A '.
AF'— AF= AF' — A'F' = AA' ^ 2a.
A'F— A'F' = A'F— AF = AA' = 2a.
Therefore, A and A' are points of the curve.
Y
X
°<l
/^
1 F'
A'
1
0
A
^l>
1 1)
~--^1k^.
--V'--
A\
/Q\
^
Fig. 67.
Fig. 68.
In FF' produced mark any point D ; then describe two
arcs, the first with F as centre and AD as radius, the second
with F' as centre and A'D as radius ; the intersections P, Q
of these arcs are points of the curve. By merely inter-
changing the radii, two more points B, S may be found.
Proceed in this way till a sufficient number of points has
been obtained; then draw a smooth curve through them.
Through 0 draw BB' J_to FF'-, since the difference of
the distances of every point in the line BB' from the foci
is 0, therefore the curve cannot cut the line BB'.
The locus evidently consists of two entirely distinct
parts or branches, symmetrically placed with respect to the
line BB', called the right-hand and the left-hand branches.
170
ANALYTIC GEOMETRY.
1 54. The point (J, halfwiiy between the foci, is the Centre.
The points A, A', where the line passing through the
foci meets the curve, are called the Vertices.
The line AA' is the Transverse Axis.
The transverse axis is equal to the constant difference
'Ja, and is bisected by the centre (§ 153).
Fig. CO.
The line BB' passing through 0 perpendicular to AA'
does not meet the curve (§ 153) ; but if B, B' are two points
whose distances from the two vertices'^, ^' are each equal
to c, then BB' is called the Conjugate Axis, and is denoted
by 2h.
Since AAOB = A A OB', OB=OB'=l>; that is, the con-
jugate axis is bisected by the centre. ;
In the triangle AOB, OA^=a, OB = h, AB^^i^;^ hence,
c" = a' -\- Ir.
The chord passing through either focus perpendicular to
the transverse axis is the Latus Rectum, or Parameter.
Note. Since a and b are equal to the legs of a riglit triangle, a
may be greater than or less than b; hence the terms '■'■major'''' and
^^ minor'''' are not appropriate in the hyperbola.
THIC HVPKKBOI.A. 17 J
155. By proceeding as in the case of the ellipse (§ 119),
using r' — r = ± 2(i instead of /•' + r = 2a, and substituting
i^ for c^ — a^, we obtain as the equation of the hyperbola
^'-1?=1. [40]
Thus the equations of the ellipse and hyperbola differ
only in the sign of h^; that of the ellipse is changed into
that of the hyperbola by substituting — V^ for + V^. Hence,
Any formula deduced front the equation of the ellipse is
changed to the correspondincj formula for the hyperbola by
merely changing -\-b- to — Ir, or b to b\l — 1.
The lengths r, r' of the focal radii for any point (a*, //) are
r ^ ± (ex — a ) , r ' = ± (ex + o ) ,
in which the upper signs hold for the right-hand branch,
and the lower for the left.
156. A discussion of equation [40] leads to the follow-
ing conclusions :
(i) The curve cuts the axis of x at the two real points
{a, 0) and(— «, 0).
(ii) The curve does not cut the axis of //. The imaginary
intercepts are ± Z* V — 1.
(iii) No part of the curve lies between the straight lines
x^-\-a and x = — a.
(iv) Outside these lines the curve extends without limit
both to the right and to the left.
(v) The greater the abscissa, the greater the ordinate.
(vi) The curve is symmetrical with respect to the axis
of a-.
(vii) The curve is symmetrical with respect to the axis
of y.
(viii) Every chord that passes through the centre is
bisected by the centre. This explains wli}- the })oint half-
way between the foci is called the centre.
172
ANALYTIC GEOMETRV.
157. An hyperbola whose transverse and conjugate axes
are equal is called an Equilateral Hyperbola. Its equation is
oc^-y--a-. [41]
The equilateral hyperbola bears to the general hyperbola
the same relation that the auxiliary circle bears to the ellipse.
Fig. 70.
158. The hyperbola that has BB' for transverse axis,
and AA' for conjugate axis, obviously holds the same rela-
tion to the axis of ?/ that the hyperbola which has AA' for
transverse axis and BB' for conjugate axis holds to the axis
of X.
Therefore its equation is found by simply changing the
signs of a^ and b^ in [40], and is
1.
(1)
The two hyperbolas are said to be Conjugate.
THE HYPERBOLA. 173
159. The straight line z/=?na;, passing through the centre
of the hyperbola -^— -, = !> meets the curve in two points,
the abscissas of which are
-\- ah — ah
\IU^ — ni^a' V^^ — m^a^
Hence the points will be real, wiarjinavy, or situated at
infinity, as h- — m^a? is positive, negative, or zero ; that is,
as rn? is less than, greater than, or equal to —•
The same line, y = mx, will meet the conjugate hyperbola
— r — TZ = — 1 in two points, whose abscissas are
a' ¥
-\- ah, — ah
Hence these points will be imaginary, real, or situated at
h^
infinity, as m^ is less than, greater than, or equal to —^•
Whence,
If a straight line through the centre meets an hyperhola in
iraaginary points, it will meet the conjugate hyjjerhola in real
points, and vice versa.
160. An Asymptote is a straight line that passes through
finite points, and meets a curve in two points at infinity.
We see from § 159 that the hyperbola
x"^ //
^ "~ Ij" ~
has two real asymptotes which pass through the centre of
the curve, and which have for their equations y = -\--x and
b
X ; or,
1^ = 0. [42]
y = — -x; or
^2
a2
174 ANALYTIC GEOMETRY.
Exercise 36.
What is the equation of an hyperbola, if :
1. Transverse axis = IG, conjugate axis = 14 ?
2. Conjugate axis = 12, distance between foci = 13 ?
3. Distance between foci = twice the transverse axis?
4. Transverse axis = 8, one point is (10, 25) ?
5. Distance between foci = 2c, eccentricity = V2 ?
6. Prove that the latus rectum of an hyperbola is equal
2lr
to Also 2a : 20 ■.•.2b: latus rectum.
a
7. The equation of an hyperbola is 9x-^ — 16^/ = 144;
iind the axes, distance between the foci, eccentricity, and
latus rectum.
S. Write the equation of the hyperbola conjugate to the
hyperbola 9x-^— !()//-= 144, and find its axes, distance
between its foci, and its latus rectum.
9. If the vertex of an hyperbola bisects the distance
from the centre to the focus, find the ratio of its axes.
10. I'rove that the point (.*•, y) is ivitlumt, on, or ivithin
the hyperbola, according as —^ — 'j^—^ is nerjatioe, zero, or
positive.
11. Find the eccentricity of an equilateral hyperbola.
12. Find the points that are common to the hyperbola
25x-2— 9/=22r), and tlie straight line 25.^+12^ = 45.
13. The asymptotes of an hyperbola are the diagonals
of the rectangle CDEG (Fig. 70, p. 172).
14. Find the foci and the asymptotes of the hyperbola
16a;2 — 9/ = 144.
THE HYPERBOLA. 175
15. The asymptotes of an equilateral hyperbola are per-
pendicular to each otlier. Hence the equilateral hy})erb()la
is also called the rectan<juJar hyperbola.
16. Two conjugate hyperbolas have the same asymi)totes.
17. Find tlie length of the perpendicular dropped from
the focus to an asymptote.
18. Prove that the squares of any two ordinates of an
hyperbola are to each other as the products of the segments
into which they divide the transverse axis externally.
Tangekts and Normals.
Note. The results stated in the following six sections may be
established in the same way as the corresponding propositions in the
ellipse, or the first five may be obtained by § 155.
161. The slope of the tangent at (x^, i/i) is —^, and its
a i/i
equation is
a^_2/LM^l. [43]
162. The equation of the nnrninl at (.Vi, y{) is
O-Jl I
[-14]
63.
The subtangent =^~ ^ the stdjnormal
b'ix.
64.
The straight line whose equation is
y = mx ± \lm'a'^ — b^
is a tangent for all values of m (§ 135).
165. The equation of the director circle of an hyperbola
is a-2_|_y/2 = ,,2 — /;2 (^ 130).
166. The tangent and the nornaal at any jioint of an
hyperbola bisect the angles formed by tlie focal radii of
the point (§ 133).
176 ANALYTIC GEOMETRY.
Exercise 37.
1. Find the equations of tlie tangent and of the normal
at the point (4, 4) of the hyperbola 16a;^ — %^= 112. Also
lind the lengths of the subtangent and the subnormal.
2. Show that in an equilateral hyperbola the subnormal
is equal to the abscissa of the point of contact.
3. The equations of the tangent and the normal at a point
of an equilateral hyperbola are ox — 4?/ = 9, 4x ■i-5ij = 40.
What is the equation of the hyperbola, and what are the
coordinates of the point of contact ?
4. For what points of an hyperbola, is the subtangent
equal to the subnormal ?
5. To draw a tangent and a normal to an hyperbola at
a given point of the curve.
6. If an ellipse and an hyperbola have the same foci,
prove that the tangents to the two curves drawn at their
points of intersection are perpendicular to each other.
7. Prove that the asymptotes of an hyperbola are the
limiting positions of tangents to the infinite branches.
8. Prove that the length of a normal in an equilateral
hyperbola is equal to the distance of the point of contact
from the centre.
9. Find the distance from the origin to the tangent
through the end of the latus rectum of the equilateral
hyperbola x^ — y^= a^.
10. "What condition must be satisfied in order that the
X ?/ x^ ?/
straight line — h - = 1 may .touch the hyperbola '— — 7^=! ?
11. When is the director circle of an hyperbola imaginary?
12. Find the locus of the foot of the perpendicular
dropped from the focus of an hyperbola to a tangent.
THE HYPERBOLA. 177
Exercise 38. (Review.)
1. The ordinate through the focus of an hyperbola,
produced, cuts the asymptotes in P and Q. Find PQ and
the distances of P and Q from the centre.
2. In the hyperbola 9x^ — 16?/^ = 144 what are the
focal radii of the points whose common abscissa is 8 ?
What other points have equal focal radii ?
3. What relation exists between the sum of the focal radii
of a point of an hyperbola and the abscissa of the point ?
4. Prove that in an equilateral hjq^erbola every ordinate
is a mean proportional between the distances of its foot from
the vertices of the curve. Hence, find a method of con-
structing an equilateral hyperbola when the axes are given.
5. In an equilateral hyperbola the distance of a point
from the centre is a mean proportional between its focal radii.
6. In an equilateral hyperbola the bisectors of the
angles formed by lines drawn from the vertices to any
point of the curve are parallel to the asymptotes.
7. If e, e' are the eccentricities of two conjugate hyper-
bolas,
8. Through the positive vertex of an hyperbola a tangent
is drawn. In what points does it cut the conjugate hyperbola?
9. The sum of the reciprocals of two focal chords per-
pendicular to each other is constant.
10. Through the foot of the ordinate of a point in an
equilateral hyperbola a tangent is drawn to the circle
described upon the transverse axis as diameter. What
relation exists between the lengths of this tangent and the
ordinate of the point ?
178 ANALYTIC GEOMETRY.
11. In an equilateral hyperbola find the equations of
tangents drawn from the positive end of the conjugate axis.
12. From what ])oint in the conjugate axis of an hyper-
bola must tangents be drawn in order that they may be
perpendicular to each other ?
13. What condition must be satisfied that a square may
be constructed whose sides sliall be parallel to the axes of
an hyperbola and whose vertices shall lie on the curve ?
14. Find the equation of the chord of the hyperbola
lG.r^ — %- = 144 that is bisected at the point (12, 3).
15. Find the equation of the tangent to the hyperbola
16.X- — 9//^=144 parallel to the line y = 4x — 3.
16. Find the product of the two perpendiculars let fall
from any point of any hyperbola upon the asymptotes.
17. A chord of an liyperbola that touches the conjugate
hyperbola is bisected at the point of contact.
SUPPLEMENTARY PROPOSITIONS.
Note. Many of the following propositions are closely analogous to
propositions already establislied for tlie ellipse ; hence the proofs are
omitted, and references given to th(i chapter on the ellipse.
167. 7'iao distinct, two coincident, or no tangents can he
drawn to an hyperhola through any point (Ji, k^, according as
the point is without, on, or within the curve (§ 137).
168. TJie equation of the chord of contact of the tiro tan-
gents draum from an external point (Ji, /.•) to the hypterhola
'-,-f=l, is^-fe=l. (§138)
«/ b<, a- b^ ^ ■'
Til 10 HYl'KlUiOl-A. 179
169. The equatloti of tlie polar of the pole (Ji, k) icltk
regard to the hijperbola is
lix kii . ^ . ^ ,,,
— -7T = 1. (§139)
a^ O' ^ ^
The tangent and chord of contact are particular cases of
the polar, and the i)ruposition of § 74 holds true for poles
and polars with regard to the hyberbola.
170. The equation of a diameter of ait hupc.rhula is
y = ,;?;'. (S"i)
in which in is the slope of its chords.
171. If m' is the slope of the diameter bisecting the
chords parallel to the diameter // = nix, then (§ 142)
iiini' — -^,' r4ol
a- >- -■
Since ??i and m' are alike involved in [45], it follows that
If 0716 diameter bisects all chords par allel to another, the
second diameter will bisect all chords parallel to the first.
Two such diameters are called Conjugate Diameters.
172. From [45], the slopes of two conjugate diameters
must agree in sign ; hence,
Ttvo conjxKjate diameters of an hyperbola lie on the same
side of the coiijagate axis, and their included angle is acute.
Also, if ni in absolute magnitude is less than -, then in'
b "
must be greater than-- But the slope of the asymptotes is
b ^^
equal to ± — Therefore,
Two conjugate diameters lie on opposite sides of the asymp-
tote in the same quadrant : and of two conjugate diameters,
one meets the curve in real points and the other in imaginary
jtoints (§ 159).
180 ANALYTIC GEOMETRY.
173. The length of a diameter that meets the hyperbola
in real points is the length of the chord between these points.
If a diameter meets the hyperbola in imaginary points,
that is, does not meet it at all, it will meet tlie conjugate
hyperbola iu real points (§ 159); and its length is the length
of the chord between these points. But from § 159 we
know that if a diameter meet one of the hyperbolas in the
imaginary point (AV— 1, k^J—V), it will meet the other in
the real point (Ji, k); hence, the length of the semi-diameter,
which is V/r + k^, is known from the imaginary coordinates
of intersection.
174. The equations of an hyperbola and its conjugate
differ only in the signs of a^ and b^. But this interchange
of signs does not affect the equation
, b'
mm = — •
Therefore, if tivo diameters are conjugate with respect to
one of two conjugate hyperbolas, they ivill be conjugate with
respect to the other.
Thus, let POP' and QOQ' (Fig. 71) be two conjugate
diameters. Then POP' bisects all chords parallel to QOQ'
that lie icithin the branches of the original hyperbola and
betioeen tlie branches of the conjugate hyperbola; and QOQ'
bisects all chords parallel to POP' that lie within the
branches of the conjugate liyperbola and between the
branches of the original hyperbola.
From the above theorem it follows immediately that
If a straight line meets each of two conjugate hyperbolas in
two real jioints, the two jjortlons of the line contained between
the hyperbolas are equal (thus, BD^= B'D', Fig. 71).
175. The tangent di'atvn through the end of a diameter is
parallel to the conjugate diameter (§ 143).
THE HYPERBOLA.
181
176. Having given the end (^j, ?/i) of a dianiete?-, to find
the end, (./•.,, y.,) of the conjugate diameter.
Fig. 71.
Let (xi, ]/i) be on the given hyperbola, then (arg, 2/2) is on
the conjugate. The slope of the tangent at (xi, 1/1) is -7-^;
hence, the equation of the diameter conjugate to the diame-
ter through (xi, 1/1) is
y=^x. (1)
a-i/i
Now (xo, 1/2) is on the diameter (1) and also on the con-
jugate hyperbola ; lience, we have
2/2 =
1.
a'l/i a^ 0-
Solving equations (2) for Xo and 7/2, we obtain
a b
(2)
X2 '■
l/u
y.2 = ±-xy.
a
182 ANALYTIC GEOMETRY.
The positive signs belong to one end, and the negative
signs to the otlier end, of the conjugate diameter.
177. To find flic equation of an liijperlmla referred to any
initr of eoirjiigate dimneters as axes of coordinates.
From the symmetry of the curve with respect to each, of
the new axes, tlie required equation must be of the form
Denoting tlie intercepts of the curve on the new axes by
a' and b' V— 1 (§ 172), we obtain
Whence, ^^--^^l (1)
is tlie required equation, in which a' and V are semi-conju-
gate diameters.
Since the form of equation (1) is the same as that of the
equation referred to the axes of the curve, it follows that
all formulas that have been obtained without assuming
the axes of coordinates to be at right angles to each other
hold good when the axes of coordinates are any two
conjugate diameters. Por example, the equation of the
asymptotes of the hyperbola represented by equation (1) is
0 9
and the equation of the tangent is
178. The tangents through tlie ends of two conjugate
diameters meet in the asymptotes.
The equations of these tangents referred to the conjugate
diameters are
a; = ± «', y = ±b'.
THE HYPKKBOLA. 183
Hence, tlieir intersections are («', b'), (a', — b'), ( — a', b'),
and ( — a', — b'). Hut these points evidently lie upon the
asymptotes, or the locus of (2) in § 177.
179. //' 6 denotes tlie aiKjle formed by two covj agate semi-
diametei's, and a' and b' their lengtlis, then sin ^ = -y-,-
Substituting ^V — 1 for b, and i'V — 1 for ^' in equation
(1) of § 147 and cancelling the imaginary factorS; we obtain
the above result.
CoR. 1. Since Irt'i^'sin ^ = 4«^, the parallelogram SES'R'
(Fig. 71) equals the rectangle on the axes of the curve.
CoR. 2. The length of the perpendicular from 0 upon
the tangent SPIi= OF sin OFS ^a' sin e = '-^-
b
Cor. 3. From §§ 145, 155, 177, we have
a'-^ — b'''^a'--b\
180. If a straight line cats an hyperbola and its asymp-
totes, the portions of the line intercepted between the curve and
its asymptotes are equal.
Let CC (Fig. 72) be the line meeting the asym})totes in
C, C'and the curve in. 7*, B', and let the equation of the
line be
y = vix-\-c. (1)
Let M be the middle point of the chord BB' ; then
(§ 170) the equation of the diameter through vi is
^ = ^.- (2)
By combining equation (1) with the equations of the
asymptotes, we obtain the coordinates of the points C and
C ; taking the half-sum of these values, we get for the coordi-
nates of the point halfway between C and C the values
184
ANALYTIC GEOMETRY.
05 =
U^ — m^d^ '
y-
V'c
U' — m^a^
These values satisfy equation (2) ; therefore, the point
halfway between C and C" coincides with M\ therefore,
MC= MC. And since MB=MB\ therefore, BC= B'C.
Dy
Fig. 72.
Cor. Let CC be moved parallel to itself till it becomes
a tangent at P, meeting the asymptotes in R, S; then the
points B, B' coincide at F, and we have PB = PS. Hence,
The portion of a tanr/ent intercepted by the asymptotes is
bisected by the p)oint of contact.
181. The following method of showing that an hyperbola
has asymptotes, and finding their equations, is more general
than the method given in §§ 159, 160.
THE HYPERBOLA. 185
The abscissas of the points where the straight line y = mx
-\- e meets an hyperbola are found by solving the equation
x^ (mx -f cy
a^ b''
1,
Now, from Algebra we know that as the coefficients of
x^ and X in (1) approach zero, both roots of (1) increase
without limit. Hence, each root becomes infinity when
V^ — nvii? = 0, and Imc = 0,
or when m = ± -, and c ^ 0.
a
Therefore, ij^^±i-x are asijmptotes to the hyperbola.
If only b- — m^a^ = 0, then m ^ ± -, the line is parallel to
an asymptote, and one root of (1) is infinity, while the
, . h' + c"
other IS
Zmc
Hence, a right line parallel to an asymptote meets the
hyperbola in only one finite point.
182. To find the equation of an hyperbola referred to the
asymptotes as axes of coordinates.
Let the lines OB, OC (Fig. 73) be the asymptotes, A the
vertex of the curve, and let the angle AOC=^a.
Let the coordinates of any point P of the curve be x, y
when referred to the axes of the curve, and x\ y' when
referred to OB, OC as axes of coordinates.
Draw PMl_to OA, PN\\ to CO ; then
X = ON cos a -\- NP cos a = {x' + //') cos a,
y = NP sin a — ON sin a= (y' — x') sin a.
186
ANALYTIC GEOMKTKY.
Fig. 73.
Hence, by substituting in [40], we obtain
(.rJ -{- //')- cos- a ill' — x'y sin^tt
b-
1.
But
AD
cos a:
OD Vft2 + i2
OA a
OB -sJa'-\-b'^
Substituting these values, and dropping accents, we have
4:XU = H^ + bK [4G]
Cor. 1. The equation of the conjugate hyperbola is
4xy=i — (rt- + Z--).
2ab
CoR. 2. Sin COT? = sin 2^ = 2 sin « cos a= , , ,,•
a' + ¥
It a = b, sin COB = 1 ; therefore, COB = ^ir.
Cor. 3. Let {x^, y^) denote P (Fig. 72), referred to the
asymptotes ; then
OS X OR = 20ffX 2 IIP = 4a-i//i = a^ + b\
That is, tJie product of the intercepts of a tangent on the
asymjjtotes is equal to the sum of the squares of the semi-axes.
THK HYPERBOLA.
187
CoK. 4. In Fig. 72, the area of tlie triangle ROS equals
h OS X OB sin ROS = h (a' + b') ^?^.. = ah.
That is, the area of the triangle formed by any tangent
and the aaymptotes is equal to the product of the sevii-axes.
183. The polar of the focus («e, 0) is
Fig. 74.
Hence, if OD is taken so that
OF: OA=OA:OD,
^hen DN perpendicular to OF is the polar of F, and is
called a Directrix of the hyperbola. In like manner we may
construct i>'X', or the directrix corresponding to the focus F'.
Cor. As in § 150 we may prove that
_PF
188 ANALYTIC GEOMETRY.
Whence, the hyperbola may be defined as
The locus of a jJoint whose distances from a fixed point and
a fixed straight line hear a constant ratio greater than unity.
1 84. To find the polar equation of an hyperbola, the left-
hand focus being taken as pole.
If X is reckoned from the centre, and we write
p=^ex-\- a, (1)
p will be positive or negative according as the point is on
the right or left-hand branch.
Now x^ p cos 6 — c^=p cos 6 — ae.
Whence, by substitution and reduction,
e cos 6 — 1 "- -■
From (1) we know that a point is on the right or left-
hand branch, according as p in [47] is positive or negative ;
that is, according as cos ^ > or <! —
If 0 = 0, p = ae^a = F'0+OA = F'A (Fig. 70).
If e cos 0 — 1 = 0, or ^ = cos~' -, p =y:>, as it should,
e
since in this case the radius vector is || to the asymptote.
If 6 = ^TT, p ^ — a (e- — 1)= — semi-latus rectum.
If e = ^, p= a — ae = — F'A'.
Exercise 39.
1. What is the polar of the point (—9, 7) with respect
to the hyperbola 7x- — 12/ = 112 ?
2. Find the equations of the directrices of an hyperbola.
3. Find the angle formed by a focal chord and the line
that joins its pole to the focus.
THE HYPERBOLA. 189
4. Find the pole of the line Ax-]- B)/-{- C = 0 with re-
spect to an hyperbola.
5. Find the polar of the right-hand vertex of an hyper-
bola with respect to the conjugate hyperbola.
