BOUGHT WITH THE INCOME OF THE
SAGE ENDOWMENT FUND
THE GIFT OF
HENRY W. SAGE
1891
arV19523
Analytic geometi
Cornell University Library
3 1924 031 218 971
oiin.anx
The original of tliis book is in
tine Cornell University Library.
There are no known copyright restrictions in
the United States on the use of the text.
http://www.archive.org/details/cu31924031218971
ANALYTIC GEOMETRY
BY
MARIA M. ROBERTS
Professor of Mathematics in Iowa State College
AND
JULIA T. COLPITTS
Associate Professor of Mathematics in Iowa State College
FIBST EDITION
NEW YORK
JOHN WILEY & SONS, Inc.
London: CHAPMAN & HALL, Limited
1918
Copyright, 1918, by
MARIA M. ROBERTS
AND
JULIA T. COLPITTS
Stanbope ipcess
F. H. GILSON COMPANY
BOSTON, U.S.A.
PREFACE
This book is the result of several years of experience in
teaching mathematics to students of engineering and science.
Since at the outset, analytic geometry opens to the student
an entirely new method of approaching mathematical truth,
much stress is placed on the first two chapters in which the
student is building the concepts on which the future chapters
rest. Emphasis has also been placed on those portions of
analytic geometry in which experience has shown the student
of calculus to be most frequently deficient. In this con-
nection, in particular, polar coordinates have received more
than usual attention and transcendental and parametric
equations considerable space. The exercises are numerous
and varied in character, and the teacher will thus be enabled
to select from them those which best emphasize the points
which he considers important.
The book has been used for two years in mimeographed
form in the class room both by the authors and their col-
leagues, and many valuable suggestions arising from such
use have been incorporated into the final form of the text.
The material is so arranged that the first ten chapters
together with a portion of Chapter XIII include those sub-
jects ordinarily offered to such freshman classes as cover in
the first year the three subjects, college algebra, trigonometry
and analytic geometry. The addition of Chapter XIV will
round out a good course of five hours a week for a semester.
The entire book should easily be covered in a three hour
course throughout a year.
iv PREFACE
The authors take pleasure in expressing their thanks to
their colleagues in the department of mathematics of the
Iowa State College, for their assistance in reading proof and
solving problems as well as for their many helpful suggestions.
MARIA M. ROBERTS,
JULIA T. COLPITTS.
Ameb, Iowa, January, 1918.
CONTENTS
CHAPTER I
CARTESIAN COORDINATES
Art. Page
1. Definition 1
2. Directed lines 1
3. Position of a point in a plane 2
4. Rectangular coordinates 3
5. Notation 3
6. Distance between two points 5
7. Inclination and slope ^ 7
8. Point of division 11
9. Area of a triangle 15
CHAPTER II
LOCI
10. Equation of a locus 18
11. Locus of an equation 22
12. Plotting the locus of an equation 23
13. Discussion of an equation 27
14. Points of intersection of two curves 40
15. Locus by factoring >, 42
16. Loci through intersections of two given loci 44
CHAPTER III
THE STRAIGHT LINE
17. Conditions which fix a line 47
18. Equation of a Une in terms of point and slope 47
19. Equation of a line in terms of slope and y-intercept 48
20. Equation of a line in terms of two given points 50
21. Equation of a line in terms of the intercepts 51
22. Locus of equation of first degree 52
v
VI CONTENTS
A»T. Page
23. Plotting straight lines 53
24. Normal equation of a straight line 54
25. Reduction to normal form 56
26. Perpendicular distance from a line to a point 58
27. Bisectors of the angles between two lines 61
28. The angle which a line makes with another line 64
29. Systems of straight lines 67
CHAPTER IV
POLAR COORDINATES
30. Definition 73
31. Equation of a locus: polar coordinates 76
32. Locus of an equation: polar coordinates 77
33. Equations of the form p = a sinks and p = a cos fee 82
34. Di£Sculties arising from the multiple representation of points
in the polar system 83
35. Spirals 85
36. Intersections of curves 87
CHAPTER V
TRANSFORMATION OF COORDINATES
37. Definition 90
38. Translation of axes 91
39. Rotation of axes 92
40. Degree of equation not changed by translation and rotation 94
41. Simplification by transformation 94
42. Transformation from rectangular to polar coordinates and
vice versa 97
CHAPTER VI
THE CIRCLE
43. Definitions lOO
Equation of circle, center at Qi, k) and radius r 100
Equation of circle, center at origin and radius r 100
44. General form of equation of circle 101
45. Radical axis 103
46. Circle determined by three conditions 104
CONTENTS vii