6. Find tlie distance from the centre of an hyperbola
to the point where the directrix cuts the asymptote,
7. If (xi, iji) and (x.2, ijo) are the ends of two conjugate
diameters, then
X1X2 jh!h _ ^
8. The equation of a diameter in the hyperbola 2ox^ —
16^/^^400 is 3>/ = x. Find the equation of the conjugate
diameter.
9. In the hyperbola 49a'^ — 4y" = 196, find the equation
of that chord which is bisected at the point (5, 3).
10. Find the length of'the semi-diameter conjugate to
the diameter y = 3x in the hyperbola 9x- — 4^^ = 36.
11. Two tangents to an hyperbola at right angles intersect
on the circle
a-^ -[- ?/^ = «^ — b^.
12. Tangents at the extremities of any chord of an hyper-
bola intersect on the diameter which bisects that chord.
13. Prove that FQ (Fig. 71) is parallel to one asymptote
and bisected by the other.
14. An asymptote is its own conjugate diameter.
15. The conjugate diameters of an equilateral hyperbola
are equal.
16. Having given two conjugate diameters in length and
position, to find by construction the asymptotes and the
axes.
17. To draw a tangent to an hyperbola from a given
point.
190 ANALYTIC GKOMKTKY,
18. Find the equation of a tangent at any point (x^, v/i)
of the h^^perbola Axt/ = (r-{- Ir.
19. Find the equation of an hyperbola, taking as the
axis of ij
(i) the tangent through the left-hand vertex ;
(ii) the tangent through the right-hand vertex.
20. Find the polar equation of an hyperbola, taking the
right-hand focus as pole.
21. Find the polar equation of an hyperbola, taking the
centre as pole.
22. To find the centre of a given hyperbola.
23. The distance from a fixed point to a fixed straight
line is 10. Find the locus of a point which moves so that
its distance from the fixed point is always twice its distance
from the fixed line.
24. Show that the locus of x^ — \if — 2x— 16y — 19 = 0
is an hyperbola ; find its centre and axes.
25. If A and B have unlike signs, prove that the locus
of Ax^ -\- Bif + I)x + E[i + F^^ 0 is in general an hyperbola
whose axes are parallel to the coordinate axes ; and deter-
mine its semi-axes.
26. Through the point (— 4, 7) a straight line is drawn
to meet the axes of coordinates, and then revolved about
this point. Find the locus of the point midway between
the axes.
27. A straight line has its ends in two fixed perpen-
dicular lines, and forms with them a triangle of constant
area d\ Find the locus of the middle point of the line.
28. The base «. of a triangle is fixed in length and posi-
tion, and the vertex so moves that one of the base angles
is always double the other. Find the locus of the vertex.
CHAPTER VIII.
LOCI OF THE SECOND ORDER.
185. The loci represented by equations of the second
degree that are not of the first order are called Loci of the
Second Order.
In the preceding chapters we have seen that the circle,
parabola, ellipse, and hyperbola are loci of the second order.
We will now inquire whether there are other loci of the
second order besides the four curves just named ; in other
words, we will determine what loci may be represented
by equations of the second degree.
For this purpose we shall write the general equation of
the second degree in the form
Ax^ -H Bir + Cri/ + Z)a; + % + F= 0, (1)
and shall assume that the axes of coordinates are rectan-
gular. This assumption will in nowise diminish the gener-
ality of our conclusions ; for if the axes were oblique, we
could refer the equation to rectangular axes, and this change
would not alter the degree of the equation or the nature of
the locus which it represents (§ 91). •
186. To find the condition that the rjeneral equation of
the second degree may represent two loci of the first order.
To do this let us solve (1) with respect to one of the
variables. Choosing y for this purpose, we obtain
Cx + E 1 , .
r y = ^^^^^Lx^ + Mx^N, (2)
where L=C^ — iAB, iI/= 2 ( CE — 2BD) , N=^ E- — \BF.
192 ANALYTIC GKOMKTRY,
If Lx^ -\- Mx -\- N is a perfect square, then tlie locus of
(2), or (1), will be two loci of the first order.
Now, from Algebra, we kuow that the condition tliat
Lx^ + 3fx + N should be a perfect square is
or, substituting the values of L, M, and N, we have
(CE-2BDy-{C^ — 4.AB) (E'-4:BF) = 0,
or F(C^ — 4:AB)-\-AE^ + BD^—CI)i;=0. (3)
The quantity on the left-hand side of equation (3) is
usually denoted by A, and is called the Discriminant of
equation (1).
This same result was obtained by a more general method
in § 57 ; hence,
Whenever A = 0, eqiiation (1) represents two loci of the
first order. These loci may be readily determined by
resolving (1) into two simple equations in x and y.
CENTRAL CURVES. 2 NOT ZERO.
187. A centre of a curve is a point that bisects every
chord passing through it. Loci are classified as Central
and Non-Central, according as they have or have not a
definite centre. The circle, ellipse, and hyberbola belong
to the first class, the parabola to the second.
188. To find the equation of the central loci represented,
by eq^iation (1) referred to their centre.
To do this let us change the origin to the point {h, k),
and then so choose the values of h and k that the terms
involving the first powers of x and y will vanish. Making
the change by substituting in (1) x-\-h for x, and y-\-k for
y, we find that the coefficients A, B, and C remain unaltered,
and we may write the transformed equation
Ax^ + By"" + Cxy + D'x + E'y = R, (4)
LOCI OF THE SECOND ORDER. 193
wh ere D' = 2 A h + Ck + D,
E' = 2Bk+Ch-\-E,
R=— [A/i" + Bk- + Ckk + Dh + m- + F^.
The values of h and k that will make 1)' and E' vanish
are evidently found by solving the equations
2Ah+Ck-{-D = 0,
2Bk-\-Ch-\-E=0,
CE-2BB j_CD — 2AE
and are ^'^-4^^_C2' ''-AAB-C'
If 4:AB — C^, denoted by 2, is not zero, these values of h
and k are finite and single, and equation (4) may be writttui
Ax^ + Bi/^-\-Cxi/ = E. (5)
From the form of (5) we see that if (x, y) is a point in
its locus, so also is ( — x, — //) ; that is, the new origin (Ji, k)
is the centre of the locus. Hence,
Whe7i 2 is not zero, equation (1) can he reduced to the
form of (5), and represents central curves.
When, however, 2 = 0, the values of h and k become
infinite or indeterminate, and the locus of (1) lias no defi-
nite centre. Hence,
When 2 = 0, (1) cannot be reduced to the form of (5), and
rejiresents no7i-central curves.
The value of R can be reduced to the following useful
form, which shows also that R and A vanish together.
R = — [Ah' + Bk- + Ch k + l)h + Ek + F^
= -i[(2Ah+Ck + D)h
+ (2Bk + Ch -\- E)k + Dh + Ek + 2/^]
= — i {D'h + Ek + Dh + Ek + 2F)
= -^(I)h-\-Ek-\-2F)
_ 2BD'—CI)E-\-2AE?—CnE-\-2F(C'-AAB)
~ * C' — 4:AB
_ A
~ 2'
194 ANALYTIC GKOMETRY.
189. To reduce (5) to a known form hij rausinf/ the tern„
in xy to disappear.
For this purpose we change the direction of the axes
through an angle 0, keeping the origin unaltered, and then
determine the value of 0 by putting the new term that
involves xy equal to zero.
The change is made by substituting for x and y, in
equation (5), the respective values (§ 86),
X cos 6 — y sin 6,
X sin 6-]r y cos 6 ;
and equation (p) now beco]nes
Px''-^ny-+C'.ry = B,
where P = A cos- 6 -f- J^ sin^ 0-}- C sin 0 cos 6, (&'^
Q=^ A sin- 6-\- B cos- d — C sin 6 cos 6, (7)
C' = 2{B — A) sin 9 cos 0 + C(cos- $ — sin^ 6). (8)
Putting C'=^0, we obtain, by Trigonometry,
(A — B) sin 26 — C cos 20 = 0, (9)
or tan 2^ = —^-. (10)
A — x>
Since any real number, positive or negative, is the tan-
gent of some angle between zero and tt, equation (10) is
satisfied by some value of 9 between zero and •^tt. In what
follows we shall use the simplest root of (10).
I>y this transformation, equation (5) is reduced to the
form p^.2^^y._^^ (jl)
of which the discussion will be found in the next section.
Cor. 1. The values of P and () in terms of A, B, C may
be found as follows :
From (G) and (7), by addition and subtraction,
P^Q = A + B, (12)
P — ^ = (^ — J?) cos 2^ + C sin 26. (13)
LOCI OF THE .SECOND ORDER. 195
Equation (9) may be written
Oz=(A — B) sin 26 — 0 cos 26. (14)
Adding the squares of (13) and (14), we have
(P-Qf = (A-Br+C', (15)
or P-Q^±\/{A-Bf+C'i (IG)
Whence, from (12) and (16),
P = ^[A + B± V(^ - By + C% (17)
Q = i[.l + P rp V(.4 - By + C'l (18)
These values of F and Q are evidently always real.
Cor. 2. By squaring (12) and subtracting (15), we obtain
4FQ = 4.AB — C = X (19)
Hence, I* and Q have like or unlike sir/tis, accordintj as %
is positive or negative.
Cor. 3. In applying formvdas (17) and (18), the question
arises which sign before the radical should be used. If in
(13) we substitute for cos 26, its value obtained from (14),
we have
p ^^[(^-/?)^+C'^]sin2g
C/
Since the numerator of the fraction is always positive,
P — Q must have the same sign as 6'; that is, the upper
or lower sign in (16) must be taken according as C is posi-
tive or negative.
Hence, the upper or lower signs in (17) and (18) are to he
taken accordinr/ as C is positive or negative.
190. The nature of the locus represented by equation
(11) depends upon the signs of /*, Q, and R. There are
two groups of cases, according as 2 is positive or negative,
and three cases in each group.
196 ANALYTIC GEOMETRY.
Group 1. 2 Positive.
In this group, P and Q must, by (19), agree in sign.
Case 1. If R agrees in sign with P and Q, then, by § 124.
the locus is an ellipse whose semi-axes are a/— and \-7^-
li P= Q, the locus is a circle.
Case 2. If R differs from P and Q in sign, no real values
of X and ij will satisfy (11), so that no real locus exists.
Case 3. If 2t = 0, the locus is the single point (0, 0).
Group 2. % Negative.
In this group, P and Q, by (19), must have unlike signs.
Case 1. If ^agrees in sign with P, we may, by division
(and by changing the signs of all the terms if necessary),
put equation (11) into the form of equation [40], page 171.
Therefore, the locus is an hyperbola, with its transverse,
axis on the axis of x, and havinar for semi-axes
«=V]
Case 2. If R agrees in sign with Q, we may, by divisioi.
(and by change of signs if necessary), put equation (11)
into the form of equation (1), page 172. Therefore, the locus
is an hyperbola, with its transverse axis on the axis of y.
Case 3. If i? = 0, the locus consists of tioo straight lines,
intersecting at the origin, and having for their equations
NON-CENTRAL CURVES. 2 = 0.
191. To determine the locus of (1) ivhen A and 2 are loth
zero.
LOCI OF THE SECOND ORDER. 197
If A = 2 =^ 0, then from the first form of A in § 186 we
must have ^j,j -2BD = 0. (20)
Hence, L = 3I=0, (§186)
and (2) becomes
2Bi/ +Cxi-JSzf '\lE^ — 4.BF= 0, (21)
which represents two parallel straight lines, two coincident
straight lines, or no locus, according as E^ — 4:BF';>, =,
or<0.
When 4:AB — C^ = 0, if C is not zero, neither A nor B
can be zero; if C = 0, either A or B must be zero, but both
cannot be, since if A=^B = C = 0, (1) is no longer of the
second degree.
When C=^B=0, by solving (1) for x, and introducing
the above condition, we should obtain, instead of (21) its
corresponding equation.
2 Ax -\-Ci/-\-D:f \/l)^—4:AF= 0, (22)
whose locus is also two parallel straight lines. Hence,
When A and 2 are both zero, equation (1) represents two
parallel straight lines, two coincident straight lines, or no locus.
Cor. Eliminating B between 2 = 0 and (20), we obtain
CD — 2AE=0. (23)
In like manner (20) follows from 2 = 0 and (23).
From these results, and the values of h and k, we learn that
(i) When A=2=0, h and k are both indeterminate,
and conversely.
(ii) The values of h and k are indeterminate together.
192. To determine the locus of (1) ivhe7i 2 is zero and A
is not zero.
We simplify (1) by first making the term in ary disappear
by proceeding exactly as in § 189; that is, by turning the axes
19<S ANALYTIC GEOMETRY.
through an angle 9, the value of which is determined by the
equation f-i
^^^^^ = jZZb- (24)
If F, Q, U, Frei)resent the new coefficients of x^, if, x, y,
respectively, P and Q will have values identical with those
of P and Q given in § 189, and
U= JD cos e + i: sin 9, (25)
V= — D sin 9 + 1!: cos 9. (26)
Since C-=4:AB, from (17) and (18) we have, when C is
positive, p_^^^ g_0.
When C is negative,
F = 0, Q = A-\-B,
and (1) assumes the form
Qf + Ux +Vy + F=0. (27)
To further simplify, we divide by Q, and obtain
U V F
^^+^^- + ^^+^ = ''
If we now take as a new origin the point
4:QF-V^ V
^UQ 2Q
equation (28) becomes
U
which represents a parabola whose axis coincides with the
axis of X, and which is situated on the positive or the negative
side of the new origin, according as U and Q are unlike or
like in sign (§ 94).
liOCI OF THE SECOND OIIDEK. 199
The vertex of the parahuhi is the new origin, and the
parameter is equal to the coeiiicient of x in the equation
of the curve.
This hist transformation is possible except when fJ^=0 ;
but, when U=^ 0, (27) evidently represents two parallel
straight lines.
Suppose that C is positive. Then the general equation
becomes
Fx^-i- Ux-i-V>/+F=0. (29)
And this, by changing the origin to the point
4:PF— U^ U
(■
4.VP IP
becomes cc^ = — ^ y.
This represents a parabola having the axis of y for its
axis, and placed on the j/oslf ire or the neyafire side of the
new origin, according as V and P are unlike or like in sign.
It should be noted that the value of P or Q, when not
zero, is A + P- The values of U and V* can be found from
(25) and (26).
* We may obtain the values of U and T" iu terms of the origmal
coefficients, as follo%YS :
From (24) we find, by Trigonometry,
-(A- B) ± -J (A - BY- + C2
tan 6 = jf •.•
Introducing the condition iAB — C^, we obtain
2A
tan 6 = —5 if C is negative ;
2B .. ^ .
= -^1 if C IS positive ;
whence, if C is negative,
2^ -C
sin e
V442 -I- (>2 V4^2 + 02
And if C is positive,
200
ANALYTIC GEOMETRY.
If C^A = 0, the given equation is of the form of (27),
its locus is a parabola and can be found as that of (27). If
6' = ^ = 0, the given equation is of the form of (29), and
its locus is a parabola.
193. The main results of the investigation are given in
the following Table :
Loci
Represented by the General Equation of the
Second Degree.
Ax^ + By^-+ Cxy + Dx + i:y+ F=0.
CLASS.
CONDITIONS.
LOCL'S.
I.
Loci
having a
centre.
S positive, A not zero.
S positive, A = 0.
2 negative, A not zero.
S negative, A = 0.
Ellipse, or no locus.
Point.
Hyperbola.
Two intersecting straight lines.
II.
Loci not
having a
centre.
2 = 0, A not zero.
S = 0, A = 0.
Parabola.
r Two parallel straight lines,
] One straight line,
V Or no locus.
Thus it appears that there are no loci of the second order
besides those whose properties have been studied in the
preceding chapters.
sm 6 =
2B
cos e =
V452 + C2
By substitution, we obtain from (25) and (26)
if C is negative.
V4B2 + C2
if C is positive,
U =
U =
2AE - CD
2 BE + CD
^JIiF+~c^
v =
v =
2AD+ CE
CE-2BD
■\J4B^ + C^
(30)
(31)
• LOCI OF THE SECOND ORDER. 201
194. Examples. 1. Determine t?ie nature of the locus
ox^-\-57/-{-2xi/ — 12x — 12ij = 0, (1)
transform the equation and construct it.
Here A = 5, B=o, C=2, D=-12, E=-12,F=0.
Whence, 2 = 96, A ==1152.
Hence, the equation represents an ellipse or no real locus.
Therefore, the equation of the locus referred to new
parallel axes through the centre (1, 1) is (§ 188)
5x'- + 2xi/ + 5f- = 12. (2)
To cause the term in x)/ to disappear, we have
Whence, 2$ = 90°, or ^ = 45°.
(We use the upper signs m the values of P and Q by § 189.)
Hence, by § 189 the equation of the ellipse referred to
its own axes is
60.2 + 4^^-12, or 1 + 1 = 1. (3)
Whence, a = VS, b = V2, and a lies on the axis of i/.
To construct the equation, draw the axes OiXi,OiYi (Fig.
75); locate the centre (1, 1). Through this point O2 draw
the second set of axes, O^X^, 0^ Y„. Through O2 draw the
third set of axes OaA'g, 02^3, making X0O0X3 equal to 45°.
Lay off O2A' = O.A = Vs, and 0^0^ = O^B = V2. The
ellipse having BO^ and A A' as axes will be the required locus.
202
ANALYTIC GEOMETRY.
The equation of the locus referred to the axes OiX^, Oi Yi
is (1); its equation referred to O^Xo, O^Y^ is (2); and its
equation referred to O^X^, O^Y^ is (3). In constructing
the locus it is not necessary to draw the second set of axes
C/a JCj, (/a jfa*
(1)
Fig. 75.'
2. Determine the nature of the locus
x^-\-?f- 5x1/ + Sx- - 20^ + 15=0,
transform the equation and construct it.
Here 2 = -21, A = — 21.
Tlierefore, the locus is an hyperbola.
i? = A^2 = l, A = -4, k = 0.
Hence, the first transformed equation is
x^-\-y" — 5xi/ = l.
In the second transformation,
^-45°, P = -|, Q = i.
Hence, the equation of the curve referred to its own axes is
3x' — 7/ = — 2,
from which we see that a = Vf , b = V§, and a, or the trans-
verse axis, lies on the axis of y.
(2)
LOCI OF THE SECOND ORDER.
203
To construct the equation, draw the axes OiXi, OjFi,
locate the centre ( — 4, 0), and through it draw the third set
Fig. 76.
of axes O2X3, O2YS, making A'lOaA's^ 45°. Then lay off
02A'=02A=\/^, and O^B = O^B' = ^ ^ ; and draw the
hyperbola having AA' and BB' as its transverse and conju-
gate axes respectively.
3. Determine the nature of tlie locus
a;^ + 2.ry-y2-f-8.r + 4^-8 = 0, (1)
transform the equation and construct it.
Here 2 = — 8, A= — 176; hence, the locus is an hyperbola.
^ = 22, h = — 3, /.• = — !;
hence, the first transformed equation is
x' -\- 2x7/ — if- = 22. _ (2)
0 =22|-°, P=\l2, Q = — \/2;
hence, the equation of the curve referred to its own axes is
V2a;2— V2// = 22. (3)
The hyperbola is equilateral, and its transverse axis lies
on the axis of x. (Let the reader construct it.)
4. Determine the nature of the locus
x^-\-,f — 2x!i-\-2x — i/ — l=0, (1)
transform the equation and construct it.
204
ANALYTIC GEOMETRY.
Here 2 = 0, A is not zero.
Therefore, the locus is a parabola.
0 = 4:5°, F = 0, Q = 2, U=^-\l2, V= — %\l2-
hence, by revolving the axes through an angle of 45°, the
equation becomes
2^" + i V2a; — %-^2ij -1 = 0,
or t/2-fV2^+(tV2)^=-iV2^ + M.
or (2^_3V2)2 = -iV2(x-f|V2). _ (2)
Passing to parallel axes whose origin is (f|V2, |V2),
(2) becomes
y^ = -iV2^, ■ (3)
the locus of which is a parabola whose latus rectum, or
parameter, is ^\l2.
Fig. 77
To construct the equation, draw the original axes O^X^,
OxYi, then draw the second set of axes O1X2, O1Y2, making
XiC>iXj = 45°.
Locate the new origin O3, (f|V2, |V2), and through it
draw the third set of axes O^Xz, O3 Pg, to which (3) refers
the locus, which is now easily drawn.
5. Determine the nature of the locus
x^ — ixy + iy^ — 6x-\- 12y =■ 0.
LOCI OF THE SECOND ORDER.
205
Here 2 = 0 and A = 0.
Factoririg the first member of the equation, or solving
the equation for x or y, we obtain as the equations of these
lines
x — 2ij = 0, x — 2y — Q = 0.
Hence, the locus is two parallel straight lines.
6. Determine the nature of the locus of
if' — xy — Q>x- — 3.x + ?/ = 0.
Here 2 is negative, and A = 0.
Hence, the locus is two intersecting straight lines. Their
equations are
y + 2a; + 1 = 0, and ?/ — 3x == 0.
195. A Conic is the locus of a point whose distance from
a fixed point bears a constant ratio to its distance from a
fixed straight line.
196. To find the equation of a conic.
Y
N
Fig. 78.
Let F be the fixed point, and YY' the fixed straight line.
Through jPdraw XO perpendicular to YY, and use OA'and
OFas axes of reference. Let^:> denote the distance OF, and
20G ANALYTIC GEOMETRY.
e tlie constant ratio ; then FP -=r NP = e, P being any point
(a-, I/) in the curve.
Now FP^ = Fir' + MP\
But FP = eX NP = ex,
FM=x—p, MP = ij.
Therefore, e^x'^ = (x — j'Y + if^
or ( 1 - e^) a:^ + / - 2px + /^^ = Q, (1)
which is the equation required.
CoR. In equation (1), which is of the second degree,
2 = 4 (1 — tj-), and A = 4/vV.
Hence, when the fixed point is without tlie fixed line, A
is not zero, and
If e <C 1, 2 > 0, and the conic is an ellipse.
If e = 1, 2 = 0, and the conic is a parabola.
If e >• 1, 2 <C0, and the conic is an hyperbola.
When the fixed point is in the fixed line, A ^ 0, and
If e < 1, 2 > 0, and the conic is the point (0, 0).
If e = 1, 2 = 0, and the conic is a right line.
If « > 1, 2 < 0, and the conic is two intersecting right
lines.
If e^O, by § 61, the conic is a cii'cle or a point, accord-
ing as p is not or is zero.
From §§ 92, 150, 183, it follows that the fixed point is a
Focus, the fixed right line a Directrix, and the constant
ratio the Eccentricity, of the conic.
Exercise 40.
Determine the nature of the following loci, transform
their equations, and construct them :
1. ^x'-{-2if—2x-\-y-\=0.
2. 3a;2-f-2j7/ + 3/ — ir.// + 23 = 0.
LOCI OF TlIK SKCOND ORDER. 207
3. X- — I O.v.v + //" + •'• + // + 1 = 0.
4. x^ -f- xi/ -{- //'^ -\- X -{- 1/ — 5 = 0.
5. if — X- — y = 0.
6. l+2::c + 3^2 = 0.
7. y-~2xi/-\-x^ — Sx-\-16 = 0.
8. ar — 2x1/ -\- 1/'^ — Gx — 6^ + 9 = 0.
9. 7/_2x — 8// + 10 = 0.
10. 4a;2 + V + 8.r + 36// + 4 = 0.
11. 52cc=^ + 72x// + 73/ = 0.
12. 9/ — 4x2 _ 8.^ _|_ 18^ + 41 = 0.
13. y^ — xy — 5x-\- 5i/ = 0.
CHAPTER IX.
HIGHER PLANE CURVES.
197. An Algebraic Curve is one whose rectilinear equa-
tion contains only algebraic functions. A Transcendental
Curve is one whose rectilinear equation involves other than
algebraic functions. Thus, the loci of 7j=^a^, ?/ = tan x,
y = {a — X) tan {jttx -i- a), y = sin ~^x are transcendental
curves. Transcendental curves and all algebraic curves
above the second order are called Higher Plane Curves.
Let the symbol F (x, y) denote any rational integral func-
tion of x and y, of the third or higher degree. If F (x, y)
breaks up into simple or quadratic factors in x and y, the
locus of F (x, y) = 0 consists of lines of the first or second
order. If F (x, y) does not break up into rational factors in x
and y, the locus of F(x, ?/) ^= 0 is a higher plane curve whose
order is of the degree of the equation. Thus, the locus of
y^ — x^^{y — x) (y^ + a-y + x^) = 0,
consists of the right line y — x = 0 and the ellipse y- -\- xy
-\- x^ = 0 ; the locus of
y*-\-xy-{-2y^—2x*—5x"—3=(y'-\-2x^+S)(y^—x^-l)=0,
consists of the ellipse y- -\- 2x^ + 3^0, and the hyperbola
y^ — x^ — 1=0; while the locus of y^ — ax -\- x'^ ^ 0 is a
higher plane curve of the third order.
In this chapter we shall consider a few of the higher
plane curves, some of which possess historic interest from
the labor bestowed on them by the ancient mathematicians.
198. The Cissoid of Diodes. Let X^(rig. 79) be a
tangent to the circle XSO at the vertex of any diameter
HIGHEK PLANE CURVES.
209
OX ; let OR be any right line from 0 to XH cutting the
circle at S, and take 0F = ES; then the locus of P, as Oli
revolves about 0, is the Cissoid.
To find its equation referred to the rectangular axes OX
and OY, let OM=x, MP=ij, and OC=CX= CD = r.
Now, MP : OM :: NS : ON. (1)
Since OP = ES, OM=NX.
Hence, 0N= OX— NX= OX— 0M= 2r — x,
and NS= -^ONXNX^ \l{2i — x)x.
Substituting these values in (1), we obtain
£C3
2/^ V(p;
iC)X
which is the equation sought.
or y^
2r-ac'
[48]
210 ANALYTIC GEOMETRY,
A discussion of [48] leads to the following conclusions:
(i) The curve lies between the lines x = 0 and x = 2r.
(ii) It is symmetrical with respect to the axis of x.
(iii) It passes through the extremities of the diameter
perpendicular to OX.
(iv) It has two infinite branches to which x = '2r is an
asymptote.
CoR. 1. To find the polar equation, let 6 = X0F and
P=OF;
^ OX OS
then '^'^=0e'''0X'
Hence, p= OF = SE= OE- 0S=-^ — 2r cos 6
^ cos 0
sin'^O
cos 9
Therefore, p ^ 2r sin 6 tan 6,
which is the equation sought.
The cissoid was invented by Diodes, a Greek mathema-
tician of the second century b.c, for the solution of the
problem of findlvr/ tivo mean proportionals, of which the
dupUcation of the cube is a particular case.
Cor. 2. To duplicate the cube, in Fig. 79, take CB = 2r,
and draw 5X cutting the cissoid in ^; then, smceCX=^CB,
EX=^EQ. But from tlie o(] nation of the cissoid, we have
W == ^' = r^; therefore, W = 20E'.
Let c denote the edge of any given cube ; take ^i so that
OE: EQ::c:c,, or OE' : W - «' = «i'-
But W = 2 7a£^' ; th eref ore, Cj' = 2c^ ;
that is, ci is the edge of a cube double the given cube in
volume. In like manner, by taking CB = mr, we can find
the edge of a cube m times the given cube in volume.
HIGHER PLANE CURVES.
211
199. The Conchoid of Nicomedes. The Conchoid is the
locus of a point P such that its distance from a fixed line
XX', measured along the line through F and a fixed point
A, is constant. A is the Pole, .\'A'' the Directrix, and the
constant distance BP, denoted by b, is the Parameter.
P>^
Y
JM
\k' m
0 /r
~^_
Y
A
Y
-■■B
Fig. 80.
To construct the conchoid by points, through A draw any
line AP cutting XX' in P. La}' off PP^^h on both sides
of XX'. In like manner locate the points P', and any
number of others, and trace the branches through them.
To find the equation of the curve referred to XX' and a
line through A perpendicular to A'A'' ; let
0M= x, MP = y, AO = a.
Now, AB : BP:: RM : MP,
and BM= ^'rF - HlF = ^iW^f.
Therefore, x \ y-\-<t ■.: ■\/lr' — if : //.
Therefore, x'^y^ = {y + af- {b^ - y'), [49]
which is the equation of both branches of the conchoid. A
discussion of [49] leads to the following conclusions :
(i) The curve lies between the lines // = ^' and //^ —h.
(ii) The curve is symmetrical with respect to the axis of y.
212
ANALYTIC GEOMETRY.
(iii) The axis of x is an asymptote to each branch of the
curve.
If 6 > a, the lower branch has an oval or loop, as in the
figure.
If i = a, the lower branch passes through A and is some-
what like that in the figure, without the loop below A.
If 6 < a, the upper and lower branches are like the dotted
lines in the figure.
If a = 0, the conchoid becomes a circle.
CoR. 1. If A is the pole and A Fthe polar axis, we have
p = AB'±B'F' = asec 6 ±h,
which is the polar equation of the conchoid.
The conchoid, invented by Nicomedes, a Greek mathema-
tician of the second century b.c, was, like the cissoid, first
formed for solving the problem oi finding ttvo mean propor-
tionals or duplicating the cube. It is more readily applicable,
however, to the trisection of an angle, a problem not less
celebrated among the ancients.
Cor. 2. To trisect any angle, as CAP (Fig. 81), on ^C
lay off AB any length ; through B draw BK perpendicular
to AP, and take KF = 2AB. Construct a conchoid with
^ as a pole, XX' as a directrix, and KP as a parameter.
HIGHER PLANE CURVES. 213
At R erect a perpendicuhir to XX' intersecting the cou-
choid in D ; then DA will trisect the angle RAF.
For bisect DB in S ; then
RS = SD = iKF = AR.
Therefore, Z RAS= Z ^'S'^ = 2Z RDS= 2Z DAP.
Therefore, Z DAP = iZ RAF.
/
200. The Lemniscate of Bernoulli. The Lemniscate is
the locus of the intersection of a tangent to a rectangular
hyperbola with the perpendicular to it from the centre.
To find its equation we proceed as follows :
The equation of the tangent to the equilateral hyperbola
x^ — y^ = ci^ at the point (xi, i/{) is
a'la;— yi?/ = «^- (1)
The equation of the perpendicular from the origin to (1) is
y^-^:r, or - = -^. (2)
Xl Xi i/i
Solving (1) and (2) for x^ and v/i by multiplying each
term of (1) by one of the members of (2), we obtain
a"x a'i/
X + V == — =
^1 y\
^, . ti^x ahi
Therefore, cr, = ^r-, — 7/ ?/i ^ ;— r — ^•
But, since (a^i, y^) is on the hyperbola, we have Xx — y\=^o^\
hence, by substitution, we obtain
(^2 + ^•2)2 = „2(ar;2 _ 2,2)^ ^5QJ
which is the equation sought.
214
ANALYTIC (ilOoMlCTHV.
From \_iiO^ it follows tliat the curve is syiimietrical with
respect to both axes. The form of the curve is given in
Fig. 82.
Fig. 82.
CoR. 1. Substituting p ens 6 for x, anrl p sin 6 for y in
[50], and remembering that cos'^ ^— sin^ ^=cos 20, we obtain
p^:=a^ cos 20
as the polar equation of the lemniscate.
(2)
CoR. 2. From (2), p = ±a Vcos 20.
Hence, when 0=^0, p = ±a. AsO increases from 0° to 45°,
p changes from ± « to ± 0, and the portions in the lirst and
third quadrants are traced. As 0 increases from 45° to 135°,
cos 20 is negative and p is imaginary. As 0 increases from
135° to 180°, p changes from ± 0 to ± a, and the portions
in the second and fourth quadrants are traced. Therefore,
(i) The curve consists of two ovals meeting at the pole 0.
(ii) The tangents to the curve at 0 are the asymptotes
to the equilateral hyperbola.
HIGHER PLANP: CURVES.
215
201. The Witch of Agnesi. Let 17/ be a tangent
to the circle OKYnt the vertex oi' the diameter OY ; let
OB be any line from 0 to Ylf cutting the circle in K;
produce the ordinate I>K, and make DP'^^YE; then the
locus of P is the Witch.
To find its equation, let the tangent OX and the diameter
or be the axes; let 0F=2randPbe any point; then
OD=ij, DP = x, and
OD : OY :: I)K : DP{= YE), (1)
9, '~
or y -.'Zr :: \ly(2r — y)'-x.
Therefore, oc-y = 4r-(2/- — y)
is the equation of the Witch.
[51]
CoR. 1. Since a; = ± 2j' V(2/ — y) -r-y, it follows that
(i) The curve is symmetrical with respect to the axis of y-
(ii) The curve lies between the lines y=0 and y = 2r.
(iii) The axis of x is an asymptote to eacli infinite branch.
CoR. 2. From (1) it follows that corresponding abscissas
of the circle and the Witch are proportional to the ordinate
and the diameter of the circle.
The Witch was invented by Donna Maria Agnesi, an
Italian mathematician of the eiQ:hteenth oontnrv.
216
ANALYTIC GEOMETRY.
202. The Cycloid. A Cycloid is generated by a point P in
the circumference of a circle RFC, which rolls along a right
line OX. The curve consists of an unlimited number of
branches, but a single branch is usually termed a cycloid.
The right line 0X\?> called the Base ; the rolling circle RFC
the Generatrix ; and P the Generating Point. If 0K=^ KX,
the perpendicular KH is the Axis, and JI the Highest Point.
OM
To find the equation of the curve, first take as the axis
of X the base OX, and as the origin the point 0, where
the curve meets the base. Let r denote the radius of the
generatrix RPC, and $ the angle PCR; then arc PR
equals OR over which it has rolled, and 6 ^= a,vG PR -^r.
Denote the coordinates of P by x and y ; then
X = 0M= OR — MR = arc PR — PN= r6 — r sin d.
y = MP = RC — NC=r— r mse.
Therefore,
X =^r (6 — sin 6)
y=^r (1 — cos 6)
Equations (1) taken as simultaneous are the equations of
the cycloid.
To eliminate 6 between these equations, from the second,
we obtain
(1)
HIGHER PLANE CURVES. 21'
cos 0 = : therefore, sin ^ = ± — — :
/• /•
and vers ^=ri —cos ^1 = - ; or^ = vers'^--
L -* r r
Substituting these values of 6 and sin 9 in the first of
equations (1), we have
x = r vers '^ a? y. T ^2ri/ - yS [52]
the equation of the cycloid in the more common form.
In the value of sin 6, and therefore in equation [52],
the upper or lower sign is used according as ^ < or > tt ;
that is, according as the point is on the first or second half
of a branch.
From 1/ = r (1 — cos $) it follows that the locus lies
between the lines i/ = 0 and ?/ = 2r.
For ?/ = 0 in [52], x = 0, ± 2iri', ± 47rr, ; hence, the
locus consists of an unlimited number of branches like
OHX, both to the right and to the left of 0 Y.
203. Let the highest point 0 (Fig. 85) be taken as the
origin, and OX parallel to the base as the axis of x; then
OY, the axis of the curve, will be the axis of y. Let 6
denote the angle HCK.
The point K was at Y when P was at 0, and arc KH=
YH. Hence,
X = 0M= Yir+ BP = re + r sin 6.
y = - MP = — XC + BC= — r -\- r cos $.
Hence, the equations are
a- — ;■ (^ + sin 6) ") ^^^
y=^r (cos ^ — 1) ) '
or a; = r vers ^ ( — •- \ ± V— 2rij — y\ (2)
218
ANALYTIC GEOMETRY,
R Y
Fig. 85.
Tlie invention of the cycloid is usually ascribed to
Galileo. After the conies, no curve has exercised the
ingenuity of mathematicians more than the cycloid, and
their labors have been rewarded by the discovery of a
multitude of interesting properties. Thus, the lengtli of
the branch ROA is eight times the radius, and the area
BOA is three times the area of the s:eneratins[ circle.
Exercise 41.
1. Prove that the cissoid is the locus of the intersection
of a tangent to the parabola if^= — 8ra; with the perpen-
dicular to it from the origin.
By § 99, the equation of a tangent to if^ — Srx is
(1)
y = mx ■
— )
m
(2)
and the perpendicular to it from (0, 0) is
y = X : therefore, ?m =
m y ^
Eliminating m between (1) and (2), we obtain
a-'
•'^"=27^-
2. At the centre of any circle C (Fig. 86), erect CH _L
to the diameter OA'; and on A'O produced layoff 0^4 =
OC^i'. Let LQii be a rectangular ruler of which the leg
QR equals 2r. If the ruler is moved so that the leg LQ
HIGHKIl PLANK CURVES.
219
passes through A, while the end li sli.Ies along C'll, ])rove
that the locus of P, the middle point of Qli, is a cissoid.
Fig. 8fi.
Let OA^be the axis of a^, 0 the origin, and AQK any
position of the ruler ; then OM^x, MP^=^y,
EQ = MN^ CM= X — r, A X= 2 0M= 2x.
EP^ = PQ" — JiJif = r' — (.T - ry = 2rx — x'\ (1 )
y = MP=NQ-\-EP.
NQ-.AN.-.EQ.EP,
or NQ : 2.r : : .r — r : \l2rx — x'
From (1), (2), and (,']), we obtain
\j2rx — .1
(-0
(3)
y
+ yj2rx-x%
or !f-
:^r — X
This method of deseribing tlie oissoid by continuous
motion was invented by Sir Isaac Newton.
220 ANALYTIC GEOMETRY.
3. In Fig. 79 prove that jVS and OJV are two mean pro-
portionals between OM and iV7 ; that is, prove
OM-.NS:: ON: NI.
NS'' = NXX 0N= OMX ON.
The right line 01 will pass through K; hence,
Z OJN^ Z VOI= ^ arc 0K= ^ arc SX
^ZSOX;
therefore, NS: ON:: ON: NI, etc.
4. If in the lemniscate (Fig. 82) 0F'=0F=^a^2,
prove that FF X F'F is constant, F being any point on the
curve; and hence that the lemniscate may be defined as
the locus of a point the product of whose distances from
two fixed points is constant.
5. Construct the logarithmic curve y^=a^, or x^log^y.
Prove that every logarithmic curve passes through the
point (0, 1), and has the axis of x as an asymptote.
6. The Trochoid is the curve traced by any point in the
radius of a circle rolling on a right line. If r denotes the
radius of the circle, h the distance of the generating point
from its centre, and Q denotes the same angle as in § 202,
show that the equations of the trochoid are
x = r6 — b sin $ > ,.y.
y=^r — b cos 6 )
When b<ir, the trochoid is called the Prolate Cycloid;
and when b> r, the Curtate Cycloid. When b = r, the
curve is the cycloid.
Spirals.
204. A Spiral is the locus of a point whose radius vector
continually increases, or continually decreases, while its
vectorial angle increases (or decreases) without limit.
HIGHER PLANK CURVES.
221
A Spire is the portion of the spiral traced during one
revolution of the radius vector.
The Measuring Circle is the circle whose centre is the
pole and whose radius is the value of p when 6^2Tr.
205. The Spiral of Archimedes. If the radius vector
of a moving point has a constant ratio to its vectorial
angle, that is, if
p = ce, ■ (1)
the locus is the Spiral of Archimedes.
From equation (1), we have
when^ = 0, Itt, |7r, |7r, ir, |7r, |7r, |7r, 27r, |7r,
p=0, \itC, lire, fTTC, ttC, ^ttC, &TrC, JttC, 2TrC, ^wC,
Hence, to construct the Spiral of Archimedes, draw the
radial lines OH', OA, OB, OG, including angles of ^n;
on these lay off Oa = J 7rr, Ob = |7rr, Oc = ^irc, , OH^ 27rc,
and trace the first spire OabcdefgH through these points.
Any number of other spires are easily constructed by noting
that the distance between two spires, measured on a radius
vector, is equal to 2irc. Thus, by taking a A. bB, cC, dD, eE,
fF, gG, HH', each equal to 2irc, we obtain points of the
second spire ; and so on. Any number of a(^ditional radial
lines may be drawn to locate ]ioints in tlie curve.
222
ANALYTIC GEOMETRY.
The spires, being everywhere equally distant along
radial lines, aie said to be parallel. The measuring circle
is JIMS, whose radius is OH or 2Trc.
206. The Reciprocal or Hyperbolic Spiral. If the radius
vector of a point varies inversely as its vectorial angle,
that is, if pd=^c, (1)
the locus is the Reciprocal Spiral.
Since p=^c^6, we have
when 6 = 0, lir, lir, ^tt, tt.
4 " ' 2 " ' 4"}
( V V V r, V V V
p = CC,i^^ > 4 ^, « — , J — , A — , 4 8
ZtT wTT wTT -iTT ^TT iiTT ZtT ZtT
Hence, the radius of the measuring circle is c-f-27r, and
its circumference is c.
Fig. 88.
To construct the curve, draw the radial lines OX, Oa, Ob,
Oc, Od, Oe, Of, ();/, including angles of Jtt. Take OH^ e
-^ Stt, and lay off Oa = 8 X OJf, Ob ^ 4 X OH, Or =f^X OH,
0(1 = 2 X OH, etc.; and through the points a, b, c, d, etc.,
trace the curve. In like manner, any number of spires may
be drawn. From (1) it is evident that p approaches zero,
as 0 approaches infinity ; that is, the curve continually
approaches the pole without ever reaching it.
Since pd = e, it follows that the arc FK described with
the radius vector of any point /'is constant and equal to c.
HIGH KM PLANK CUUVKS.
223
Now, as p appi'oaclies infinity, tiiis arc a,|)[n-();i(',lies a perpen-
dicular to OX. Hence, the line parallel to OX at the distance
of c above it is an as}- niptote to the infinite brancli of the
spiral.
207. The Lituus. If the square of the radius vector of a
point varies inversely as its vectorial angle, that is, if
irO = c,
the locus is the Lituus.
Fig. 80.
Let the student construct tlie curve from its equation
and show that
(i) The curve continually approaelies the pule without
ever reaching it, as 6 increases withoiit limit.
(ii) The polar axis is an asymptote to the infinite branch.
208. The Logarithmic Spiral. If tlie radius vector of
a point increases in a geometrical ratio, while its vectorial
angle increases in an arithmetical ratio ; that is, if
p = a^, or 6=:log„p, (1)
the locus is the Logarithmic SpiraL
Since p = l when 6 = 0, every logarithmic spiral passes
through the point (1, 0).
To construct a logarithmic s])iral, let r/ = 2; then p = 2".
When^ = 0, 1(=57.,r), 2(= 114.0°) '2^.
P = l, 2 ,4 _ , , 77.88.
224 ANALYTIC GEOMETRY.
In Fig. 90 let XOb = 57.3°, XOc = 114.6°, Oa = l,0b = 2,
Oc = 4:, then a, h, c are three points on the spiral. As 6
increases, p increases rapidly, but it becomes infinity only
when B does ; and hence only after an infinite number of
Fig. 90.
revolutions. As 6 decreases from zero, p decreases from unity.
Since p approaches zero as 9 approaches negative infinity,
the curve approaches the pole without ever reaching it.
209. The Parabolic Spiral. If in the equation if = ^)x,
the values of x are laid off from A (Fig. 91) on the circle
AH, and those of y on its corresponding radii produced, the
locus of the point thus determined is the Parabolic Spiral.
To find its equation, denote the radius OA by r, and let
P be any point ; then
X = AmH ^= rO.
y = HF=OF—OH=p — r.
HIGHER PLANE CURVES. 225
Siibstitutiug these values of x and // in y^ = 4^jx, we ob-
tain the polar equation
Fig. 91.
The curve consists of two branches beginning at A ; the
one determined by the positive values of y is an infinite
spiral lying entirely without the circle; the other branch
passes through the pole, forms a loop, and passes without
the circle when p = — /•, and ^ = ?• -|-^j.
Note. Among the ancients no problems were more celebrated than
the "Duplication of the Cube" and the "Trisection of an Angle."
Hii^pocrates of Chios reduced these two problems to the more general
problem of finding two mean proportionals between two given lines.
Thus, if c is the edge of the given cube, and x and y are the two mean
proportionals between c and 2c, we have
c : X = X : y = y : 2c.
Therefore, (~^ - - X - X ^^- = i, or x^ - 2c^
\x/ x y 2c
Hence, x, the first of the two mean proportionals between c and 2c,
is the edge of a cube double the given cube in volume.
After years of vain endeavor to solve these problems by the right
line and the circle, the ancient geometers began to invent and study
other curves, as the conies and some of the higher plane curves. The
invention of the conies is credited to Plato, in whose school their
properties were an object of special study.
PART 11. — SOLID GEOMETRY.
CHAPTER I.
THE POINT.
210. The position of a point in space may be determined
by referring it to three fixed })lanes meeting in a point.
The fixed planes are called Coordinate Planes, their lines
of intersection the Coordinate Axes, and their common
point tlie Origin. In what follows we shall employ coordi-
nate planes that intersect each other at right angles.
Let XOY, YOZ, ZOX, be three planes of indefinite ex-
tent intersecting? each other at risrht anerles in the lines
A'A', YY', ZZ'. These coordinate planes are called the
planes xy, yz, zx, respectively; the axes XX\ YY\ ZZ^ are
called the axes of x, y, z, respectively; and their common
point 0 is the origin.
TIIK POINT. 227
The coordinate i)laiies divide all space into eight portions,
called Octants, wliich are numbered as follows: The First
Octant is 0-XYZ, thi- Secomi, 0-YX'Z, the Third, 0-X'Y'Z,
the Fourth, 0-Y'A'Z,i\ni Fifth, 0-XYZ', the Sl.rth, 0- YX'Z',
the Seventh, 0-X'Y'Z', the Fiffhth, 0-Y'XZ'. The lifth
octant is below the first.
Let F be any point in space, and througli it pass three
planes parallel respectively to the three coordinate planes,
thus forming the rectangular parallelopiped F-OJtXM.
The position of F will be determined when we know tiie
lengths and directions of the lines LP, QP, NP. These
three lines are called the Rectangular Coordinates x, y, z,
of the point P, which is written {i-, ij, z).
A coordinate is jjositive when it has the direction of OX,
OY, or OZ; hence, it is ner/atlve when it has the direction
of OX', OY', or OZ'.
Thus the coordinate x is positive or negative according
as it extends to the right or to the left from the plane //.~;
1/ is positive or negative according as it extends to the front
or to the rear of the plane zx; and z, according as it ex-
tends upward or downward from the ]ilane x//. Hence, the
octant in which a point is situated is determined by the
signs of its coordinates. Since the first octant has the posi-
tive directions of the axes for its edges, tlie coordinates of
a point in the first octant are all i)Ositive. If (a, b, e) is a
point in the first octant, the corresponding point
in the second octant is ( — a, h, r),
in the third octant is ( — (t, — b, c),
in the fourth octant is (k, — b, r),
in the fifth octant is (a, b, — r),
in the sixth octant is ( — a, b, — <•),
in the seventh octant is ( — a, — b, — c),
in the eighth octant is {(i, — b, — c).
228 ANALYTIC GEOMETRY.
The point (x, y, 0) is in the plane xy.
The point (x, 0, 0) is in the axis of x.
The point (0, 0, 0) is the origin.
The lines OM, OR, OS, or OM, MN, NP, may be taken
as the coordinates of P, for they have the same length and
direction as LP, QP, NP, respectively. To construct P,
(x, y, z), we take OM^x, draw MN parallel to OY, take
MN=y, draw NP parallel to OZ, and take NP = z.