chapter vii
abt. the parabola P^,,
47. Conic sections 110
48. Equation of parabola, vertex at origin, axis on a;-axis Ill
Latus rectum of parabola 113
Equation of parabola, vertex at origin, axis on y-axis 113
Equation of parabola, vertex at Qi, k) and axis parallel to the
a>-axis 114
Equation of parabola, vertex at (h, k) and axis parallel to the
^-axis 114
49. Construction of the parabola 117
50. General equation of a parabola, axis parallel to one of co- ■^
ordinate axes 118
CHAPTER VIII
THE ELLIPSE
51. Equation of ellipse, major axis on a>axis, center at origin.. . 120
52. Second focus and directrix 123
53. Latus rectum of ellipse ■ 124
54. Equation of ellipse, major axis on y-asia, center at origin. . . 124
Equation of eUipse, major axis parallel to the a>axis, center
a.t(h,k) 124
Equation of eUipse, major axis parallel to the 2/-axis, center
a,t (h,k) 125
55. Construction of an ellipse 128
56. General equation of eUipse, axes parallel to coordinate axes 129
CHAPTER IX
THE HYPERBOLA
57. Equation of hyperbola, transverse axis on a;-axis, center at
origin 132
58. Latus rectum of hyperbola : . . . 135
59. Equation of hyperbola, transverse axis on y-ajda, center at
origin 135
Equation of hyperbola, transverse axis parallel to the avaxis,
center a,t (h,k) 135
Equation of hyperbola, transverse axis parallel to the y-axis,
center at (h,k) • 135
Viii CONTENTS
Abt. P*°=
60. Construction of hyperbola 138
61. General equation of hyperbola, axes parallel to coordinate
axes 1""
62. Asymptotes to the hyperbola 140
63. Conjugate hyperbolas 142
64. Equilateral or rectangular hyperbola 143
CHAPTER X
TANGENTS AND NORMALS
65. Definitions 146
Equations of tangent and normal at given point on curve . . 146
66. Equation of tangent to the curve represented by general
equation of second degree 150
67. Lengths of tangents and normals, subtangents and sub-
normals 152
68. The equation of the tangent when slope is given 155
CHAPTER XI
POLES, POLARS, DUMETERS AND CONFOCAL CONICS
69. Harmonic division 158
70. Pole and polar 159
71. Diameters 163
72. Conjugate diameters 167
73. Conf ocal conies 169
CHAPTER XII
GENERAL EQUATION OF SECOND DEGREE
74. Removal of xy-term from general equation of second degree 171
75. Test for distinguishing the type of conic when equation con-
tains the xy-term 174
76. Conies through five points 175
CHAPTER XIII
TRANSCENDENTAL AND PARAMETRIC EQUATIONS
77. Loci of transcendental equations 177
78. The exponential curve y = a" 177
79. Relation between natural and common logarithms 179
CONTENTS IX
Art. Page
80. The logarithmic curve, y = logc x 179
81. The sine curve, j/ = sin x 181
82. Periodic functions 183
83. The curve, y = asvuhx 183
84. The tangent curve, y = tan x 184
85. Loci of parametric equations ^ 186
86. Derivation of parametric equations. ; 190
87. The cycloid 190
88. The epicycloid 191
89. The hypocycloid 192
90. The hypocycloid of four cusps 193
91. The path of a projectile 193
92. The witch of Agnesi 194
93. The cissoid 195
CHAPTER XrV
SOLID ANALYTIC GEOMETRY
94. Rectangular coordinates in space 197
95. Distance between two points 198
96. Point of division 200
97. Orthogonal projections 201
98. Polar coordinates 203
99. The angle between two directed lines 205
100. The equation of a locus • 206
101. Cylindrical surface with elements parallel to one of the co-
ordinate axes 207
102. Spherical surface 208
103. Surface of revolution 209
104. Equations of a curve 211
105. The locus of an equation 212
106. Quadric surfaces 214
107. The ellipsoid 214
108. The hyperboloid of one sheet 215
109. The hyperboloid of two sheets 216
110. The eUiptic paraboloid 217
111. The hjrperbolic paraboloid 218
112. The cone 219
113. The normal form of the equation of a plane 220
114. The general equation of first degree 221
X CONTENTS
Abt. Page
115. Plane determined by three conditions 222
116. The equation of a plane in terms of its intercepts 222
117. The angle between two planes 223
118. The distance from a plane to a point 223
119. The general equation of a straight line 225
120. The equations of a straight line through a given point and in
a given direction 225
121. The equations of a straight line through two given points . . 226
122. The projection form of the equations of a straight line 226
123. Direction angles of a line 227
ANALYTIC GEOMETRY
CHAPTER I
CARTESIAN COORDINATES
1. Analytic geometry differs from other geometry mainly
by the introduction of two new ideas : first, that a point in a
plane is determined by its distances from two fixed inter-
secting lines in that plane, and second, that an equation in
two variables completely represents a geometric locus.