211. The Radius Vector of a point is the line drawn to
it from the origin. Thus, OP (=p) is the radius vector of P.
From the rectangular parallelopiped in Fig. 92, we have
op" = Oil' -f MN"^ + NP".
Hence, denoting the coordinates of P by x, y, z, we have
That is, the square of the radius vector of a point is equal
to the sum of the squares of its rectangular coordinates.
212. By the angle between two non-intersecting straight
lines is meant the angle between any two intersecting
straight lines that are parallel to them. Thus, any line
parallel to OP (Fig. 92) makes the angles XOP, YOP, ZOP
with the axes of x, y, z, respectively.
The angles which a line makes with the positive direc-
tions of the coordinate axes are called its Direction Angles ;
and the cosines of these angles are called the Direction
Cosines of the line.
The direction angles of a line are always positive and
cannot exceed ir, or 180°.
213. Let a, fS, y denote the direction angles of OP (Fig.
92), or any line parallel to it, and x, y, z the coordinates of
P, and let p = OP; then evidently
THE POINT. 229
ac= p cos a, y = p COS p, s = p cos y. [54]
Squaring and adding [54], and substituting in [53], we
obtain
cos2 a + cos2 p + cos2 7=1. [55]
That is, the sum of the squares of the direction cosines of
a line is equal to unity.
Cor. Whatever are the values of x, y, z in [54], if
each is divided by p, or Vx^ + 2/^ + «^ the quotients are the
direction cosines of the radius vector of the point {x, y, z).
Hence, if any three real quantities are each divided by the
square root of the sum of their squares, the qiwtients will be
the direction cosines of some line.
Exercise 42.
1. In what octants may (oc, y, z) be, when x is positive ?
when ic is negative ? when y is positive? when ?/ is negative ?
when z is positive ? when z is negative ?
2. Inwhatoctantis(— 2,4, 6)? (2,4,-3)? (—2,4,-1)?
(-2,-3,-1)? (-2,-3,3)? (2,-3,1)? (2,-1,-3)?
Construct each point.
3. In what line is {a, 0, 0) ? (0, 0, c) ? (0, b, 0) ?
4. In what plane is {a, b, 0) ? {a, 0, c) ? (0, b, c) ?
5. Find the length of the radius vector of (3, 4, 5),
(2, — 3, —1), (7, — 3, —5). Find the direction cosines of
the radius vector of each point.
6. The direction cosines of a line are proportional to
1, 2, 3 ; find their values. What is the direction of the line ?
7. What is the direction of the line whose direction
cosines are proportional to A, B, C? Wliat are the values
of its direction cosines ?
230 ANALYTIC GEOMETRY.
8. Two direction aiii;los of a lino aro C>0° and 45°, what
is tlie third ? If two are (U)° and 30°, wliat is tlie third?
If two are 135° and G0°, what is the third ?
214. Projections. The 2)roJectio7i of a point upon a right
line is the foot of the perpendicular from the point to the
line ; or it is the intersection of the line with the plane
throngh the point perpendicular to the line. Thus 31, JR., S
(Fig. 92) are the projections of the point P upon the axes
of X, y, z, respectively. Here and in the following pages,
by projection is meant the orthogonal projection.
The 2^^'ojection of a limited right line on another right
line is the part intercepted between the projections of its
extremities. Thus, 031, OB, OS (Fig. 92) are the projec-
tions of OF on the axes of x, y, z, respectively.
That is, the coordinates of anypnhU are the i^rojeetlons of
its radius vector o7i the three axes.
215. The projections of any line PQ on parallel lines are
equal ; for these projections are parallel lines included
between parallel planes through P and Q. Now tlie projec-
tion of any straight line on another that passes tlirough one
of its extremities is evidently equal to the })roduct of its
length into the cosine of their included angle.
Hence, the projection of any limited strait/ht live on any
other straiyht line is equal to its length nmdtiplied hy the
cosine of the angle between the lines.
216. Let AT) (Figs. 93, 94) be a straight line, and ABCB
any broken line in, space, connecting the points A and D, and
let A\ B', C", D' be the projections of A, B, C, D upon OX,
whose positive direction is OX. Denote the angle RADhy
<!>, the lengths of the lines ^47?, BC, CD by l^, l^, 4, and the
angles which they make with the positive direction of OX
by ai, tta, Us ; then in Fig. 93 we have
TIIK I'OINT.
231
A'jy = .17/' + 7;'c" + c'lf.
Therefore, AI>vms^=Ii cosai + /a cos o.^ + ^3 cosa^. [5(j]
111 Fig. 04, CD' is negative ; but ^3 cos a, is also negative,
since ug, or SCV, is obtuse 5 hence foriuuhi [oG] holds true
in all cases.
D ])
B' C D' '" " A' D' D'
Fig. 93. Fig. 94.
That is, the ulgebralc sum of the ijrnjertiDnx on a r/iren line
of the jifU'ts of (1)11/ broken line connecthuj aiii/ two points is
equal to the projection on the same line of the straiijht line
joining the same two points.
2,\1 . To find the angle between two straigJit lines in terms
of their direction cosines.
FiR. 95.
Let OP and OF' be ])aralh'l respectively to any two given
lines in space. Let 6 denote their included angle, and u. /?. y
232
ANALYTIC GEOMETRY.
and a, (3', y' their direction angles, respectively. Let OM,
MN, NP' be the coordinates of P' ; then the projection of
OP' on OP equals the sum of the projections of OM, MN,
NP' on OP ; that is,
OP' cos 6 = OM cos a + MN COS (i + NP' cos y.
But 0M= OP' cos a', MN=OP' cos ^', NP'=^ OP' cos y'.
Hence, OP' cos 6 = OP' cos a' cos a + OP' cos /3' cos (3
-\- OP' cos y' COS y,
or cos 9 = cos a cos a' + cos p cos P' + cos 7 cos y, [57]
which is the required formula.
' 218. To find the distance between two points in terms of
their coordinates.
Fig. 96.
Let Pi be the point (cci, y^, z^), and Pg the point {x^^y^, «2).
Through P^ and Pj pass planes parallel to the coordinate
planes, thus forming the rectangular parallelepiped whose
diagonal is P1P2, and whose edges PiL, LK^ KP^ are
parallel to the axes of x, y, z, respectively.
Then 1\P' = 1\V ^Tk' + KPI (1)
THE POINT.
233
Now P^L is the difference of the distances of P^ and Pg
from the plane yz, so that P^L=^x^ — x-^. For like reason,
LK= 1/2 — yi, and KP^ = ^2 — ^i- Hence, denoting the dis-
tance P1P2 by D, we have, by substituting in (1),
D = V(X2 - iCiY' + {»/2 - yif + (^2 - si)S [58]
which is the required formula.
Cor. 1. Since PiL, LK, KP^ are equal to the projections
of the line P1P2 on the coordinate axes, it follows, from
[58], that
The square of any line is equal to the sum of the squares
of its 2)roJections 011 the axes.
CoR. 2. If a, ^, y are the direction angles of the line
P1P2, we have
X2 — Xi^ D cos a, ?/2 — yi=:D cos /?, «2 — «i = -E> cos y.
219. Polar Coordinates. Let XOY be a fixed plane,
OX a fixed line in it, and OZ a perpendicular to it at the
fixed point 0. To P, any })oint in space, draw OP, and
through OP pass a plane perpendicular to A'OY, intersect-
ing the latter in OJV; then tlie distance OP and the angles
ZOP and WON determine the point P, and are called its
234 ANALYTIC GEOMETRY.
Polar Coordinates ; Ol', denoted b^^ p, is the Radius Vector ;
and tlie angles ZOP and MON, denoted by 6 and <^, re-
spectively, are the Vectorial Angles. The point F is
written (p, 6, cf>). <^ determines the plane ZOX, 6 determines
the line OF in that plane, and p locates P in OP.
CoR. If" XOY is a riglit angle, the rectangular coordi-
nates of P are OM, MX, XP. To express these in terms
of the polar coordinates of P, we have
X = OM = O^'^cos <^ = OP sin 6 cos 4>^P ^i^^ ^ ^''^^ ^■
y = MX= OX&\\ifi>=^ OP sin ^ sin <^ = /J sin ^ sin ^.
z = XP^ OP cose=p cos 9.
We readily obtain also
tan Q = !-^ ^ tan <i>=~-
z X
220. The Projection of a p(jint on a plane is the foot of
the perpendicular from the point to the plane. The per-
pendicular itself is the Projector of the point. Thus, the
point iV(Fig. 97) is the i)rojection of P on the plane xy,
and PTVis its projector.
The jjt'ojection of a limited straight line on a jilane is the
straight line joining the projections of its extremities. The
Inclination of a line to a plane is the angle it makes with
its projection on that plane. The projection of a limited
line is evidently equal to its length multiplied by the
cosine of its inclination. Thus, 0X= OP cos XOP.
The p7-ojectio7i of any curve upon a plane is tlie locus of
the projections of all its points. The Projecting Cylinder
of a curve is tlie locus of the projectors of all its ])oints.
In the case of a right line this locus is the Projecting Plane.
THK POINT. 2.35
Ex^cise 43.
1. Find tlie distance between the point.s (1 , 2, 3), (2, 3, 4);
(2, 3, 4), (3, 4, 5); (1, 2, 3), (3, 4, 5).
2. Prove that tlie triangle formed by joining tlie three
points (1, 2, 3), (2, 3, 1), (3, 1, 2) is equilateral.
3. The lengths of the projections of a line on the three
coordinate axes are 3, 4, />, respectively; fin<l the length of
the line.
4. Find the direction cosines of the radius vector of
the point ( — 3, — 4, 5).
5. Wliat lines have direction cosines proportional to 3,
— 2, — 5 ? Find the values of these direction cosines.
6. Find the angle between two straight lines whose di-
rection cosines are proportional to 1, 2, 3 and 2, 3, 6, respec-
tively.
7. Find the angle between two straight lines whose di-
rection cosines are proportional to 1, 2, 3 and 5, — 4, 1,
respectively.
8. Find the polar coordinates of the point ( V3, 1, 2 V3).
9. Find the rectangular coordinates of (4, ^tt, -^tt).
10. \i (.r,?/,.~)bisects tlie line joining (^'i,yi,r:i)to(.r^, t/.i. :^o),
prove that x = ^(xy + .r^), y = hiiu + Z/a), '- = i(-i + ■-2)-
11. If (x, 1/, z) divides the line joining (xy, //,, Zi) to
(x^, 1/2: ^2)1 ii^ the ratio viy : vi„, prove that
irioXy -}- niyXo ?/?oVi + viyf/n nu~i -\- miS^
X = — = — . -^ y = —=^ 1 ^5 ,- = — — ,
12. Find the coordinates of the point that divides the
line joining (3, —2, 4) and (1, 3, — 2) in the ratio 1 : 3.
13. Find the point that divides the line joining ( — 2,
— 3,-1) and (—5, —2, 4) in the ratio 0 : 2.
CHAPTER II.
THE PLANE.
221. To find the equation of a plane in terms of the length
of the perpendicular from the origin, and its direction cosines.
Fig. 98.
Let OF be the perpendicular to the phme ABC from the
origin 0; denote its length by^, and its direction angles by
a, )8, y. Let P be any point in the plane, OP its radius vec-
tor, and OM, MN, NP its coordinates, x, y, z. Then the
projection of OP on OF is equal to the sum of the projec-
tions of OM, MN, NP on OF. But as the plane is perpen-
dicular to OF, p is the projection of OP on OF; and the
projections of OM, MN, NP on OF are respectively x cos a,
y cos (B, z cos y ; hence,
j7cosa + 2/cosP + scos-y = 2>> [59]
THE PLANE. 237
which is the equation required. Equation [59] is called
the normal equation of a plane.
Cor. 1. When the plane is perpendicular to one of the
coordinate planes, the plane xyiov example, Oi^lies in the
plane xy ; hence, y = •j-ir, cos y =0, and equation [59] becomes
X co^ a-\- y CO?, [i^p. (1)
CoR. 2. When the plane is parallel to one of the coordi-
nate planes, as the plane yz, Oi^lies in the axis of x ; hence,
cos a = l, cos ^ = 0, cos y==0, and [59] becomes
x=p. (2)
CoR. 3. Since OF is perpendicular to the plane ABC,
and OX to YOZ, the dihedral angle J-i?C-0 = angle FOX.
For like reason, B-CA-0 = FOY, and C-BA-0 = FOZ.
222. The lociis of every equation of the first degree between
three variables is a plane.
A general form embracing every equation of the first
degree between x, y, z is
Ax + By+Cz = D, (1)
in which D is positive.
Dividing both members of (1) by V--1" + B^ + C'\
we obtain
A , B , C
'J-
= , ^ , (2)
in which the coefficients of x, y, z are the direction cosines
of some line (§ 213, Cor.). Thus (2) is in the form of [59]
§ 221 ; hence, the locus of (2), or (1), is a plane.
CoR. 1. The length of the perpendicular from the origin
upon plane (1) equals the second member of equation (2), and
238 ANALYTIC GEOMETRY.
the direction cosines of tliis pfrpcndicular are the coeflficients
in (2) of X, I/, z, resi^ectively. These direction cosines are
evidently proportional to A, B, C.
Hence, to construct equation (1), draw tlie radius vector of
the point {A, B, C) ; the plane perpendicular to this line at
the distance , from theoritrin is the locus of (1 ).
Cou. 2. To reduce any simple equation to tlie ttur/nal
form, put it in the form of A.r + Jli/-\- Cz = I), in which U is
positive, and divide both niembers by '\/A^-\- J J'^-\- €''■'.
CoR. 3. If a simple equation contains only two variables,
its locus is perpendicular to the corresponding^ coordinate
plane ; if only one variable, its locus is perpendicular to
the corresponding coordinate axis.
223. To find tlie e(/'Ucftlo)i of a plane in terms of its inter-
cepts on the axes.
Let a, b, c denote respectively the intercepts on the axes
of the plane whose equation is
Ax-\-n!/+Cz = B. (1)
Making ?/ = ,^•=:0, and therefore x^^a, (1) becomes
Aa = IJ, or A=: D-T- a.
Making x = z^^ 0, and therefore // = b, (1) becomes
m=D, or B^D-^h.
Making a; = //:= 0, and therefore ,t' = (", (1) becomes
Cc — I>, or C—D^c.
Substituting these values in (1) and dividing by Z>, we have
^+^^+5=1, [60]
which is the required equation. Equation [60] is called
the sijniinetrical equation of a plane.
THK PLANK. 239
224. To find the (nKjJe hetween any two planes.
The angle included lietween the two planes
Ax-\rB'y^-Cz = D\
Ax-\-By+ Cz = D,
is evidently equal to the angle included between the perpen-
diculars to them from the origin. l>ut the direction cosines
of these perpendiculars are respectively (§ 222, Cor. 1),
ABC
ylA' + B- + V' V.i- + B"" -[- C ^/A' + B' + C'
A' B' C"
Substituting these values in [;">"], we have
AA'^BIi'^CC
cos 0 =
[Gl]
in which 6 erpials tlie angle included between the planes.
Cor. 1. If the planes are iiaralld to each other, 6^0
and cos ^ = 1. Putting cos 6=^1 in [<)1], clearing of frac-
tions, squaring, transposing, and uniting, we obtain
{AB' - BA 'f + {A a — CA')' -\-(BC- CB'f = 0.
Each term being a square, and therefore ])ositivt\ this
equation can be satisfied only when each term equals zero,
giving us
AB'=BA, AC"= CA', BC'= CB\
^_B^_C
^^ A~ B'~ C'
Hence, if two planes are parallel, the coefficients of x, ij, z,
in their equations, are proportional, and eonversely.
Cor. 2. If the planes are 2)erpendiciilar to each other,
cos ^ = 0 ; and hence,
AA'-\-BB'+CC' = 0,
and converseli/.
240 ANALYTIC GEOMKTKY.
225. To find the perpendlcidar distance of a given 'point
from a given pUnie.
Let the equation of the given plane be
X cos a.-\-y cos /8 + « cos y =-p, (1)
and let (cci, y^i ^-^ be the given point. Let the plane
X cos a 4" ,'/ cos y8 + 2; cos y =^:»', (2)
which is evidently parallel to the given plane, pass through
the given point (a^i, ?/i, z^ ; then we have
a-i cos a + 2/i cos ^-\-z^ cos y = js'. (3)
Hence, x^ cos a + yi cos ^^z-^ cos y — p ^=p^ — p.
But ^9' — ^9 equals numerically the distance between the
planes (1) and (2), and is therefore the required distance.
Hence, to find the distance of any point from the plane
x cos a + y cos ^ + « cos y =_p,
substitute the coordinates of the point for x, y, z in the
expression
X cos a-\- y cos /3 + 2: cos y — p.
CoR. If the equation of the plane is Ax-\- By-\- Cz=^D, and
d denotes the distance of (xi, y^, z-^) from this plane, we have
Ja-i + By^ -\-Cz, — D
The distance as given by the formulas will evidently be
pjositive or negative, according as the point and origin are on
opposite sides of the plane, or on the same side. The sign
maybe neglected if simply the numerical distance is required.
Exercise 44.
1. To which coordinate plane is 3y — 4,? = 2 perpendicular?
X — 8-? — 7 = 0? X — 2y = 2? x = mz-\-p? y^nx-\-q?
What is the locus of .- = 5? y = — 7? y = 4:? z = — 2?
x = 0? y = 0? z = 0 ?
THE PLANE. 241
2. Reduce to the normal form
3x — 2y + ;2 = 2; 5iK — 4 // + .- = 4.
Wliat is the distance of each of these planes from the
origin? What are tlie direction cosines of the perpendicu-
lars to each ? Wliich of the eight octants does each truncate?
3. Find the intercepts on the axes of 3x — 2?/ + 4,^ — 12
= 0; of 6x — 4.>/ — 3z + 24: = 0; of 5a; + 7?/ + S.v + 35 = 0.
Which of the eight octants does each truncate? Reduce
each equation to the symmetrical form.
4. What is the equation of the plane at the distance 7
from the origin, and perpendicular to the line whose direc-
tion cosines are proportional to 2, — 3, and V3 ?
5. What is the equation of the plane whose intercojjts
on the axes are respectively 4, — 3, — 7? — 1, — 2, — 5?
6. Find the equation of the plane passing through the
points (1, 2, 3), (0, 4, —1), and (1, —1, 0).
7. Find the angle between the planes
2x^z-,j = 3,
z-\-x-{-2i/ = 5.
8. Find the angle between the planes
3z + 5x — 7>/ = — l,
3z — 2x— y = 0.
9. Find the angle that the plane Ax + Bij -\- Cz = D
makes with each of the coordinate planes.
10. Find the distance from (2, —3, 0) to the plane
V3;? + 2a; — 3^ = 4.
11. Show that the two points (1, — 1,3) and (3, 3, 3) are
on opposite sides of, and equidistant from, the plane
5x + 2>j — lz^Si^0.
242 ANALYTIC GEOMETRY.
12. If, in Fig. 92, OM=a, OE = b, OS=r, find the
equation of the phine through the points 31, 1', A*. Find
the length of the perpendicular from S ujwu thi.s plane.
13. Prove that the plane
A(x - X,) + B(i/ - y,) + Ciz - ,.0 = 0
passes through the point (a-i, t/^, z-^, and is parallel to the
plane Ax -f- By -\- Cz = D.
14. Find the equation of the plane passing through the
point (3, 4, — 1), and parallel to the plane 2x + 4y — ,^ = 2.
15. What three equations must be satisfied in order that
the plane Ax -\- By -\- Cz = D may pass through the two
points (xi, yi, z-^), (x^, y^, z^), and be perpendicular to the plane
A'x + B'y-{-C'z = I>'?
16. Find the equation of tlie plane passing througli the
points (1, 1, 1), (2, 0, — 1), and perpendicular to the plane
x-\-y — z = 3.
17. What three equations must be satisfied in order
that the plane Ax -f- By + Cz = D may pass through the
three points (x^, v/i, z^), (x., y^, z^), (x^, y^, Zs)?
18. Find the equation of the plane that passes through
the points (1, 2, 3), (3, 2, 1), (2, 3, 1), and find the distance
of this plane from the origin.
19. Find the equation of the jdane through (2, 3, — 1)
parallel to the plane 3x — 4// -|- 7z = 0.
20. Find the equation of the plane that passes through
the point (1,2,3), and is perpendicular . to each of the
planes a; + 2* = 1, y -\- 5z = 1.
CHAPTER III.
THE STRAIGHT LINE.
226. To find the equations of a straight line.
The coordinates of any point on the line of intersection
of two planes will satisfy the equation of each of these
planes. Hence, any two simultaneous equations of the first
degree in a-, y, and ,-.- represent some straight line. Of the
Fig. 99.
indefinite number of pairs of planes that intersect in, and
therefore determine, a straight line, the ecpiations of its
projecting planes on the coordinate i)lanes are the simplest,
and two of them are taken as the etpiations of the line.
Thus, let FPl'21l and W US be the projecting planes of any
244 ANALYTIC GEOMETRY.
straight line PP' on the coordinate planes xz and yz, respec-
tively; and let the equations of these projecting planes be
X = mz -\-p, (1)
y = nz-\-q; (2)
then are (1) and (2) the equations of the line PP'.
Cor. 1. Let RE and SH be the projections of the line
PP' on the planes xz and yz, respectively. Since the line
RE lies in the plane PP'RE, equation (1) expresses the
relation between the coordinates x, z of every point in RE \
hence, (1) is the equation of RE referred to the axes ZZ'
and OX. For like reason, (2) is the equation of the pro-
jection SH referred to the axes ZZ' and OY.
Hence, rti = tan ZA E = slope RE ;
■p = OP = intercept of RE on 0X\
TO = tan ZBff^ slo^e SH;
q = 0S=^ intercept of SH on 0 Y.
Eem. The locus of (1) in space is the plane PP'ER,
while its locus in the plane xz is the line RE. Similarly,
the locus of (2) in space is the plane PP'HS, while its
plane locus is SH. The locus in space of (1) and (2), con-
si9.ered as simultaneous, is the line PP'.
CoR. 2. Eliminating z between (1) and (2), we obtain
n ( np
y=i — x-\-\ q
•^ m \^ VI
whose locus in space is the projecting plane of PP' on the
plane xy, and whose locus in the plane xy is the projection
of PP' on that plane.
CoR. 3. Making ?,' = 0 in equations (1) and (2), we obtain
x=p, y=q\
hence, the line PP' pierces the plane xy in the point
(/>, q, 0). This is evident also from the figure.
THE STRAIGHT LINK. 245
In like manner, we find tliat the line })ierce.s the planes
xz and yz respectively in the points
np — mq 0, _ 1 Y ({), ^g ~ '^P^ _ 21
n ^^/ \ ^ ^"'
227. To find the symrtietrical equatioiis of a rujltt line.
Let a, (3, y be the direction angles of any right line,
(^15 l/u ^i) some fixed point in it, and (x, y, z) any other
point of the line. Let r denote the distance between these
two points. Then by § 218, Cor. 2, we have
X — Xi=^r cos a, y — yi = r cos (3, z — s^ ^ ?■ cos y. (1)
TTtn 2C — Xl y—y\ Z — Zl rn.,-,
Whence, - = '^ — ^ = > rCol
cos a COSp COSY "■ -*
which are the symmetrical equations of a right line pass-
ing through the point (xi, y^, z^.
Cor. If [63] passes through a second point (x.^, y,, '-'2);
its coordinates must satisfy [63] ; hence, we have
^2 ^1 1/2 Vl ~2 ^l ,n-.
cos a cos /? cos y ^ ^
Dividing each member of [63] by the corresponding
member of (2), we obtain
xz~x\ 2/2-2/1 zt—zx L -•
which ai"e the equations of a right line through the two
points (xi, ?/i, Si) and (a-g, y^i z^).
228. If we divide the denominators in any equations of
the form
x — xi^y — yi^z — zi
L M N ^^
by \l L^-\- ]\P-{- N'^, the denominators will then be the
direction cosines of some line (§ 213, Cur.), and the equa-
tions will be in the form of [Qo^-
246 ANALYTIC GEOMETKY.
Hence, to rcdvice equations in the form of (1) to the sym-
metrical form, divide each denominator hij the square root of
the sutn of the squares of the denominators.
Cor. The locus of equations (1) is the line through
(xi,7ji,z{) parallel to the radius vector of the point (L, M, JV).
229. To find the diKjle between the lines
L ^ M ^ N '
and — ^ — ^- — -^■
L' M' N'
By § 228 the direction cosines of these lines are respectively
L M N
L' M' N'
^L'-'-^-lI'^+N"" \/L"'-\-M"'-\-N'^ \/Z'2+J/'2+^'2
Substituting these values in [57], we obtain
„ LL'+ MM'+ NN< ^^^^
cos e = , --' , [651
Cor. 1. If the lines are parallel, —=-—=—, and con-
versely.
Cor. 2. If the lines are perpendicular,
LL' -f MM< + NN' = 0,
and conversely.
230. To find the inclination of the line
x — Xi^y — ih^z — Zy_
L M N ^ >
to the plane Ax -f- Bij -j- Cz =^ D. (2)
THE STRAIGHT LINE, 247
The equation of the perpeiidiciilar from (x^, iji, z-^) to the
plane IS a: — a^i _ y — ?/i _z — z^
A ~ B ~ C ' ^'^^
Now, the inclination of line (1) to plane (2) is evidently
the complement of the angle between the lines (1) and (3).
Denote this inclination by v; then sin t> = cos ^, 0 being
the angle between the lines (1) and (3). Hence,
AL + BM+ C]>f^ ^,.^^
CoR. 1. If the line is parallel to the plane, sin v ^ 0,
and, therefore, AL-\- BM-\- C'i\^=0, and conversely.
Cob. 2. If the line is perpendicular to the plane, sin v =: 1,
and, therefore, ^ = f=f, and conversely.
Cor. 3. If line (1) lies in plane (2), then
AL-{-BM-\-CN=Q,
and Ax^ -\- By^ + Cz^ = D,
and conversely.
Exercise 45.
1. Determine the position, direction cosines, and direc-
tion angles of the intersection of the planes x-{- y — z + 1
= 0, and4a; + y/ — 2,~ + 2 = 0.
Eliminating successively y and z between the equations,
X 1/ z — 1
we obtain ox — .~+l =0 and 2x — '/ = 0: or 7 = !. = — r; —
1 -J o
From the last form we know that the line passes through
the point (0, 0, 1), and is parallel to the radius vector of the
point (1, 2, 3). The direction cosines are found by divid-
ing the denominators 1, 2, 3, by Vl4; and the direction
ancrles are found from their cosines.
248 ANALYTIC GEOMETRY.
2. Determine the position and direction cosines of the
intersection of x — 2y = 5 and 3x + ?/ — 7z = 0.
Here — - — = '- = " ^ , whence the line passes through
the point (5, 0, -y), and is parallel to the radius vector of
(2, 1, 1).
3. Determine the position of the line
5a; — 4?/ = 1, '6y — bz = 2.
4. Wliat is the position of the line a; = 3, 2/^=4? Of
the line y = 4:, z = — 5? Of the line x = — 2, z = 3?
5. Find the equations of the right line passing through
the points (1, 2, 3), (3, 4, 1).
6. Find the points in which the line of Example 5
pierces the coordinate planes.
7. Two of the projecting planes of a line are x-\-?/ = 'i
and 2x — 5z = — 2 ; find the third.
8. A line passes through (2, 1, —1) and (—3, —1, 1);
find the equations of its projections on the coordinate planes.
9. Show that the lines t=9=t and 3-=^^= :|^ are
at right angles.
10. Show that the line4a; = 3y = — z is perpendicular
to the line 3x= — ?/ = — 4«.
11. Find the angle between the lines
X ?/ z X ?/ z
12. Find the angle between the right lines
y:=OX-\-3, Z:=3x-\-5,
and y^=2x, z = x-\-l.
13. Find the angle between the lines
y = 2.r + 2, z = 2x + l,
and y^4:X-\-l, z= x-\-5.
THE STRAIGHT LINE. 249
14. Show that the lines
3x + 2i/-{-z — 5 = 0,x + y—2z — 'S = 0,
and 8x — 4i/ — 4:z = 0,7x-\- 10// — 8z = 0
are at right angles.
15. Find the equations of the line through ( — 2, 3, — 1)
parallel to the line y=^ — 2x + 1, z = 3x — 4:.
16. Find the equations of the line through (3, — 7, — 5),
its direction cosines being proportional to — 3, o, — 6.
17. Find the equations of the line through (2, — 4, — 6)
perpendicular to the plane 3x — 6i/-\-2z = 4:.
18. Find the inclination of the line
x — 4 ?/ + 2 z — 5
3 ~ — 2 ~ — 4
to the plane 2x — Ay-\-3z^=l.
19. Reduce the equations, x = mz -\-p, y= nz -\- q, to the
symmetrical form, and thus find the direction cosines in
terms of m and ?i.
20. Show that the formula for the angle included between
the lines
x = mz +p, y = nz -\- q,
and 2"= vi'x -\-p', y = 7i'z -\- q'
mm' -\- nn' + 1
IS cos 6 = — = — ■ •
Vm'+?i2_^l Vm'^ + ?i'2 + l
21. Prove that the lines in Example 20 are perpendicular
if vim' -\- nn' -\- 1 = 0, and conversely. Prove that they are
parallel if vi = m', and n = n', and conversely.
22. Prove that two lines are parallel if their projections
are parallel, and conversely.
250 ANALYTIC GEOMETRY.
SUri'LEMENTAllY PROrOSlTlONS.
231. The Traces of a plane are its lines of intersection
with the coordinate planes. Thus AB, JW, CA (Fig. 98)
are the traces of the plane ABC.
232. To find the equations of the traces of the ylane
Ax-\-By+Cz = D. (1)
For every point in the plane xy, z^O ; hence, putting
« = 0 in (1), we obtain
Ax^Bu = D, (2)
whicli is tlie equation of the trace AB (Fig. 98) on the
plane xy. For like reason,
By-^Cz=D - (3)
and Ax + Cz = D (4)
are the equations of the traces BC and CA on the planes
yz and xz, respectively.
CoR. The perpendicular from the origin to (1) is
x // z
A~B^C'
and its projections on the coordinate planes are
Bx = Ay, Cy=Bz, Cx = Az. (5)
By comparing coefficients, we see that lines (5) are per-
pendicular to (2), (3), and (4), respectively. Hence,
If a line in sjxice is 2)(ir2Jendicular to a 2)lane, its projections
are jierpendicular to the traces of the plane.
233. To find the conditioii that the right lines
X = niz +;y, y — nz + q,
and X = m'z + /'', y = n'z -\- q',
may intersect, and to find their points of intersection.
THE STRAIGHT LINE. 251
Equating the two values of x and y, we have
jj' — p q' — q
z = -) z=- ■•
7ft — nv n — 71
If the two lines intersect, these two values of z must be
p' — /' 7' — Q • •
equal ; hence, ^ = -. is the equation of condition
m — 111 n — 7i'
that the two given lines in si)ace intersect.
When this condition is fultilled, the values of x and /j
may be found by substituting either value of z in the
equations of either line.
234. To pass a plane through the point (.Tj, ?/o, z^) and
the right line
L ~ 31 ~ N ' ^"^
If the plane Ax + By + Cz = B (1)
passes through the point (.Tg, ?/„, .-o), we have
Ax^^By.,^Cz. = D; (2)
and if line (a) lies in plane (1), we have
Ax^^By,-^Cz,=D, (3)
and AL^BM\CN=^. (4)
The equation of the required plane is found by eliminat-
ing A, B, C, D from (1), (2), (3), and (4).
To simplify the process of elimination, (1) might be written
in the form A'x-\- B'y-\- C'z = 1, but the solution would be
less general, as it would not embrace the case when Z> ^ 0.
235. From the forms x = mz -{-j), y = nz-\-q, show that
the equations of a line passing through (xi, y^, z^ are
\ .r — .r, == m (z — z^),
CHAPTER IV.
SURFACES OF REVOLUTION.
236. It has been shown that a single equation of the
first degree between three variables represents a plane
surface, and that two such equations in general represent a
right line. It is evident, moreover, that in general three
such equations determine a point common to their loci.
Thus, if in Fig. 92, OM=a., OE = b, and OS=c, then the
equations x = a, y=^h, z^c determine the point P, and are
called its equations. We proceed to show that.
In general, any single equation of the form f (x. y, s ) = 0
represents a surface of some kind; two such equations represent
a curve, and three determine one or more pohits.
(i) Let two of the variables be absent ; for example, let
the equations be / (x) = 0. Now / (a;) = 0 may be written
in the form
{x — ai) (a- — a^) {x — a^) (x — «.„) =0, (1)
in which a^, Oo, a^, «„ are the n roots of/(a;)=;0. The
locus of (1) is evidently the n parallel planes a? = «!, a; = a^,
, x=-a^. Similarly, the equations/ (?/) =0,/(s)=:0 repre-
sent planes perpendicular to the axes of y and z, respectively,
(ii) Let one of the variables be absent ; for example, let
the equation be / {x, y) = 0. The locus of /(re, y')=0 in the
plane xy is some plane curve. Through P, any point in this
curve, conceive a line parallel to the axis of z ; then the
coordinates x, y of all points in this line will equal those of
P, and hence satisfy the equation / (x, y) = 0. Hence, the
SURFACES OF REVOLUTION. 253
locus in space of / (x, y) = 0 is the surface generated by a
right line which is always parallel to the axis of z, and
which moves along the plane locus oif(x, y) =0.
That is, the locus in space of f (x, y) = Q is a cylindrical
surface whose elements are parallel to the axis of z, and whose
directrix is the pjlane locus off (x, y) = 0.
Similarly, the equations / (x, z) = 0 and / (y, z) = 0 rep-
resent cylindrical surfaces whose elements are parallel to
the axes of y and x, respectively.
(iii) Let the equation be / (x, y, z) = 0. If in this equa-
tion we putx = a and y=^b, the roots of the resulting
equation in z will give the points in the locus that lie on
the line through (a, b, 0) parallel to the axis of z. But as
the number of these roots is finite, the number of points
of the locus on this line is finite. Hence, the locus which
embraces all such points for different values of a and b
must be a surface and not a solid.
(iv) Two equations considered as simultaneous are satis-
fied by the coordinates of all the points of intersection of
their loci ; that is, they represent the curve of intersection
of two surfaces.
(v) Three independent simultaneous eqiiations are satis-
tied only by the coordinates of the points in which the curve
represented by two of them cuts the surface represented by
the third ; hence they determine these points.
Cor. 1. From (ii) it follows that x"^ -\- if = i^ is the equa-
tion of a cylinder whose axis is the axis of z, and whose
radius is r. Also,
f = ^p:c,~^- = l, and -.-^=1
are equations of cylindrical surfaces whose elements are
parallel to the axis of ,~, and whose directrices are respec-
tively the parabola, the ellipse, and the hyperbola.
254 ANALYTIC GEOMETRY.
CoK. 2. I (■ F (x, i/)=0 is the equation obtained by
eliminating z between the two equations / (a-, //, s) = 0 and
/i ('^'j l/> ^) = ^j then the locus in space of F (.r, y) = 0
is the projecting cylinder on the plane xi/ of the curve
represented by the two equations. The j/lane locus of
F(_x, y):=0 is the projection of this curve on tlie plane xi/.
If the curve is parallel to the plane of jn-ojection, the curve
and its projection are equal.
The equation obtained by eliminating x ov y between the
two equations has evidently a like interpretation.
237. The Traces of a Surface are its intersections with
the coordinate planes.
If/(.^;, 11, 0)=0 denotes the equation obtained by mak-
ing z = 0 in f(x, y, z) = 0, then the plane locus of / (x, y, 0)
= 0 is evidently tlie trace of the surface f (x, y, z) =0 on
the plane xy.
Surfaces of Revolution.
238. A Surface of Revolution is a surface that may be gen-
erated by a curve revolving al)out a fixed straight line as an
axis. The revolving curve is called the Generatrix ; and
the fixed right line, the Axis of Revolution, or simply the
Axis, A section of the surface made by a plane passing
through the axis is called a Meridian Section. From these
definitions, it follows that
(i) Every section made by a plane perpendicular to the
axis is a circle, whose centre is in the axis.
(ii) Any meridian section is equal to the generatrix.
239. To find the general equation of a surface of revolution.
Let the axis of z be the axis of revolution, and let P be any
point in the meridian section made by the plane xz. Let PHR
be a section through P jterpendicular to the axis of z, and
denote the radius CH, or CP, of this circular section by r.
SURFACES OF UEVOLUTION.
25/
Now for all points in this circular section, we have »•- +
y2=r /-^^ and z = MP. The value of r^, in terms of •:, is ob-
tained by substituting r for cc in the equation of the me-
ridian section made by the plane zx. Denoting this value
of r* \iy f(z), and equating tlie two values of r^, we have
Qc'^+y^ = f{z), [G7]
Fig. 100.
which expresses the relation between the coordinates cr, y, z
of all points in the section PHB. But as P is any point
in the meridian section NP, [67] is the general equation
of a surface of revolution whose axis is the axis of z.
240. Paraboloid of Revolution. A Poraholnid of Perolu-
tion is a surface that may be generated by a parabola
revolving about its axis.
In this case the equation of the meridian section in the
plane zx is a;^ = 4/>;s ;
hence,
^^ = 4.pz=f(z).
256 ANALYTIC GEOMETRY.
Substituting in [07], we obtain
X"- + !/- = 4rpz, [68]
which is the equation of the j^ctt'cil^oloid of revolution.
If in [08] we put x = m, we obtain
which is the equation of the projection, on the plane yz, of
the section of the paraboloid made by a plane parallel to
the plane yz, and at a distance from it equal to m. Now
the plane locus of (1), for all values of in, is a parabola ;
hence, every plane section of the paraboloid parallel to the
plane yz is a parabola. If in [68] we put y^n,we obtain
x^ = 4pz — n^. (2)
From (2) we learn that all plane sections parallel to the
plane xz are also parabolas. From definition, we know
that all plane sections parallel to the plane xy are circles.
241. Ellipsoid of Revolution. Xn Ellipsoid of Revolu-
tion, or Spheroid, is a surface tliat may be generated by an
ellipse revolving about one of its axes. It is called Oblate
when the revolution is about the minor axis; and Prolate
when about the major axis.
(i) "When the revolution is about the minor axis, the
equation of the meridian section in tlie plane xz is
7^2 + 7i=l5 lience, r-'^aH l--^ =/(^)-
Substituting in [07], and reducing, we obtain
S + S + g=l. [69]
which is the equation of the ohlate spheroid.
Cor. 1. If a = h, [69] becomes
x' + lf + ^^a:^, (1)
which is the equation of a sphere whose radius is a.
SURFACES OF REVOLUTION. 257
Cor. 2. If in [69] we put x^=in, we obtain
^4---!- — . (2-)
Since (2) represents an ellipse, a point, or no locus in the
plane yz, according as w? <C, =, or > a^, the surface lies
between the two tangent planes a; = « and a; = — a, and all
plane sections parallel to the plane yz are ellipses.
. If in [69] we put y = n, we obtain
Since (3) represents an ellipse, a point, or no locus in the
plane «cc, according as v? <, =, or !> a^, the surface lies
between the two tangent planes y^a and y = — a, and all
plane sections parallel to the plane zx are ellipses.
If in [69] we put z = q, we obtain
x-^-\-f = a'^(l-^j\ (4)
Equation (4) represents a circle, a point, or no locus in
the plane xy, according as ff <, =, or > h"^. Hence, the sur-
face lies between the tangent planes z = b and z^ — b, and
all plane sections parallel to the plane xy are circles.
(ii) When the revolution is about the major axis, the
equation of the meridian section in the plane xz is
^2+1=1; hence, ,-=^^(^l-^,^=/<.).
Substituting in [67], we obtain
f5+|.+ f,= l. [70]
which is the equation of the ^jrolate spheroid.
If in [69] we interchange a and b, we obtain [70].
Hence, we interchange a and b in the discussion of [69].
258 ANALYTIC GEOMETRY.
242. Hyperboloid of Revolution. An Htjperholokl of
Reiwlutkm is a suriace that ina,y be generated by an hyper-
bola revolving about one of its axes. It consists of one or
two nappes, or sheets, according as the hyperbola revolves
about its conjugate or transverse axis.
(i) If in [69] we substitute — V^ for h"^, we obtain
^+'4-f^=l, [71]
a^ a- b^ L J
which is the equation of the hijperholoid of one nappe.
If in [71] we put x^=vi, we have
2 9 9
whose plane locus is an hyperbola for all values of m.
Hence, all plane sections parallel to the plane yz are hyper-
bolas. The transverse axis of any one of these hyperbolas
is evidently parallel to the axis of y or z, according as rn^
■< or > a^. If m'^ = a- (1) becomes
a
Hence, the sections of [71] made by the planes a; = ±a
are each two intersecting riglit lines.
If in [71] we put .y = ?', we have
X z ir
Hence, all plane sections of [71] parallel to the plane xz
are hyperbolas, whose transverse axes are parallel to the
axis of X or z, according as «-<[ or > a^- and the sections
made by the planes i/=±a are eacli two intersecting right
lines.
If in [71] we put z = q, we obtain
x^ y^_ rf
a^ a^ If
SURFACES OF REVOLUTION. 259
whose plaue locus is a circle for all values of «/. This cir-
cle is smallest when y ^ 0. This smallest circle, which is
the trace of the liyperboloid on the plane a"//, is called the
Circle of the Gorge.
(ii) If in [70] we substitute —V^ for V^, we obtain
f?+f?-4=-l, [T2J
h- b' a^ '- -■
which is the equation of the hyperholoid of two nappes.
The discussion of [72] for parallel plane sections is left
as an exercise for the student.
243. The Centre of a surface is a point that bisects all
chords passing through it.
Central Surfaces are such as have a centre.
The ellipsoids and hyperboloids of revolution are central
surfaces. For, from their equations, it is evident that, if
{x\ y', z') is a point in any one of these surfaces, ( — x', — y',
— z') is also a point in the same surface. But the chord
joining these two points is bisected by the origin, which is,
therefore, the centre of the surface.
244. Cone of Revolution. A Cone of Revolution is a
surface that may be generated by a right line revolving
about an axis which it intersects.
Here the equation of the meridian section in plane xz is
z = mx -\- c ;
therefore, ?-^ = ( j =f(z).
Whence, in-{x- + */-) - {z- <■)- [73]
is the equation of the cone of rerolufiun.
260
ANALYTIC GEOMETRY.
In this equation c is the distance of the vertex from the
origin and m = tan XDB.
Fig. 101.
If c = 0, [73] becomes
im?{oi? -\- ]f^ = z^. (1)
From (1) it is evident that the cone is a central surface.
If in (1) we put y = n, we obtain
z^ 3?
ri?if)i} IT? '
whose plane locus is an hyperbola for all values of n. Hence,
all plane sections of the cone parallel to the plane zx are
hyperbolas whose transverse axes are parallel to the axis of
the cone. In lilie manner, we find that all plane sections
parallel to the plane yz are hyperbolas. If ^ = 0, s = ± mx,
whose locus is two intersecting right lines. Hence, any
plane section of a cone parallel to its axis is an hyperbola,
and any section containing the axis is two intersecting right
lines.
A Conic Section is the section of a cone made by a plane.
245. To determine the nature of any conic section that
is not parallel to the axis of the cone, we find the equation
of any such section referred to axes in its own plane.
SURFACES OF REVOLUTION.
261
Let NFJ^ be any section of the cone VB YN passing
through the axis of y ; then this section will be perpendic-
ular to the plane xz. The cone, and therefore the section,
is symmetrical with respect to the plane xz. Refer this
section to ON and 0 Y as the axes of x and y respectively.
Let (cc, y, z) be the point P referred to the coordinate planes,
and {x\ xj) be P referred to ON and OY. Let XON= cf>
and 0DV^=6. Draw PJf perpendicular to ON; then it
will be perpendicular to the plane xz, and we have
y = y', 0B= OM cos <l>, or a; = a:'cos <f> ;
BM= Oil/sin <^, or s^cc'sin ^.
Substituting these values of x, y, z in [73], we obtain
tan^ 6 (cc'2 cos^ <^ + y'^) = (x' sin <f> — cf.
Omitting accents and performing indicated operations,
we have
?/^tan-^+a'-(cos-</) tan - 9 — sin^<;i)) -\-2cx sin ^ — c-=0.
Substituting cos^ </> tan- ^ for sin- ^, we obtain
2/2 tan2 9 -t-ic^ cos2 <}> (tan2e - tau2 (j>)+ 2cac sin <}> - c- = O, [74]
262 ANALYTIC GEOMETRY.
which is the equation of the conic ^ViV/ referred to ON
and 0 Y as axes.
By giving to c all values between 0 and oo , and to <^ all
values between 0° and 90°, equation [74] is made to rep-
resent any section of a cone except those parallel to its axis,
which have already been considered.
Discussion of equation [74].
Here 2=4 cos^ <^ tan^ 9 (tan- 0 — tan- cf>),
A= 4c^ [cos- <j) tan- 6 (tan- ^— tan^ </>) -f-tan^ 6 sin- ^].
(i) Fh'.sf su2)j)ose c not equal to zero.
Let (f><iO ; then tan- <^ <C tan'^ 6, 2 is positive, and A is
not zero ; lience the section is an ellipse.
Let <^ = ^ ; then tan^ cj> = tan^ $, 2 = 0, and A is not zero ;
hence the section is a parabola.
Let <ji'>0; then is tan" <^ > tan- ^, 2 is negative, and A
is not zero ; hence, tlie section is an hyperbola.
Hence, when the cutting ])lane does not pass through tlie
vertex of the cone, the section is an, ellipse, a i>arahola, or
an hyperholn, according as the angle ibhicli the cutting plane
makes with the base of the cone is less than, equal to, or
greater than tlutt made hij an element.
(ii) If c = 0, A =: 0 ; henee, when the cutting plane passes
through the vertex, the elliptical section reduces to a point,
the parabolic to a straight line, and the hyperbolic to two
intersecting right lines.
If </) = 0, the cutting plane is perpendicular to the axis
of the cone, and equation [74] becomes
rf + .r- ^ c^ eot^ Q,
whose locus is a circle.
If c = a3, the cone becomes a cylinder, and the section
made by a plane parallel to an element is two parallel lines
or a single rigrht line.
SURFACES OF REVOLUTION. 263
Bxercise 46.
1. What is the locus in space of x^ -{- 3x^ — 6a; — 8 = 0 ?
of 2/3— 2/ — 5y/ + 6 = 0? of s;^ + m,t; = 0 ?
2. What is the locus in space of //-=:^Sx? of 4a;--|-9iy-
= 36 ? of 9^2 — 16 //2 z= 144 ? of (2a — z) {f — h'-) = 0 ?
of z'-\-x- = i^?
3. Find the equations of the projecting cylinders of the
curve x" + 3i/ — 2,^' = 8, x' + 2f + 3,^^ = 16 ?
4. Find the equations of the projections of the curves
^^ + y^ + 2^~16, 9(.r^ + /)+4^2 = 36.
5. Find the semi-axes and eccentricity of the ellipse
4a;2 + 9/ + 4,^2 = 37, ;i' = i.