These inventions are due to Rene Descartes (1596^1650)
who published his discoveries in 1637. In honor of his
name, this branch of mathematics is often called Cartesian
geometry, and the system of coordinates here used, Carte-
sian coordinates.
2. Directed lines. — On a fixed line X'X, let a fixed point 0,
called the origin, be chosen from which to measure distances.
^ B' o B
It is customary to call distances measured to the right
positive, and those measured to the left negative.
Let some unit of length be applied to OX, and suppose
OB is 7 imits long, then +7 is represented by OB. If the
same measure is applied to OX', and OB' is 6 units long,
then —6 is represented by OB'. Moreover, while OB meas-
ured to the right is 4-7 units, BO measured to the left is
—7 units.
1
2 CARTESIAN COORDINATES
From the above definition, it is evident that ii A, B,
and C are three points on a line, then AB + BC = AC.
A B C
B [ C ^
c .4_J B
Construct three other figures also showing that AB + BC
= AC.
Locate four points A, B, C, and D on a line and show in
three different figures that AB + BC + CD = AD.
3. Position of a point in a plane. Cartesian Coordinates.
— If it is known that a point is located on a given fine, it is
only necessary to know one nvunber in order to locate the
point, namely, that nimiber which represents the distance
and direction from the origin.
If the point is located anywhere in a plane, its position is
fully determined by two numbers.
Let X'X and Y'Y be two straight lines intersecting at 0,
and let the point P be the given
P point. Draw NP and MP through
// P parallel to X'X and Y'Y respec-
_x J/'^ ^ tively.
^'' X The position of the point P is
ftilly determined if MP and NP
are known.
The fine X'X (usually horizontal) is called the x-axis.
The Une Y'Y is called the 2/-axis and their point of inter-
section is called the origin.
X = OM = NP is the abscissa of the point.
The abscissa of a point is its distance from the y-axis meas-
ured parallel to the x-axis.
y = ON = MP is the ordinate of the point.
The ordinate of a point is its distance from the x-axis meas-
ured parallel to the y-axis.
NOTATION
The two intersecting lines are called the coordinate axes
and the two numbers which locate the position of the point,
the Cartesian coordinates of the point.
Abscissas are taken as positive or negative according as
they are measured to the right or left of the origin, and
ordinates as positive or negative according as they are
measured above or below the a;-axis.
4. Rectangular coordinates. — The coordinate axes may
intersect at any angle, but results are usually simpler if
the axes are perpendicular, in which case, Cartesian coor-
dinates are called rectangular coordinates. Cartesian
coordinates when not rectangular are called oblique coor-
dinates.
Unless otherwise specified, rectangular coordinates will
always be used.
5. Notation. — The point whose coordinates are x = a,
and y = b, is usually written P = (a, h) or P (a, b). This
is read " P whose coordinates are a and b."
Variable points are in general represented by P (x, y),
and fixed points by Pi (xi, y,), Pi {x^, 1/2), etc.
To plot a point in Cartesian coordinates choose any con-
venient unit of measure, lay off from the origin on the
X-axis a number equal to the abscissa, and from the extremity
of this hne, and on a parallel to
the 2/-axis, a number equal to the
ordinate.
Thus to plot the point P (— 4, 5),
lay off OM = -4 on OX and draw
MP = 5 parallel to OY.
The use of coordinate paper will
be found to be of decided advan-
tage in a rectangular system of coordinates. Such paper is
constructed as in the figure in which the above point has
been located.
*"
J y
M. C
__ " _i
CARTESIAN COORDINATES
EXERCISES
1. Plot accurately the points: (5, 6), (-2, -3), (0, 2), (-5, 0).
2. Let the axes OX and 07 be inclined at an angle of 45°. Plot the
points given in Ex. 1.
3. Draw the quadrilateral whose vertices are (3, 2), (— 4, 2), (— 4, — 1),
(3, — 1). Prove the figure is a rectangle and find the lengths of its sides.
4. Where are the points whose ordinates are 07 Whose abscissas
are 0? Whose abscissas are 2?