6. Find the nature of the curves
x'^-if^^z' = 2b,l{x'^>f)-l^ = m
7. Find the traces of tlie surface 2x--\-bif — 7.^' = 9;
of the surface x^ -\- 3//'^ = 8z.
8. Find the equation of the surface of revolution Avhose
axis is the axis of z, and one of whose traces is ,-;= ± 3a' + 5 ;
find its trace on the plane xi/.
9. Find the equation of a cone of revolution one of
whose traces is 3c^-\-y^ = 9, and whose vertex is (0, 0, 5).
10. Find the equation of the paraboloid of revolution one
of whose traces is 2x^^^oz -{- 5.
11. Find the equation of the paraboloid of revolution one
of whose traces is i/'-^= 8x.
12. Find the equation of the cone of revolution whose
axis is the axis of z, and one of whose traces is 2^/ ^±.~ + 6 ;
find its vertex.
264 ANALYTIC GEOMETRY.
13. Find the equation of the surface of revolution whose
axis is the axis of z, and one of whose traces is ^3^-\-^z'^ = 36.
14. Find the equation of the surface of revolution whose
axis is the axis of z, and one of whose traces is 16y^-\-9z'^
= 144.
15. Find the equation of the surface of revolution whose
axis is the axis of z, and one of whose traces is dz"^ — 4?/^
= — 36.
16. Find the equation of the surface of revolution whose
axis is the axis of z, and one of whose traces is z^x = l;
also when one trace is z^ ^= 2if.
17. Each element of a cone makes an angle of 45° with
its axis ; find the semi-axes of the section made by a plane
cutting the axis 5 below the vertex and at an angle of 60°.
SUPPLEMENTARY PROPOSITIONS.
246. To find the (/eneral equation of the sphere.
Let r denote the radius of any sphere, (a, b, c) its centre,
and (x, y, z) any point on its surface. Then, since r is the
distance between the points {a, b, c) and (x, y, z), we have
{X - a)2 +{y- 6)2 + (s - c)^ = »-2, [75]
or x'^+f-\-z' — 2ax — 2by — 2cz=^r''—a'' — b^ — c\ (1)
which is the general rectangular equation of the sphere.
If the origin is at the centre, then a = 5 = c = 0, and [75]
becomes ^2 _^ ^2 _^ .2 ^ ^-2^ (2)
From (1) it follows that any equation of the form
x''-\-y^ + z'^Gx + Hy + Iz = K (3)
is the equation of a sphere.
Any equation of the form of (3) can readily be reduced to
the form of [75], from which the centre and radius of its
locus become known.
THE SPHERE. 265
Since (3) or [75] contains four arbitrary constants, a
sphere may in general be made to pass through any four
given points.
247. The intersection of two spheres is a circle.
Let the equations of the two spheres be
x'+f + z'-^Gx + Hy +Iz=K, (1)
and a;2 + y2 _|_ ^2 ^ q^^ _|_ fj,y _|_ jr^^ j^, ^2)
Subtracting (2) from (1), we obtain
{G —G')x^ (B — H')i/^ (/- T) z = K—K'. (3)
Hence, the intersection of the spheres (1) and (2) lies in
the plane (3), and is the same as the intersection of (1) and
(3). But the plane section of a sphere is a circle. Hence,
the intersection of the two spheres is a circle.
248. To find the equation of the tanr/ent plane to a sphere
at a given point.
Let the given point be (cci, y^, z^; then the equation of
the radius to this point, that is, of the line passing through
(a, h, c) and (x^, y^, z^)] is
x^:^Xi _ y — //i _z — Zi
a — Xi h — y^ G — z-^
Now the tangent plane is perpendicular to (1) at the
point (ccj, yi, z-^\ but the equation of the plane through
(xi, yx, Zi) perpendicular to (1) is
(a-x,) (x - X,) + (b- y,) {y - y,) + (c-z,) (z - z,) =0, (2)
which is, therefore, t/ie equation of the tangent jdane.
If the origin is at the centre, a= ?; = c =0, and (2) becomes
266
ANALYTIC GEOMETRY.
Transformation of Coordinates.
249. To cJiange the orirjin of coordinates without changing
the direction of the axes.
Let (m, n, q) be the new origin referred to the old axes.
Let X, y, z be the okl, and .r', y\ z' the new coordinates
of any point F; then, evidently, we have
x^m-^ x', 1/ = n-\- y', z = q-\-z'.
Hence, to find the equation of a locus referred to new
parallel axes whose origin is (vi, n, q), substitute m-\-x,
n + //, cmd q + z, for x, y, and z, resjjectively.
250. To change the direction of the axes ivithout changing
the origin.
Let ai, /3i, yi ; ag, /iJg, 72 5 "s, /^s, ys be, respectively, tlie direc-
tion angles of the new axes 0X\ 0 Y\ OZ' referred to the
Fig. 103.
old axes OX, 0 Y, OZ. Let x, y, z be the old, and x', y\ z' the
new coordinates of any point P. Draw PN perpendicular
to the plane X'OY', and JVJf perpendicular to OX; then
OM^^x', MN= y', and NP^z'. Now the projection of
QUADRICS. 2G7
OP on OX { = ^) is equal to the sum of the projections of
OM, MN, and NP on the same line ; hence,
.'C=:x'c0S ai + ^'C0Sa2 + ;s'C0S ttj. (1)
In like manner, we obtain
?/ = CC' COS ^1 + i/' cos ySo + s;' COS ySs, (2)
and z=^x' cos yi + y' cos yo + z' cos yj. (3)
Hence, to change the direction of the axes witliout
changing the origin, stibstitiite for x, y, and z their values
as given in equations (1), (2), and (3),
Since the values of x, y, z are each of the first degree in
x', y\ z\ any transformation of coordinates cannot change
the degree of an equation. (§ 91.)
251. Quadrics. The locus of an equation of the second
degree that contains three variables is called a Quadric.
Thus the general equation of a quadric is
Ax'^By' + C^ + Dxy + Exz + Fyz -[-Gx+Hy
-\-Iz + K=0. (1)
Putting z = q in (1), we obtain
Ax^ + D^r->, + /,y ^(^Eq + G)x + (Fq +H)y
+ (tVr + /'/ + A')-0. (2)
Since the locus of (2) in the plane xy is a conic, and since
the coefficients A, D, B are tlie same for all values of q, all
plane sections of the quadric (1), parallel to the plane a-y, are
similar conies. Now the axis of coordinates may be so
changed that the new plane xy will be one of any system of
parallel planes cutting the quadric. But, as this transforma-
tion does not change the degree of the equation, it follows that
All parallel plane sections of any quadric are similar conies.
268 ANALYTIC GEOMETRY.
252. By transformations of coordinates the general equa-
tion (1) of § 251 may be reduced to one of the two following
simple forms : *
Px'-\-Qu^-\-Rz'=S. (1)
Px^+Qy'=Uz. (2)
Now whatever be the values or signs of P, Q, R, S, equa-
tion (1) evidently represents central quadrics. But the loci
of (2) have no centre ; for if tliey had, and the origin were
changed to that centre, the first power of z would disappear
from the equation. But no expression of the form q -}- z,
when substituted for z, can cause z to disappear.
Hence, (2) represents non-central quadrics.
253. Central Quadrics. If no one of the coefficients
P, Q, R, S is zero in (1) of § 252, we have
S^P^ S^Q^ S-^ R
which can be written in the form
* By changing the direction of the axes the general equation can in
all cases be reduced to the form
Px2 + Qy^ + Rz^ + G'x -f H'y + I'z- K-O. (1)
This transformation is analogous to that in § 189.
(i) If no one of the three coefficients P, Q, R is zero, by a change
of origin, as in § 188, we obtain
Px2 -f Q?/ + Rz"- = S. (2)
(ii) If any one of these coefficients is zero, for example R, by a
change of origin, we obtain
Px2 -I- Qy^ = Uz. (3)
If two of these coefficients are zero, (1) can be reduced to a form
embraced in (3) by first changing the origin and then the direction of
the axes.
QUADRICS. 269
5 + f!-|=l. (B)
according as S-i-F, S~Q, S-^B are all positive, two
positive and one negative, or one positive and two negative.
[If all three are negative there is no real locus.]
If S is zero, we have
Px'-\-Q!/' + Bz' = 0. (D)
If P, Q, or B is zero in (1), its locus is a cylindrical
surface by (ii) of § 236.
254. A discussion of (A) discovers the following proper-
ties of its locus :
(i) Its traces on each of the coordinate planes are ellipses,
(ii) All plane sections parallel to either coordinate plane
are similar ellipses.
(iii) The quadric is included between the tangent planes
X = dz a, 1/ ^^ ± b, z ^^ ± c.
The quadric (A) is called an Ellipsoid. If a =b, the ellip-
soid is the oblate or prolate spheroid, according as a ^or <C e.
The ellipsoid may evidently be generated by a variable
ellipse moving parallel to the plane 2:3/ with its centre in the
axis of z, its axes being chords of the traces of the quadric
on the planes yz and zx.
255. From a discussion of (B), we learn that :
(i) Its trace on the plane xij is an ellipse, while its traces
on the planes ijz and zx are hyperbolas, whose transverse
axes lie on tlie axes of y and x respectively.
(ii) All plane sections parallel to the plane a-// are ellipses,
while all plane sections parallel to the plane i/z and zx are
270 ANALYTIC GEOMETRY.
hyperbolas. The smallest elliptical section is the trace on
the plane xy. The semi-axes of this ellipse are a and h.
The locus of (B) is called the Hyperboloid of One Nappe.
If a=^b, the locus of (B) is an hyperboloid of revolution.
The hyperboloid can evidently be traced by a variable
ellipse parallel to the plane xy, whose centre moves along
the axis of z, and whose axes are the chords of the traces
of the quadric on the planes yz and zx.
256. From a discussion of (C), we learn that :
(i) Its traces on the planes yx and zx are hyperbolas
whose tranverse axes are on the axis of x.
(ii) The plane sections parallel to the plane zy are
ellipses, and no portion of the quadric lies between the
tangent planes a; = ± a.
(iii) The plane sections parallel to the planes yx and zx
are hyperbolas whose transverse axes are parallel to the
axis of X.
The locus of (C) is called the Hyperboloid of Two Nappes.
257. If the coefficients of (D) are all positive or all nega-
tive, its locus is the point (0, 0, 0). If two coefficients are
negative and one positive, by dividing by — 1, two become
positive and one negative. Hence, we need discuss only the
form represented by
2 2 2
X II Z
from which we learn that :
(i) All plane sections parallel to the planes yz and zx are
hyperbolas whose transverse axes are parallel to the axis of z.
(ii) All plane sections parallel to the plane xy are
ellipses, the trace on this plane being a point.
(iii) The traces on the planes yz and zx are each two
right lines intersecting at the origin.
QUADRICS. 271
(iv) All plane sections throngh the axis of z are two
right lines intersecting at the origin.
For, denote any plane through tlie axis of z by
y = mx. (1)
Eliminating y between (1) and {D'), we obtain
z = ±^yjb'' + a'm\ (2)
ab
Now the intersections of (1) and (D') are the same as
the intersections of (1) and (2), which are evidently two
right lines passing through the origin.
Hence, the locus of (D') is a cone whose axis is the axis
of z, and Avhose directrix is an ellipse. If a = b, it becomes
a cone of revolution.
258. Non-Central Quadrics. If no one of the coefficients
P, Q, U is zero in (2) of § 252, we have
^^ I // ^
which can be written in the form
1-1="' w
according as P and Q have like or unlike signs.
A discussion of (E) discovers the following properties:
(i) Plane sections parallel to the plane xy are ellipses,
and the surface lies above the tangent plane ?.' = 0.
(ii) All plane sections parallel to the plane yz or zx are
parabolas, and the traces on these planes are parabolas,
having the axis of z as their common axis and their con-
cavities upward.
272
ANALYTIC GEOMETKY.
By a discussion of (F) we learn that :
(i) The traces on the planes yz and zx are parabolas
whose axes lie on the axis of z, and whose concavities are
in opposite directions.
(ii) Plane sections parallel to the planes yz and zx are
parabolas whose concavities are in opposite directions.
(iii) Plane sections parallel to the plane xy are hyper-
bolas whose transverse axes are parallel to the axis of x, or
y, according as z is positive or negative. The trace on
this plane is two intersecting right lines.
DiAGUAMS.
Note. These figures are taken, by permission, from W. B. Smith's
Geometry.
Elliptic Paraboloid.
FA = 2a and GB = 26 are
half -parameters .
Simple Hyferboloid.
AB'is, Ellipse of the Gorge.
EOT) is the Asymptotic Cone.
DIAGRAMS.
273
Hyperbolic Paraboloid.
OC and OD are Parabolas.
OA and OB are Asymptotic directions for the Hyperbolas.
^^
C
,.,,--—
--"-p- -^-----^
\ B
\\}^^
Ellipsoid.
OA =a, 0B = b, OC = c.
Double Hyperboloid.
EOD is the Asymptotic Cone.
AISTSWERS.
Exercise 3. Page 7.
1. Let xi = — 2, ?/i = 5, X2 = — 8, 7/2 = — 3. Substituiing in [1],
we have
d = V(— 6)2 + (- 8)2 = VIOO = 10.
In Fig. 3, the points P and Q are plotted to represent tliis case. If
we choose to solve the question without the aid of [1], we may neglect
algebraic signs, and we have
qR = NO —MO - 8— 2= 6 ;
PR = PM + MR = b+ Z= 8 ;
... Pq2 =^ q]^2 ^ ^2 _ o(5 ^ (54 _ iQQ^ and Pq = 10.
2.
13.
5.
8.
9.
5, 5, 6.
3.
a, b, ■\la:- + 62.
4.
10.
10.
V29, 5, 2VT0, 4V5;
5.
2 Va2 + bK
2VT0, 3VT3.
6.
25, 29, 20 V2.
11.
8 or - 16.
7.
2Vl7^ 6V2, A
/1O6.
12.
(x - 7)2 +(y + 2)2 = 121.
(X
-2)2 + (y-3)2
= (x-
_4)-2
'■ + {y-
- 5)2, which reduces to x+y—1.
Ezercise 4. ]
Page 9.
13.
1. (6, 6). 3. (2, - 2). 5. (7, 1).
2. (-1, 0). 4. (3, -1), (1, -V), (-i, -I). 6. (a, -6).
7. Take the origin of coordinates at the intersection of the two
legs, and the axes of x and y in the directions of the legs. Then, if a
and b denote the lengths of the legs, the coordinates of the three ver-
tices will be (0, 0), (a, 0), and (0, h).
10. Observe that now the distances BB and BQ will be x — x-z and
y-y2- 12. (f, i). 14. (7f, -311).
11. (6,2). 13. (8,0). 15. (13, - 1), (-11, 5), (1, - 11).
Is ANALYTIC GEOMETRY.
Exercise 7. Page 23.
1. 12, 16. 14. Locus does not cut the axes.
2. - 10, 6. 15. (5, 7).
3. ±4, ±4. 16. (2, 1).
4. ± I, ± 2. 17. (3, 4) and (- 4, .3).
5. ± I, imaginary. 18. (3, 4).
6. ± t, - 4. 19. (5, 3) and (3, 5).
7. ±b, ± a. 20. (0, 0) and (2, 4).
8. 3 on axis of x. 21. (5, — 3), (6, 4), (- 4, - 1).
9. db 3 on axis of x. 22. V61, 5, 2 V26.
10. Locus passes througli origin. 23. 3, 4, 5.
11. Locus passes through origin. i (a, b) (— a, 6),
( On axis of x, 8, and — 4. I {— a, — b) [a, — b).
I On axis of ?/, 4 ± 4 V3. 25. No.
-( On axis of X, 0 and 4, „« in
( On axis of y, 0 and 8.
Exercise 9. Page 31.
1. Let X and y denote the variable coordinates of the moving point.
Then it is evident that for all positions of the point y = 3x. There-
fore, the required equation is y = 3x, or y — 3x = 0. Does the locus
of this equation pass through the origin ?
2. X - 6 = 0, X + 6 = 0, X = 0.
3. y — i = 0, y + \ = 0, y = 0.
4. Tlie line x = 3 is tlie Une AB (Fig. 74); how is this line drawn ?
The locus of the variable point consists of the two parallels to AB,
drawn at the distance 2 from AB. Let CD, EF, be these parallels,
and (x, y) denote in general the variable point, then, for all points in
CD, X = 3 + 2 = 5, and for all points in EF, x = 3 — 2 = 1. There-
fore, the equation of the line CD is x — 5 = 0, and that of the line
EF is X — 1 = 0. The product of these two equations is the equa-
tion (x — 5) (x — 1) = 0. This equation is evidently satisfied by every
point in each of the lines CD and EF, and by no other points. There-
fore, the required equation is (x — 5) (x — 1) = 0, or x^ — 6x + 5 = 0.
Verify that this equation is satisfied by points taken at random in the
lines CD and EF.
ANSWERS. 3
5. 2/2 — 10?/ + 16 = 0, two parallel lines.
6. x^ + 8x — 9 = 0, two parallellines. 7. x+ ^^=^0, y — 2—0.
8. It is proved in elementary geometry that all points equidistant
from two given points lie in the perpendicular erected at the middle
point of the line joining the two given points. This perpendicular is
the locus required, and its equation evidently is x = 3.
r
F
B
D
0
X
E
A
C
Y
P
\
0
A X
Fig. 74.
Fig. 75.
Let US now solve this problem by the analytic method. Let 0
(Fig. 75) be the origin, A the point (6, 0), and let P represent any posi-
tion of a point equidistant from 0 and A, x and y its two coordinates.
Then from the given condition
PO = PA.
Therefore, x^ + y- = {x — 6)2 + {y — 0)2,
or x2 + 7/2 = a;2_ 12X + 36-I-2/2;
whence x = 3,
the equation of the locus required.
9. X — 1 = 0. 10. y -2-0. 11. x — Sy — l = 0. 12. x — y=0.
13. x2 + ?/2 = 100, a circle with the origin for centre and 10 for
radius.
14. Express by an equation the fact that the distance from the
point (x, y) to the point (4, — 3) is equal to 5. The equation is
(x-4)2 + (y-|-3)2=25.
15. (X + 4)2 + (y + 7)2 = 64. 16. x2 + i/2 = 81.
17. Draw AO ± to BC (Fig. 76). Take A 0 for the axis of x, and
BC for the axis of y ; then A is the point (3, 0).
Let P represent any position of the vessel, x and y its coordinates
ANALYTIC GEOMKTKV.
OM and PM. Join PA, and draw PQ ± liC, and meeting it in Q.
Then from the given condition
PA - PQ= OM.
Therefore, ¥1^ - 0M\
Now PA^ = aJi'^ + PM' - {X- 3)2 + 2/2^ ad Um'^ = x2. Sub-
stituting, we have /^ — 3)2 + w2= x^ ■
whence y- = 6x — 9.
Fig. 7G.
The locus is the curve called the parabola. We leave the discus-
sion of the equation as an exercise for the learner.
18. If BC is taken for the axis of y, and the perpendicular from A
to BC as the axis of x, the required equation is 2/2 = i2x — 3G.
19. x2 — 32/2 = 0, two straight lines.
20. x2 + 2/2 = fc2 — a2, a circle.
21. 4ax ± k- = 0, two straight lines.
Exercise 10. Page 33.
4. d = Vxi2 H- 2/1-'. 6. X + 2/ = 7.
5. (X - 4)2 + (2/ - 6)2 = 64. 7. (Y, f ) ; f ^^.
8. Take two sides of the rectangle for the axes, and let a and b
represent their lengths ; then the vertices of the rectangle will be the
points (0, 0), (a, 0), (a, ft), (0, h).
9. Take one vertex as the origin, and one side, a, as the axis of x;
then (0, 0) and (o, 0) will be two vertices. Let (6, c) be a third ver-
tex ; then (a -f- b, c) will represent the fourth.
I ANSWERS. O
10. (11, 2), (- 1, 4), (15, 16). 11. (5, - 2), (i, y), (3, - V).
12. (1, - f). 13. Vl7. 14. (5, V). 16. (6, 23).
__ /Xi + 3x2 2/1 + Sy2\ /X1+X2 yi±y2\ ( 3Xi+X2 %i-h/2\
21. 3 or — 23, 23. (8, 6) and (8, - G).
^ 3 and 2 on OX 24. (2 a, a) and (— 2 a, a).
I 6 and 1 on OY. 25. (a, 0) and (- a, 0).
26. 10, 2 V26, 2 Vl3.
27. Taking the fixed lines for axes, the equation is 2/=6x, or x=6?/.
28. Taking J. for origin, and AB for the axis of x, the equation is
x2 - 3?/2 = 0.
29. Taking the fixed line and the perpendicular to it from tlie fixed
point as the axes of x and y respectively, the required equation is
x2 + (y - a)2 = 4?/2.
Exercise 11. Page 40.
1. X — ?/ + 1 =: 0.
2. 2x — ?/ — 3 = 0.
3. X + ?/ - 1 = 0.
4. X — y — 0,
5. 3x + 2?/ - 12 = 0.
6. 2x — 3?/ + 6 = 0.
7. X + 1/ — 7 = 0.
8. 4x - 3?/ = 0.
9. y = o.
10. ?y==4.
11. 6x-2y = 0.
12. nx — m?/ = 0.
13. X - 2/ - 3 = 0.
14. V3x-2/ + 7-2V3 = 0.
15. X - ?/ + 14 = 0.
16. V3x + 3?/ + 12 - 13 V3 =
17. V3x-3i/-3V3 = 0.
18. X + 2/ — 3 = 0.
19. V3x + 2/ = 0.
20.
y + S = 0.
21.
X — 2 = 0.
22.
X - 2/ + 2 = 0.
23.
X — y + ,5 = 0.
24.
X — y — 4: — 0.
25.
X — V.S?/ — 4 V3 = 0.
26.
y + 4=0,
27.
V3x — 2/ - 4 = 0.
28.
x = 0.
29.
VSx + 2/ + 4 = 0.
30.
X + 2/ + 4 = 0.
31.
x + V32/ + 4V3 = 0.
32.
2/ + 4 = 0.
33.
3x + iy - 12 = 0.
34.
X - 32/ + 6 = 0.
35.
X + 2/ + 3 = 0.
36.
.3x - 52/ - 15 = 0.
37.
X — 22/ + 10 = 0.
38.
X - 2/ - 1 = 0.
b ANALYTIC GEOMETRY.
39. x — y — n — 0.
40. 4x 4- 2/ — 4n = 0.
41. x + 2/ -5V2 =0.
42. X-2/V3 + 10 = 0.
43. x + yVSH- 10 = 0.
44. X-2/V3- 10 = 0.
x + 7?/+ 11 =0, x-3y
+ 1 = 0, 3x + y- 7 = 0. 59. (5, -3), (6, 4), (-4, -1).
45,
46
52.
y - - X ± 6 V2.
53.
•' + ^ - 1.
11 11
3 2
55.
C 6
a = — I or 1
A m
b
_C
b'
58.
A
m = --,or-
b^
a
-i
, a; - Ty = 39, 9x - 5y = 3, 60. 9x + 2y = 0, | V85.
I 4x + ?/ = 11. 61. ?/±x = yiqiXi.
i 17x — 3?/ = 25, 7x + 9y ( (d—c)x—{b—a)y=ad—bc,
47. ^ ^ 62. ^ ^ ^ '
( =— 17, 5x-6?/ — 21 = 0. I (d-c)x + (6-a)?/=M— ac.
r5x — ?/ = 0, 5x + 6?/— 35=0, f2?/2a;+(xi— 2x2)2/— Xi7/2=0,
48. -^ 3x — 2/ = 21, 9x+4y = 0, 64. <i 2/2X+(2xi— X2)y— X]?/2=0,
[2/ = 0, 14x + 32/ = 29. U2.c-(j;i+-C2)y=0.
49. X - 2/ VS — 7i = 0. 65. m = 4.
50. 2/ = X + 3. 66. ??i = 3.
51. 2/ = a; ± 6 V2. 67. 6 = - 9.
68. f^^ = f^^ , or xi (2/2 - 2/3) + a;2 (2/3 - 2/i) + 2:3 (2/1 - 2/2) = 0.
X3 Xi X2 Xi
Exercise 12. Page 44.
1. -^3_Vl3x + T-\Vl32/ = iiVl3; p = iiVi3.
2. 3\ Vsix + /^ V34y = If V34 ; p = if V3i.
3. p = j2^Vl7. 9. Fourth quadrant.
4. p = ^"j Vl3. 10. Second quadrant.
5. p = I V26. 11. Fourth quadrant.
6. p = \^ Vi. 12. Second quadrant.
I 13. Third quadrant.
7. p =
8. p =
Ve^ + c2 14. First quadrant.
r 15. Second quadrant.
Vji- + c2 16. Fourth quadrant.
ANSWERS. 7
17. Third quadrant. 20. m =
a
18. Fourth quadrant. 21. 0 ; 8.
22. C= 12, ^ = 4, B= -1.
24. ^ = (2/2 — 2/i), -S = — (Xa — Xi), G- {xxVi — XiVx)
25. ,„^J^lI^^5:^^^yi-^i^''.
X2 Xi X2 Xi
Exercise 13. Page 46.
1. 3x — 2/ - 16 = 0. 5. X — 5 = 0.
2. 3a; - 42/ - .3 = 0. 6. x + 4?/ + 49 = 0.
3. 4x — ?/ = 0. 7. 7x — 23?/ + 193 = 0.
4. 2/ — 8 = 0. 8. ?/ = 2x.
9. 35?/ + 49x - 79 = 0.
Exercise 14. Page 47.
■n
2. tan (i> — — \. 3. tan <^ = Jg. 4. tan <^
n2 + 2
5. 90°. 6. 135°. 7. 90°. 8. 0°. 9. 30°.
11. j2/ = 5x— 10, 23_ ( ?/ — 3 = m'(x — 2), _
Ix + 5j/=28. ' ( and7n'= — (8±5V3).
12 j2/=5x+ll, [-2/ — 3= ?n'(x— 1),_
' ( X + 52/ - 3 = 0. 14. ^ ^ , 8 ± 5 ^/3
and m—
22. 2x + .3// — 31 = 0. ^ 11
23. 62x + 312/ -1115 = 0. r x - .3?/ + 26 = 0,
24. 2/-6X-27. 3°- ^5^ + 3^+ 8 = 0,
, l2x + 32/- 9 = 0.
25. 2/ = mx ± dVl + ?/i2.
31. X- 6 = 0.
26. £x = ^(2/-6). r 2x- 92/4- 12 = 0,
27. ox — 62/ = a2 - 62. 32. J lOx — 42/ + 63 = 0,
28. (a±6)2/+ (6q:a)(x-a)=0. Il8x - 402/ + 111 = 0.
rx— ?/— 6 = 0^ meeting in the point.
33. -J 2x - ?/ - 2 = 0 I (- 4, - 10).
ISx — 32/ - 10 = 0 J Distance = V85.
— -4±J5tan<^,
3^- '^-^^^= i^±^tan0 (^-^^)-
ANALYTIC GEOMETRY.
Exercise 15. Page 52.
3. 4. 4. §V5. 5. 0.
The learner should construct the
given Hnes, and observe how the sign of
the required distance gives tlie direc-
tion of tlie point from the line.
8. —0, —5, — 4, 3, 2, 1, 0, — 1. The learner should construct
the lines, and observe the change of sign of the distance, as in No. 7.
1.
iVio.
2. fVs.
7.
-¥,-¥,
16 12
'5"» "5'>
-!-, -h
0, +h
- hS - hS
9 6
5> 5>
-1, 0,
+ 1-
2 ^/a^ + 62
Exercise 16. Page 54.
1. H. 4. 40. 8. 35. 11. 26.
2. 12. 5. ah. 9. lOf 12. 96.
3. 29. 7. 26. 10. i(xi?/2 - a;22/i)- 13. 41.
14. i(a-c)(&-l). 21. 9a2. 27. ^
15. |(a-6)(a + 6-2c). g^. f ^g. ,,,
16. i(a2-62).
17. 60°, 00°, 00°; 9V3. 23. 24. ^g _^
18. 10. 24. 36.
19. j^. 25. 10. 30. 56.
20. H. 26. iab. 31. 10^
2AB
ANSWERS.
Exercise 17. Page 56.
3. 2, 00, 90°, 2, 0°. 4. 0, 0, 45°, 0, 135°.
5. iV3-2, 2V3-I, G0°, — 1^^ 150°.
6. 2, f V3, 150°, 1, 00°. 26. 4y = x + 8.
7. 2, -J^/3, 30°, 1, 300°. 27. 4?/ = Ox -24.
r- OR i 9x — 20?/ + 0(3 = 0,
8. f Vs. -2, 00°, 1, 330°. 28. / . .3, ^
* ' ' ' ' / 5x — 4y -f 32 = 0.
9. ill'J;+ 2/ = 0, 29. 88x- 121;/ + 371 = 0.
I X - 5// + 20 = 0. g^^ 5^ _ y _ 10 = 0,
10. /i V827 ' ^ + '">?/ — -8 = 0.
3x + 4// - 57 = 0, 31. j -•« + 2/ - *^ = <^'
3x + 4(/+ 0 = 0, ' X- 2// -17 = 0.
11. ■{ 12x — 5// — 30 = 0, 32 \ 4x + ?/ - 20 = 0,
12x - 5y + 24 = 0. ' I X — 4?/ - 5 = 0.
Area = 63.
12. 43.
13. x=3.
jx-y+l=0,
U+?/-7 = 0.
15. 5x + G?/ — 39 = 0.
16. 14x - 3?/ - 30 = 0.
17. 4x — 5y + 8 = 0.
18. X + // — 7 = 0.
■ X — X3 X2 — Xi
20. i 2^ = •^' 1-'^ = 5x - 1,
19?/= 5x + 7.
21. 92x +09//+ 102 = 0.
22. x + 4?/ = 34.
23. 3x + 4?/ — 5(1 = 0.
24. 3x + 4?/ = 24.
25. 2/ -2/1= -^(x-Xi).
Xi
33.
2x = y, 2?/ = X.
34.
4x + 5//+ 11 ±3 V4r = 0.
35.
2/ = (7 q: 5 -V^) (X + 2).
36.
2x — 5?/ _ I 4x + 3// — 12
V29 5
37.
^ 7X-32/+ 15 = 0,
1 3x + 7// — 03 = 0.
38.
\ 8x + 7// -19 = 0,
I 16x + 3y +17=0.
39.
135°.
40.
00°.
41.
31 V20
143
42.
. hh + al(^ — ah
\/a^ + 62
43.
c-
V/i^ + t-i
44.
±^Vl + m--'.
45.
C-.
10 ANALYTIC GEOMETRY.
xy represents the two axes.
a = 5.
x + a = 0, X — b = 0.
x + a = 0,y+ b = 0.'
The axes and x = y.
2x — y = 0, 7x + y = 0.
62. If h denotes the altitude of the triangle, and the base is taken
as the axis of x, the locus is the straight line y = h.
63. The equation of the locus is
(x — Xi)2 + {y- yi)- - {X— iCo)^ + {y — y^f.
Tills is the equation of the straight line bisecting the line joining
(Xi, 2/i) and (X2, y^)-, and _L to it.
64. The two parallel lines represented by
46.
fe2
6'
2a2 +
54.
47.
• bab + 262
6
57.
58.
48.
49.
17i.
59.
50.
59.
60.
51.
(10,5
\)-
61.
Ax+By-^C±d ^A^ + i'-' = 0.
__ , _, __ Ax + By+C , A'x+ B'y + C ,
65. x + y = k. 66. , — I ^- = k.
Vyl2 + B^ sJA'-^ + B'2
67. Let h denote the base, k"^ the constant difference of the squares of
the other two sides. Taking the base as the axis of x, and the middle
point of the base as origin, the equation of the locus is 26x = ± fc^.
Exercise 18. Page 64.
1.
lx + y- 0.
4
J X - ?/ + 8 = 0,
6.
64x - 23?/ = 59.
2.
X + 2?/ - 13 = 0.
\ x + ?/_6 = 0.
7.
44x + ?/ = 0.
3.
bx + Qy — 37 = 0.
5.
2/ = X + 3.
8.
5x + 2/ - 16 = 0.
9. {A C - A'C)x + (BC - B'C)ii - 0.
10. {BA' - AB')y + CA' - AC = 0.
Ax+By+C _ A'x + B'y + C
■ Axi + Byi+ C~ A'xi +B'yi + C'
12. 472x - 292,' + 174 = 0. 13. 2/ = x Vs + 3 - V3.
(4x+32/— 25 = 0, ^_^ — "^^~ ^
' ( 3x — 42/ + 25 = 0. 'a & ma + 6
16-18. Generally the easiest way to solve such exercises as these
is to find the intersection of two of the lines, and then substitute its
coordinates in the equation of the third line.
ANSWERS. 11
-^ - ^« ,TTi '>^" ~ '>n b" — b
19. m = 1. 20. When — = — •
in — m b — b
21. If we choose as axes one side of the triangle and the corre-
sponding altitude, we may represent the three vertices by (a, 0),
(- c, 0), (0, b).
22. Choosing as axes one side and the perpendicular erected at its
middle point, the vertices may be represented by (a, 0), (— a, 0), (6, c).
23. It is well here to choose the same axes as in No. 21.
24. Choosing the origin anywhere within the triangle, it is evident
that the equations of the bisectors in the normal form may be written
as follows :
(X cos a + y sin a. — p) — {x COS a + y sin a — p') = 0,
(x cos a + y sin a.' — p') — (x cos a." + y sin a" — p") — 0,
(x cos a" + y sin a" — p") — (x cos a + y sin a.— p) = 0.
Now, by adding any two of these equations, we obtain the third ;
therefore, the three bisectors must pass through one point. •
2g_ ( 2V2, VlO, 2V1O. 29 (x-2/ + 2 = 0
Origin within the A- (x + y — 14 = 0.
26. iJ- VlO, 1 & V.34, If V]3.
(X — 1=0
27 (^ + 2/ + 10 = 0, 30. ^_i^o:
I 7x - 7?/ + 24 = 0.
28 i ''* ~ 9?/ + 34 = 0, g^ ±(y— ?»x— b) _^ ±(y—jn'x-b')
' \ 9x + ly - 12 = 0. ■ Vl + ?n2 Vl + m'2
Exercise 19. Page 68.
1. (i) Parallel to the axis of x, (ii) parallel to the axis of y.
2. When ad = be.
3. The two lines are real, imaginary, or coincident, according as
C'^ — iAB is positive, negative, or zero. The two lines are _L to each
other when A + B = 0.
5. X + y + 1 = 0, and x — 3?/ + 1 = 0.
6. x-2y ±(y - 3) V^ = 0.
7. X — y - 3 = 0, and x — 3y + 3 = 0. 8. 45°. 9. K = 2.
10. K = - 10, or — ?./ . 11. K = 28. 12. A' = V-.
12 ANALYTIC GEOMETRY.
Exercise 20. Page 70.
1. Take the point O as origin, and the a.xiss of ij parallel to the given
lines. If the ecjiiations of the given lines are x = a, x — b, and if the
slopes of the lines drawn in the two fixed directions are denoted by
m', jn", the eiiuation of the locus is
{b — a) 1/ = vi'b (x — a) — m"a (x — b).
2. Jf a and b are the sides of the right triangle, the equation of the
locus is f(
3. Let OA = a, OB = b. Then the equation of the locus is
X + y — a + b.
4. Take as axes the base and the altitude of the triangle. Let a
and b denote the segments of the base, h the altitude. Then the equa-
tion of the locus is 2x 2?/
b — a h
This is a straight line joining the middle points of the base and the
altitude.
5. Take as axes the sides of the rectangle, and let a, b denote their
lengths. The equation of the locus is
bx — ay = 0.
Hence, the locus is a diagonal of the rectangle.
Exercise 21. Page 73.
1. X- + y-= — 2rx. 13. (4, 0), 4.
2. x-2 + 2/2 = 2ry. 14. (- 4, 0), 4.
3. x-^ + y'^= - '-iry. 15. (0, 4), 4.
4. (X - 5)- -¥ {y + ;])- = 100. 16. (0, - 4), 4.
5. X--' + (// + 2)-^ = 121. 17. (0, I), 1.
6. (X - r.)-; + y-' = 2.5. 18. (0, 0), 3A;.
7. (X + .5)-' + y- = 25. 19. (0, 0), 2k.
8. (x - 2)2 -\-{y - Zy^ = 25. 20. (0, 0), VoM^.
9. x2 + 2/2 _ .2hx - 2ky = 0. ^l. (-, o\ - VH.
11. (1,2),V5. ^'^ ^ ^.
^ ' 'h k\ VA2 + k^
12. a, I), ■V(J2. 22. (^,-),
2
ANSWERS. 13
23. When D = D' and E = E' ; in otliiT words, when the two
equations differ only in their constant terms.
24. In this case, r = 0. Hence, the equation represents simply the
point (a, b). We may also say that it is the equation of an inlinilely
small circle, having this point for centre.
{{%, I), |V2; r (i) Z>2-4c.
26. <; On OX, 3 and 2 ; 31. \ (ii) E^ = 4C.
lOn OY, 6 and 1. y (iii) 4C > D'^ and E^.
( (6, 2), 5 ; 32. x-2 + 2/2 + iqx + 10?/ + 25 = 0.
27. ^ On OX, 6±V21;
[On or, imaginary point.s. ^^- ('' "*) =^'"M8, 1).
r(2, 4), 2V5; 34. (2, 0)and(^, - ^).
28. -^' On OA', 0 and 4 ; 35. i V5.
I On OY, 0 and 8.
r(3, -2),3;
36.
V a^ + bO
37.
•2x - 2/ - 2 = 0.
38.
4.C — 5?/ — 71 = 0.
39.
3x - 5?/ - 34 = 0.
29. -^ On OX, 3 ± V5 ;
[On OF, -2.
r(-ll, 9), VhB;
30. -^ On OX, - 3 and - 10 ;
I On OY, 9±2V6.
40. Let {x, y) be any point in the required locus ; then the distance
of {x, y) from (Xi, 2/1) must always be equal to its distance from (jo, 2/2) ;
therefore, (x — Xi)2 + {y — 2/1)2 = (x — X2)2 + {y — 7/0)- ;
whence 2x (xi — Xo) + 2y (2/1 — 2/2) = (-Ci"^ + yr —x-z^ — 2/2'-).
Show that this represents a straight line _L to the line joining (xi, 2/1)
and (xo, 2/2) at its middle point.
41. 8x + 62/ + 17 = 0.
42. First Method. Substitute successively the coordinates of the
given points in the general equation oi the circle ; this gives three equa-
tions of condition, and by solving them we find the values of a, b, r.
Second Methoi>. Join (4, 0) to (0, 4) and also to (0, 4) by straight
lines, then erect perpendiculars at the middle points of these two lines ;
their intersection will be the centre of the circle, and the distance from
the centre to either one of the given points will be the radius.
A ns. K'- + y' — fix — % + 8 = 0.
43. x2 + y^ — 8x + 62/ = 0. 45. x- + y- + Sax - 6ay = 0.
44. x2 + 2/2 + Ox + 2/ = 0. 46. x2 + 2/2 + 8x + 20 y + 31 = 0.
14 ANALYTIC GEOMETRY.
48.
49.
47. x'- + ?/- — Ox — %+ 14 = 0. 51. x- + y-::f2ax^:2ay + a'^=0.
^ (X - o)--^ + (>j + 8)--2 = 169, 52. x2 + y^ =ax + by.
] (X - 22)-^ + {y- 9)- = 169. 53. (x - 1)2 + {y — 4)^ = 20.
< x-2 +2/2 _oo(a; +y) +225 =0, 54- «- + 2/' - 14x - 4?/ - 5 = 0.
Ix'+y-— (}(x+y)+ 9=0. 55. x'- + ?/2 ± V2a?/ = 0,
50. x2 + 2/- — 8x — 8y + 16 = 0. X-- + y-± \^ax = 0.
56. r?((x'+?/2)— a6= (ma— 6)x+ (m6 —a)?/. 57. x- + ?/2 = XiX + yiy.
58. (X-Xi)(X-X3)+(2/-yi)(2/-2/2)=0. 3.2_„j,+y2 = r2_ ^"
59. (l + m2)(x2+2/^) -2r(x+m2/) = 0. * ^ 2"
Exercise 22. Page 81.
1. The double sign corresponds to the geometric fact that two tangents
having the same direction may always be drawn to a given circle.
3. 2x + 3y = 26, 3x — 2y = 0; sVlS, 2VT3, -9,-4, -L3 Vl^
Xi2 — r2 r3 „„ rx2 + ?/2 = p2^
' — ^' • 22. .; , . ,
Xi2/i ( (p cos a, p sill a).
r When C = rV^lM^.
23. J rxTu Aa+Bb-C ,
] When — , — = ±r.
[ V^2 + J52
24. ax + by = 0.
25. (— a, — b).
26. (2a, b).
27. (0, 6).
28. x2 + y2 = |.
29. m = 0.
30. c = — 36 + 20V6.
14. Ax + By If W^2 + £2 := 0. 31. (x - 5)2 + {y - 3)2 = Jj-y--
15. 5x-^2/=fWI^T^ = 0. 32 ((X- 2)2+ (.y- 4)2=100,
, fi .. - . + .V2 = 0. i (X - 18)^ +{y- li>)' = 100.
Xi
5.
9x - 13y = 250.
6.
x±Sy= 10.
7.
104 1.
8.
x2 + 2/2 = 25*.
9.
14x ±6y = 232.
10.
3x + 2/ = 19.
11.
3x + 42/ = 0.
L2.
( 3x + 72/ = 93,
( 3x - 7y = 65.
13.
X = r.
16. x-2/±r\/2 = 0
17. The equation of the two taii-
33. (X- 1)2+ (2/ -6)2 = 25.
gents is {h'- - r^)y2 = r2(x - /02. 34. -=-+-■
18. x + y= ±r\f2. 35 (3.2 -\-y2^a + b + Va2 + 62)2.
19 (a; =10, -2a6(a + 6 + Va2 + &2)(x+y)
( 3x + 42/ = 50. _ -f ^252 = 0.
20. 2/ = 2x + 13 ± 6 V5. 36. x = a + r.
21. -21, -3f. 37. [4r2-2(a-6)2]i.
ANSWERS. 15
Exercise 23. Page 84.
1. iV29, (1, - 1). 11- ^' + 2/- - 5x - I2y = 0.
2. |Vrr, (- h f). 12. x2 + 2/2 - 14x - 4?/ - 5 = 0.
3. iVsi, (I, ^). 13. x2 + ?/2 + i4j. + ]4y + 49 = q.
/ 6 ab \ 14. x2 + ?/■- ;^ 2rx — 2ry + r2 = 0,
*■ ^'VVTTa^' VrTa2/ x2 + r' ± 2rx + 2ri/ + r2 = 0.
5. x"- + y^-Sl. 15. x2 + 2/2-2ax-2a2/+a-'--=0.
6. (X- 7)2 + 2/2 =9. ;Lg_ 3.2 + ^2=: 9.
7. (X + 2)2 + (y- 5)2 = 100.
o 9.0 o /o /^ \-rt 17. 5(x2+2/2)-10x,+ 30?/+49=0.
8. x2 + 2/2 — 2a(3x + 4?/) =: 0. ^ ^ ' •'
9. x2 + 2/2 + 262 4- c-2 18 H^ ~ ^^^ + ^^ ~ ^)^ ~ ^'
= 2 [(6 + c) X + (6 - c)2/] . ■ ( (X + V)' + (y + V)' = f •
10. 3a6(x2 + 2/2) + 2a6(a2 + 62) 19. x2 + y"- — 30x — Uly = 0.
= (5a2+262)6x+ {ob-^+2a-)ai/. 20. x2 + 2/2 + SOx + 882/ — 50 = 0.
2^ { x2+ 2/2 — .3Gx- 402/ +324 = 0,
I 25x2 + 25^2 _ 80x - 4942/ + 64 = 0.
22. (6,2), 5. 31. x2+2/2 = ±a2/V2,or = ±axV^
23. - 15. 32. x2 + 2/2 ± 2a(x ± 2/) = 0.
24. — 10. 33. 2(x2 — ax + 2/2 - r-) + a2 = 0.
25. 1V26. 34. X — y = 0.
26. VTO. 35. 4x + 32/=^0.
27. x,x + y,y = x,^ + 2/,2. 36. (18 ± 2V41)x -52/ = 0.
28. (i) Z»2 = A AC, (il) -B2 = 4^(7, ^'^^ a: + "^^ ± ^0 - 0.
(iii) Z>2 =:: £2 rr 4^ C. 38. X + ^ - 10 = 0.
29. r2 = 2rmc + c2. 40. +(35 + 24\/30). 43. 135°.
30. k = 40, or - 10. 44. (7, - 5) and (- 6^^, 9||).
f (X + 4)2 + (2/ + 10)2 = 85,
«. ^/^_514X2 / 670y^_8^
[\ 109/ V^ 169/ 1692
46. The circle (x — Xi)2 + (2/ — Vi)- ~ r".
47. The circle (x - a)- + (2/ — 6)2 = (r +Y)'^,
or (X - a)2 + (2/ - 6)2 ^ (r - r')2.
48. The circle (x — rt)2 + (2/ — 6)2 = 7-2 + <2.
16 ANALYTIC GKOMKTKY.
49. Take yl as orijiin, and let the radius of the circle = r; then the
locus is the circle x- + //'- = rz.
50. Take A as origin, and let the radius of the circle = r ; then the
. , „ , -, 2»irx
locus IS the circle x- + y'^ = - — , —
m + n
51. Take A as origin, AB as axis of x, and let AB = a ; then the
locus is the circle {in- — n-) (x'^ + y'^} — 2am^ + a-m- = 0.
52. Take AB as the axis of x, the middle point oi AB as origin,
and let AB = 2a ; then the locus is the circle 2{x^ + y-) — k- — 2(jfi.
53. Using the same notation as m No. 52, the locus is the straight
line 4ax = ± Ic^.
54. Taking the fixed lines as axes, the locus isthecircle4(x-+?/-) = fZ-.
55. Take the base as axis of x, its middle point as oi-igin, and let
the length of the base = 2a, and the constant angle at the vertex = d.
Then the locus is the circle x'^ + y- — 2a cot dy = a'^.
56. Take A as origin, AB an axis of x, and let AB — a, AC = h.
Then the locns is the circle (x — },a)'- + y- = —•
- ' ■ 4
4r*
57. The circle x- + y- = r-; zzi where I is the length of the chord.
58. The locns is a circle.
Exercise 24. Page 97.
1. 7x — 6?/ = 0. 2. X — ?/ = 0.
4. X + ?/ = r, 2x + o// = r, (n + b)x + (a — b) y = r^.
5. l?.x + 2>/ = 40. 6. The tangent at {h, k).
7. (i) 2x+% = 4, (ii) '^x — y = 4, (iii) x — ?/=4.
8. (i) (20, ;]n), (ii) (21, -14), (iii) (.%«, nr>h).
9. (G, 8). 18. 12x + 17;/ - 51 = 0.
^n ( --^ -JJ^1\ 19. x+ 7/ -2 = 0.