6. On what line will a point lie if its abscissa and ordinate are
equal? If equal numerically but opposite in sign?
6. The origin is the middle point of a line one of whose extremities
is (—2, —3). Find the other extremity.
7. What are the coSrdinates of a point half-way between the origin
and the point (2, 4)? Ans. (1, 2).
8. In a rectangle whose sides are 4 and 3 one of the longer sides is
chosen as the a;-axis and a diagonal as the 2/-axis. What are the coordi-
nates of the vertices and of the middle points of the sides?
9. An isosceles triangle has a base 6 and the equal sides each 5. The
base is taken as the x-axis and the perpendicular from the vertex to the
base as the 3/-axis. Find the coordinates of the vertices.
Find also the coordinates of the vertices if the base and one of the
sides are chosen as axes.
10. What are the coordinates of the vertices of a square whose side is
2 a if the origin is at the center of the square and the axes are parallel to
the sides?
What are the coordinates of the vertices if the origin is at the center,
one axis is parallel to a siide, and the other is a diagonal?
11. What are the coordinates of the vertices of an equilateral tri-
angle each side of which is o, the base being chosen as the x-axis and the
perpendicular to this base through a vertex as the j/-axis?
12. Compute the lengths of the sides of the triangle whose vertices
are (2, 1), (6, 4), and (7, 1). Ans. 5, 5, VlO.
13. Plot the points A (-1, -2) and B (2, 3). Let the horizontal
line through A cut the vertical Une through B in the point C. What are
the coordinates of CI Find the area of the triangle ABC and the length
oiAB.
14. Plot the points A (3, 2) and B (6, 6) and compute the distance
between them. Ans. 5.
15. If two points A (xi, 0) and B (%, 0) are located, show that
AB = Xj — xi whether A and B lie on the same side of the origin or on
opposite sides.
DISTANCE BETWEEN TWO POINTS 5
6. Distance between two points. — The distance d be-
tween two points Pi (xi, 2/1) and Pa (x^, yi) is given by the
formula
d = ^ixa-x,r + (ya-yir. (1)
Proof. Let Pi and P2 represent any two given points,
and let d represent the distance between them.
Wr-^
Draw the ordinates MiPi and M2P2, and through Pi
draw PiN parallel to the a;-axis, meeting at N the ordinate
MiPi (produced if necesary). :^
In the right triangle PiPaiV, PiiV = M1M2 = OM2 - OMi
= X2 — x\ and NPi = MjPz — MiPi = 2/2 — yi-_
Substituting in P1P2 = VPiiV' + NP^we get the formula
d = V{x2-xiy+iy2-yiy.
Note. — The student should notice that the above demonstration
applies equally to the two figures given, and should satisfy himself that
the proof holds good when the points are located in other positions.
Since results remain the same if the positions of the points are changed,
in future demonstrations points will be located in the simplest position
(usually in the first quadrant).
ILLUSTRATrVE EXAMPLES
1. Find the distance between the points A (—2, 2) and B (3, 4).
Solution. — Here Zi = — 2, ^1 = 2, % = 3, 2/2 = 4. Substituting in
the above formula, we have
AB = V(3 + 2)2 + (4- 2)2 = V29.
6
CARTESIAN COORDINATES
2. Find a point equidistant from the three points A (0, 1), B (5, 1),
and C (2, -3).
Solviion. — Let Pi fe, yi) be the required point.
From formula (1)
PiA = V (0 - x,Y + (T^S
PiB = V(5 - x{)^ + (1 - j/i)' ,
P,C = V(2-xi)2 + (3+j/,)2.
Since these distances are all
equal, we can make the two equa-
tions:
Va;i2+2/i2-2 2/i + l =
"^
p
,,'
^
V,
^
^
v
3
>
J'
f
Vxi' - 10 xi + j/i" - 2 yi + 26,
Va;i '+g/i'-2yi + l =
Vxi^ - 4 X. + 2/1^ + 6 2/1 + 13.
Squaring each and collecting,
10 xi = 25,
2 xi — 4 2/1 = 6.
Whence Xi = I, 2/i = — i-
It is evident that Pi is the center of the circle passing through the three
points A, B, and C.
After working each example, the student should examine his figure
carefully and satisfy himself that his answer is reasonable.
EXERCISES
1. Find the lengths of the lines joining the following points:
(a) (-1, -4), (2, 1). Am. Vsi.
(6) (3, 2), (0, -2).
(c) (o, 6), (-a, -6).
(d) (o + 6, a), (6, a + b).