"• V c ' c )
11. (4, ± 3), 4x ± ?,y = 25. ^^ ("' ~"^^^^ ~("^ -lfi)y + ac =0.
16. h:2 + /,-2 _ r2. 21. X - 7/ = 0, Vi(a + 6)^ - 4c.
17. 3. 22. (-2, -1).
ANSWERS. 17
Exercise 26. Page 109.
1. Writing x + 1 for x, and y — 'l for ?/, and reducing, we have y'^—^x.
2. x- + (/- = ?•-. 5. x'^ + ?/" = '''■'■
3. x'-^ + y- = 2rx. 6. 2xy = a^.
4. X- + y- = — 2ry. 7. x- — //- + 2 = 0.
8. (i) p = ± ((, (ii) p2 cos 2d - a^.
9. (i) p = ia tan 0 .sec ^, (ii) (a + p co.s ^)- = 4ap sin ^.
10. (i) X-' + //■- = a-, (ii) x- + y- = rtx, (iii) x^ — y- — ofi.
11. x + y = (). 15. x2 — ijxy -\-y- = 0.
12. 2x — 5y + 10 =: 0. 16. xy = .'J.
13. 12x- + Ulzy + 4i/^ +1 = 0. 17. y/- = 2(f(x V2 — a).
14. X- + y- — 25. 18. -Ix// = 25.
Exercise 27. Page 111.
1.
6V3.
12.
9x2 + 25(/2 =: 225.
2.
4 sin ^w.
13.
14.
15.
p= 8a cos e.
p = ± 4a.
p2 sin2 e — 5/3 cos 0 = -2:^.
3.
Vl3 — 12 cos w.
4.
Va- + Ij^ — 2ab cos
(^-
0).
5.
2a sin 0.
16.
p2 = 49 sec 20.
6.
2(1 cos 0.
17.
18.
P' = /i;2 cos 20.
xy - a-.
7.
aV5-2V3.
9.
2x'-2 + 2x?/ + 2/2 = 1.
19.
{x- + 2/2)2 =: 2ixy.
10.
2x2 + 2/2 = 6.
20.
^s— 2/3+ (3x— Sy— 5A;)x2/=
=0.
11.
2/ = 0.
21.
tan-i |.
22. (i
) tan
-(■
-iy
(ii) tan-'--
.4
Exercise 28. Page 117.
2. 2/2 = 4/y(x - 7/). 3. 2/- = ^^(x + p).
4. (i) ^2 = lOx, (ii) y- = lOx + 2-'., (iii) 2/2 = lOx — 25.
5. (i) 2/2 =l(3x, (ii) y- = lOx + 04, (iii) y- = lOx — G4.
18 ANALYTIC GEOMETRY.
6. (0, 0), (2, 6). 8. (4, 6) and (25, 15).
7. 6, 15, ^- 9. (12, 6).
10. The line x = 9 meets the parabola in (9, 6) and (9, — 6). The
line X =0 passes through the vertex. The line x = — 2 does not meet
the parabola.
11. The line y — 6 meets the parabola in (9, 6). The line y= —8
meets the parabola in (16, — 8).
12. p = 4. 13. (0, 0), (2, 8).
14. (i) y = 0, (ii) x = - 2, (iii) x = 2, (iv) 4x ± 3y-8=0, (v) y= - 2x.
15. (i) 4x - 5?/ + 24 = 0, (ii) x^ + y^ - 20x = 0.
16. 3j3. 17. 8pV3.
24. The latus rectum of each = 4p. The common vertex is at the
origin. The axis of x is the axis of (i) and (ii); that of y is the axis of
(iii) and (iv). Parabola (i) lies wholly to the right of the origin,
(ii) wholly to the left, (iii) wholly above, (iv) wholly beloiv. We may
name them as follows :
(i) is a right-handed X-parabola. (iii) is an upward F-parabola.
(ii) is a left-handed X-parabola. (iv) is a downward F-parabola.
Exercise 29. Page 121.
6. X — 4?/ -I- 20 = 0, 4x + y — 90 = 0.
_^ (x-?/ + 3 = 0, , <x + y-9 = 0,
7. Tangents -{ , , „ „ normals '
•^1:
x-{-y + ^ = 0; lx-y-Q = 0.
These lines enclose a square whose area = 72.
8. Tangent = V266, normal^ V95, subtangent = 14, subnormal^ 5.
9- (5, 10). 13. S^^, I^Vr^M^.
in Vl + m^ "i
14. [sjx.xo, l{y, +2/2)].
15. x + y + p = a, point of contact (p, — 2p), intercept — — p.
16. Equationsof the tangents ?/V3 = ±x±3p, required point(—3p,0).
17. For the two points whose coordinates are
X = f (1 ± vi7), 2/ = ± p Jl±^.
o \ 2
18. For the points (0, 0) and (3p, ± 2j5 VS).
19. 9x - 62/ + 5 = 0, (I, I).
20. X - 2?/ + 12 = 0, (12, 12).
4x + 2y -h3 = 0, (f, - 3).
ANSWERS.
19
21. ?/ = x(±V2- 1) + 4(± V2 + 1).
22. 4Vp(p + Xi)8
Xi
24. By the secant method we find that the equation of the tangent
at (xi, ?/i) is y— y\ _ 4
a: — a;i 2/i — 3
The points of contact are (—1,11) and ( — 1,-5) ; hence the tangents
are z — 'ly -\- 23 = 0
and X + 2i/ + 11 = 0.
r (i) 2/12/ = - 2p(x + xi),
25. ^ (ii) xix = 2i)(?/ + t/i),
[(iii) Xix= — 2i3(?/ + ?/i).
Exercise 30. Page 123.
1.
4.
7.
10.
t/2 = 24x — 144. 2. 2/2 = i(5x.
j 2?/2— lix+ 12?/ + 73 = 0,
j 2y2+ llx+ 12y-37 = 0.
2x2 = Qy. 8. I, 8x + 3
3. 2/2= -17x.
5. (2/ + 7)2 = 4(x - 3).
6. 32/2 = 4x.
0, 8x ± \by -3 = 0.
4 on OX ;
8 and — 2 on OY.
11. 4(2 ± V3)p.
r (i)2/ = x+2,
12. \ (ii) -2^,
[(ill) x+ 2/-G = 0.
13. x + y-6 = 0.
2p
14. y - 2/1 = (X - xi).
2/1
15. (8, 4), (2, 10).
16. (2, 4), (11, 10).
_ b ± \lh' + 4ap ^
20.
y"=- 9x.
21.
2/2 = 8x.
22.
4r2 - «2
23.
y2= —X.
24.
2jl2
y-- .
17. 2/
2a
Vn2 + «2
25. 2/2 = 2(2r- s)x.
26. 4pV2.
27. The equation of the circle is
(X- 3)2+ (2,-1)2=2^5.
28. (-3p, 0).
' A left-handed A'-paiabola. 29. f-, ±?pV3Y
Latus rectum =2. ^ '
30. j(^'±2p);
I 45° and 135°.
31. 4p-.
34. The parabola y-
18. <| Vertex, (—2, 0).
Focus, (— 5, 0).
Directrix, x =
19. — 2aV2.
px.
20 ANALYTIC GEOMETRY.
The loci in Nos. .35—38 are parabolas, the latus rectum in each
being half that of the given parabola. If the given parabola is y"^ = 4px,
the equations of the loci are : 39, Tj^g straiglit line y = pk.
35. ?/2 = 2px — p2. 4Q ^,^g parabola 7/^ - 4px = p-k^.
36. ?/-2 = 2px - 2p2. 4f, The straight line A:x - p.
00' K ~ ?^' , o 0 42. The circle (x - p)2 + ^2 = ^.
38. ?/2 = 2px + 2p2. ^ ^' • " jc^
43. Take the given line as the axis of ?/, and a perpendicular
through the given point as the axis of.x, and let the distance from the
point to the line = a. The locus is the parabola y- = 2a(x — -Y
Exercise 31. Page 134.
10. 3x-5y-6 = 0. fay = 2px.
11. 8y — 25= 0. 14 J T''^^ chord is parallel to the
I tangent at the end of the
12. 13x + 22y + k = 0. ,. ,
-' 1^ diameter.
13. X — 7/ — 1 = 0. 15. ?/- = 52x.
18. Writing the equation in the form (y — 3)- = 8(x — 2), and pass-
ing to parallel axes through (2, 3), we have y^ = 8x. 8, (2, 3), (4, 3),
V = 3, X = 0.
19. A numerically, ( — —, — » ^ r 2/ = — „
^' \ 4A 2) 2
B2-4C B\ B
4J~' --2>^=-2-
r.« „ ■ n / ^ A^-4C\ A
20. /> numencallv, I — „» r^:: — I' x= — — •
. .. / A A-^-^V\
21. Take the given line as the axis of y, and a perpendicular through
the centre of the given circle as the axis of x. Let the radius of the
circle = r; distance from the centre to the given line = a. There are
two cases to consider, since the circles may touch the given circle either
externally or internally. The two loci are the parabolas
?/2 = 2{a ± r) X + r- — o-.
22. Let 2a be the given base, ah the given area ; take the base as
axis of X, its middle point as origin ; then the locus is the parabola
x2 + by = a?.
Exercise 32. Page 144.
1. 5, 4, 3, |. 2. V2, 1, 1, iV2. 3. 2, ^/3, 1, \.
1_^ J^^ Is- A IB-A
■\Ia ' Vi>''
"^B V^' V^'^^^'"^^^-
SWERS.
13.
x'-i
+ 2(/2 =
; 100.
14.
8x2 + y?/2 =
= 8tt2.
16.
2 :
V3.
17
X
= y=±
«6
■ V«' + ^-
18.
(1, t)^ (-
S — 2\
31 sr
19.
(1
, -0, (1,
-2).
21
5. fVe.
6. e = iV3.
7. 4x2 + 9y2 = 144,
8. 25x2 + 1097/2 = 4225.
9. 16x2 + 252/2 = .3600.
10. 16x2 4-252/2= 1600.
11. 25x2 + 1692/2 = 4225.
12. 3x2 + 72/2=115. 20. (:!, l),(3,-l),(-3, l),(-3,- 1).
21. The equation of the locus is x2 + 42/2 = r-.
22. Taking as axes the two fixed lines, and putting AP — a, BP = 6,
the acute angle between AB and the axis of x = ^, we find that
X — a cos <p, y = b sin (p.
Therefore P describes an ellipse whose equation is
^ + ^ = 1.
a2 62
23. The two straight lines y = ± •'^\l~ r- "^^^ locus is imaginary
when y is imaginary; that is, when A and B have like signs.
29. The equations of the sides are
, ab , ab
X = ±— 1 2/=±
M
Va2+62 ^a;^+l^'
4a262
area = ,-, , •
Exercise 33. Page 151.
4x±9y = 35,
'■■\
7.
22/ = X ± 10.
8.
4x - 32/ ± Vl07 = 0.
9.
X = ± . ^ _ ' 2/ = ±
9x rp 42/ = 6.
2xif32/V3 + 12 = 0, ^^ , «'•' , ,^ ^ . ^"^ .
6xV3qi42/ + 5V3 = 0. ' Va2+62 ^ V^iH^2
r X + 42/ = 10, 10. Same answers as No. 9.
3. J 4x -2/- 6 = 0; n. ft2:a-2.
( - 8, - i.
_ a2 62 „ ft
5. — ^4- — =1. 12. X = ± — 7=' 2/ = ± — =•
7/1- H2 V2 \^
6. 9x2 + 252/2 = 225. 13 // = 4, 3x + 2y = 17.
22 ANALYTIC GEOMETRY.
14-. The equation ± Vox ± •"?;/ = 9a represents the four tangents.
15. aVl — e-^ 008^0. 16. ^(a^csc^sec^ — c^cot^).
17. The extremities of the latera recta.
19. The method of solving this question is similar to that employed
in § 136. The required locus is the auxiliary circle x- + y- = a^.
Exercise 34. Page 152.
(a;=8, 40y = 9x + 72; 8. 6x + ay qp a6 V2 = 0.
I .1,8. SJ2.
_ -,,.'!.. 9. -COS0 + T.sin^ = 1.
2. Within. a b
ab
10.
3. iV3. ■""• V^^
11. aVl — e-cos2 <f>.
12. c2:6-'.
2V3 13. a- — e*Xi2.
4. iV2.
12. c2:6-'.
1 + Vl3 _ 14. V(l- e'^)(a2-e-^Xi2).
6. X + y = ± Va2 + 52. Vl — e'^ b
„ 1 , 1. /"TT"; — ;; ,^ 15- tan 4> = = -•
7. &x 4- cy qi oVa^ + c^ = 0. e c
18. The locus is the minor axis produced.
X — j + ?/2 = f2; centre is I „, 0 j ; semi-axes ace
- and r.
20. The ellipse o'^ ( y — - j + ^^x^ = — — ; centre is (0, - j ; semi-
a , b
axes are - and -•
In 21-23 take the base of the triangle as the axis of x, and the
origin at its middle point.
21. The ellipse {.s2 — c2)x2 + s2?/2 = s2(s2 — c2).
22. The ellipse kx- + y~ = kd^.
23. The circle (x + (")- + y- = 4a2.
Exercise 35. Page 164.
1. Stt. 2. I Vs. 3. 20x + 63?/,— 36 = 0.
b. (^ ^> ^j 8. (1) jni2=-, (u)mi2=-, (m)m^2 = l.
ANSWERS.
11. 3x+8y = 4, 2x - 3?/ = 0.
12. Area = — (?n + n), m and n being the two segments (use the
2a
polar equation).
13. 2Gx + 33?/ - 92 = 0. 14. x + 2?/ = 8.
15. b'-z + a^y = 0, b^x -a^ = 0, a^y + Wx = 0,hx + ay- 0.
17. ahj^x = b^xiy. 29 ^ + ^ = ^.
23. -T 1 = 0.
« ^ _ a(l - e2)
30. /) —
24. 6xVz- — 62 ± a]/\/a;^—l^ = 0. 1 + e cos 0
25. e=^V6. g^ 62
26. See § 148. ' ^ 1 — e2cos2 6>
32. 1(5x2 + 49?/2 - 128x - 686?/ + 1873 = 0.
33. 2a = 18, 26 = 10. 34. ^' + ^2 = 5^.
144
c2 — 62 a—b
35. cos 0 = \ -^ — 7:7 • 36. tan 0 = p=-
37. Find the ratio of yi to the intercept on the axis of y.
38. h'^hx + aVcy = b^h' + am. ^^ „, ,,. x2 y2
42. The ellipse — + t:; = i.
a2 62 -
The ellipse 62x2 +a2y2 = 52^2
44.
X2
a2
+
2/2
62
= 2.
43.
(-
- 1
,1)
, a =
^,
6
= 1,
46. The ellipse 25x2 + \C^y^- - 48y = 64.
45. A/^'A/^'inwhich£:=-F+^^%^-
Exercise 36. Page 174.
s^ _ y^ _ 3. 3x2 — y2 — ;3(i2_
^^ ^^ 4. G25x2 - 84?/2 = 10,000.
4x2 ^2 _
^- 25~36~^- 5. 2.c2 - 22/2 = c2.
7. rt = 4, 6 = 3, c = 5, e = |, latus rectum = |.
24 ANALYTIC GEOMETRY.
8. Idy- — Dx- = 144, transverse axis = 0, conjugate axis = 8, dis-
tance between foci = 10, latus rectum = A/.
9. a: 6=1: Vs. 11. e = 72. 12. (5,-0^).
14. Foci, (5, 0), (— 5, 0); asymptotes, y = ± |x. 17. b.
Exercise 37. Page 176.
1. KJx -Qy = 28, Ox + Uhj = 100; |, %»-. 3. x^ - y^ = 0,(5, 4).
4. The four points represented by
x= ■ ' y
Va2 - 62 Va2 - 62
9. — =• 10. —^ ^ = 1.
11. When a is less tluin 6. 12. Tlie circle x- + y^ = (j2.
Exercise 38. Page 177.
1. 26e, ae2. 12. (0, ± Va-' - 62).
2. 14 and 6 ; (— 8, ± SVS). 13. 62>a2.
3. The sum = 2ex. 14. g4x — Oy — 741 = 0.
8. {a, 6V2), (a, - 6V2). 15. y = ix ± 8\^.
10. They are equal. a'lb-
11. y = xV2 + a. ' a2+~62
Exercise 39. Page 188.
1. 9x + \2y + 10 = 0. 6. a.
8. 75x - Wy = 0.
9. 245x- 122/ -1189 = 0.
10. fVs.
17. See § 140.
18. ^ + ^ = 2.
a;i yi
2.
x = ±--
e
3.
TT
2'
4.
/ >lffl2 562 X
1. C ' C j
5.
X + a = 0.
ANSWERS. 25
19.
X2 7/2^ 2X_^ _ ^ 5-2
(i') ^-K+'^ = 0.
21.
23. The hyperbola 3x2 — y- + 20x — 100 = 0. The centre is the
point ( — L"-, 0). Changing the origin to tlie centre, we obtain 9x2
- 3y2 = 400.
24. Writing the equation in tlie form (x —1)2 — 4(?/ + 2)2 = 4, and
x2 ?/2
changing the origin to (1, — 2) we obtain T ~ T — 1- T^^ centre i.s
(1, -2), a = 2, 6= 1.
25. Centre is ( -r-^. TTT h semi-axes are -%/ , .*/— , in whith
1)2 ^2
4J. 41?
26. Tlie locus is the curve 2xy — 7x + 4?/ = 0. If we change the
origin to the point {h, fc), we can so choose tlie values of h and k as to
eliminate the terms containing x and y. Making the change, we obtain
2x1/ + (2A; - 7)x + (2ft + A)y — Ik + 4k + 'Ihk - 0.
If we choose h and k so that 2A + 4 = 0, and "Ik — 7 = 0, that is, if
we take ft = — 2, fc = .1, the terras containing x and y vanish, and the
equation becomes xy = — 7. Hence we see (§ 182, Cor.) that the locus
is an equilateral hyperbola, whose branches lie in the second and fourth
quadrants, and that the new axes of coordinates are the asymptotes.
27. The equilateral hyperbola 2xy = cfi.
28. Taking the base as axis of x, and the vertex of the smaller
angle as origin, the locus consists of the axis of x and the hyperbola
3x2 — ?/2 — 2ax - 0.
Exercise 40. Page 206.
1. The ellipse 72x2 + 48^2 =35. 8. The parabola y- = 3x V2.
2. The ellipse 4.x2 + 2//2 = 1. 9. The parabola ?/2 = 2x.
3. The hyperbola 32x2 — 4»y-=d. 10. The ellipse 4x2 + '.)//2 = 3(5.
4. The ellip.'^e dx'^ + ?,y- - 32. 11. The point (0, 0).
5. The hyperbola 4x2 — 4//2 + 1=0. 12. 'I'ho hyperbola 4x2 _ <),/-j == ;j(]
6. The parabola y2 = — jx. 13. TlK'Straightlines.v=x,y=— 5.
7. The parabola y- = 2xV2.
26 ANALYTIC GEOMETRY.
Exercise 42. Page 229.
2. 2d ; 5th ; 6th ; 7tli ; ;3d ; 4th ; 8th.
5. 5V2; VTi; V83; O.8V2, 0.4V2, O.5V2 ; iVli, - ^\y/U,
- xVVIi ; ^V83,_- A V83, - /^V83.
6. tVVi4, ^Vi4, y\Vl4. The line is parallel to the radius vector
of the point (1, 2, 3).
7. Parallel to the radius vector of the point (A, B, C).
ABC
yjA-^ + ii2 + C2 V^2 4- £2 + c^ V^-^ + B^+ C^
8. 60° or 120° ; 90° ; 60° or 120°.
Exercise 43. Page 235.
1. V3; V3; 2V3. 3. 5V2. 4. - J-^y2, - §V2, \S.
5. Lines parallel to the radius vector of (3, — 2, — 5).
^3^V38, -tVV38, -^\V38.
6. cos-i ||Vl4. 9. (1, V3, 2V3).
7. 90°. 12. (t, - f, I).
8. (4, iTT, iTT). 13. (- -2,9-, - -V«, ¥)•
Exercise 44. Page 240.
4. 2x - 3?/ + V3z = 28.
5. ^-f-|=l,T + f + ^=-l,2x-|2/ + 62 = 0.
4 .5 / 1 z 5
6. 6x + 2/ — z = 5.
7. cos-i ^-. 12. - + ^ - - = 1 ;
8. 79° 52'. ■ a 6 c
(J — 2abc ^
^' '^^^"^ Vl^TW^C^' V6%Mr^2c2 + a262
^ 14. 2x 4- 4?/ — 2 = 23.
COS — ^ — — -' 1
V^2 + ^2 + (72 16. 3x — ?/ -f 2z = 4.
i? 18. .c + y + z = 0 ; 2V3.
cos-'i ,
V.42 + i?2 + C2 19. ?,x - iy + 7z + 13 = 0.
10. 2.25. 20. 2x + 5?/ — z = 9.
ANSWERS. 27
Exercise 45. Page 247.
3. It passes through (i, 0, — |), and is parallel to the radius vector
of (4, 5, 3).
x^zA^lL^zl^^^il. 6. (4, 5,0), (0, 1,4), (-1,0, 5).
2 2 — 2 ' 7. 22/ + 5z = 10.
8. 2x — 5?/ + 1 = 0, 2x + 5z + 1 = 0, ?/ + z = 0.
11. cos -1—0.1 x — 'P._y+7_z + o
12. 14° 57' 45". ' -3 ~ 5 ~ - 6
r- ,„x — 2w + 4z + 6
15. ?/ = - 2x - 1, z = 3a; + 5. 18. sin-i ^%.
X — p _ y — Q _ z
19.
1
Vl + »l2 + ?l- Vl + 7)1- + n- Vl + H(2 + ,j2
in which the denominators equal cos a, cos p, cos 7, respectively.
Exercise 46. Page 263.
1. The planes x=— 4, x = — 1, x = 2; the planes y = — 2, y =1,
y = 3 ; the planes z = 0, z— — m.
2. Answer to the first, the parabolic cylinder whose elements are
parallel to the axis of z, and wliose trace on the plane xy is the parab-
ola 2/2 = 8x.
3. 5x2 + 10^2 - 5(5^ 5^2 _ j^2 = 8, x2 + 13z2 = 32.
4. X- + ]/-= i, z = i: |V42 ; hence the curves are two circles in
the planes z_— ± sV42, whose centres are in the axis of z, and whose
radii are Vf- each.
5. a = 3, 6 = 2, e = iV5. _
6. Two circles whose radii are •\/l3 each.
8. 9(x2 + y"-) = (z - 5)2 ; x2 + 2/2 = 2^5.
9. 25x2 + 252/2 - 9z2 + 90z = 225.
10. x2 + 2/2 — |z — I = 0.
11. 2/2 + 22 - 8x = 0'.
12. x2 + y2 _ J22 - 3z = 9 ; (0, 0, - 6).
13. 9(x2 + 2/2) + 4z2 = 36.
14. lG(x2 + y-) + 9z2 = 144.
15. 4(x2 + 2/2) _ 9^2 = 3(5.
16. x2z* + y-z* = 1, xf + 2/2 - iz3 = 0.
17. a = 5V3, b= jV6.
WENTWORTH'S||f^°
TRIGONOMETRIES ^pitiqns
THE aim has been to furnish in these books just so much of
trigonometry as is actually taught in our best schools and
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brevity consistent with simplicity and clearness, and interesting
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love for the study. The subjects of Surveying and Navigation
are presented in accordance with the best methods in actual use,
and in so small a compass that the general student may find time
to acquire a competent knowledge of them.
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As the name implies, the work is intended for colleges and
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just enough examples to illustrate and enforce these princi-
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Every effort has been made to present in the clearest light
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