2. Find the lengths of the sides of the following triangles:
(a) (1, 1), (-2, 2), (-3, -3).
(b) (4, 2), (-3, 4), (2, -6).
(c) (a, 0), (0, -a), (a + b, a).
(d) (c + d, 0), (d, c), (d, -c).
3. Prove that the points (-3, 1), (3, 1), and (O, 1 + 3 Vs) are the
vertices of an equilateral triangle.
4. Prove that the points (4, 1), (—1, —4), and (3, 2 V2) are equi-
distant from the origin.
INCLINATION AND SLOPE 7
5. Prove that (3, 1), (2, 4), and (-2, 1) are the vertices of an isosceles
triangle.
6. Prove that the points (4, —3), (5, 4), and (—2, 5) all he on a circle
whose center is (1, 1). Find the radius.
7. Prove that (1, 1), (3, 4), and ( — 5, 5) are the vertices of a right
triangle.
8. Prove that (1, 2), (-5, -3), (1, -11), and (7, -6) are the ver-
tices of a parallelogram.
9. Prove that (0, -1), (3, 2), (0, 5), and (-3, 2) are the vertices of a
square.
10. Find a point on the ^-axis which is equidistant from (4, 0) and
(-2, -2). Ans. (0, 2).
11. One end of a Une whose length is five is at (4, 2); the abscissa of
the other end is 1. Find the ordinate. Ans. 6 or —2.
12. Find the point equidistant from (0, 2), (3, 3), and (6, 2). Ans.
(3, -2).
13. The point (x, y) is equidistant from (2, —1) and (7, 4). Write
the equation which x and y must satisfy. Ans. a; + 2/ = 6.
14. Express algebraically that the distance of the point {x, y) from
the point (2, 3) is equal to 4.
16. The angle between obhque axes is 60°. Find the distance be-
tween the points (3, 5) and (5, 1). Ans. 2 Vs.
Hint. — Locate the points, draw their coordinates and apply the law
of cosines from trigonometry.
16. The angle between oblique axes is w. Find the distance between
the points Pi (ii, yi) and Pa (x2, 2/2).
7. Inclination and Slope. — The angle which one line
makes with another is the angle not greater than 180° measured
counter-clockwise from the second to the first.
Lr
Thus, the angle which the line Li makes with another line
La is the angle <j> in the figure.
8
CARTESIAN COORDINATES
The inclination of a line is the angle which it makes with
the a;-axis. This angle is always measured from the posi-
tive direction of the x-axis. Thus <^ in the figures below
represents the inclination of the line AB.
The slope of a line is the tangent of its inclination.
Formula for slope. — The slope m of a line joining the two
points Pi {xi, yi) and Pi (a^, y^ is given by the formula
m =
Xg — Xi
(2)
Proof. — Let Pi (xi, yi) and P^ (xa, 2/2) be two points on
a hne whose inclination is c^. It is required to find m the
slope of the line.
Draw the ordinates MiPi
and M2P2 and through Pi draw
PiiV parallel to the x-axis cut-
ting MiPi in N. Then angle
= NP1P2 (why?).
From the figure, it is seen
that m = tan </> = tan NPiP^
_NP2_ yi-yy
PiN X2 — Xi
Parallel lines. — If two lines are parallel, their slopes are
equal, and conversely.
R
N
v' /t
/
■ /°
N
1, N
». '
INCLINATION AND SLOPE
9
Proof. — Let ^i and <^2 be the inclinations and mi and jwj
the slopes of the parallel
lines Li and Li. ^^^i
Then <^i = 02 (why?)
and therefore mi = mi.
The proof of the con-
verse is left to the stu-
dent.
Perpendicular lines. —
// two lines are -perpendic-
ular, the slope of one is the negative reciprocal of the slope
of the other, and con-
versely.
Proof. — Let 0i and
4>i be the inclinations
and mi and rm the
slopes of the perpen-
dicular lines Li andZ/2.
Then</.2 = 9O°-|-0i
(why?).
Therefore
tan <^2 = tan (90° + <^i) = — cot <^i = , whence ma = •
tan <^i mi
A similar proof ap-
plies when 01 is ob-
tuse. The proof of
the converse is left to
the student.
aLUSTRATIVE EXAMPLE
Prove that the line
ioining A ( — 1, 1) and B
(1, 5) is perpendicular to
the line joining C ( — 2, 3)
and D (2, 1).
/
L
j
"V
■^
c
t
^
y
/
■^
^
p
'/
■V
>
/
"V
^
/
^
/
10 CARTESIAN COORDINATES
5 — 1
Solution. — From formula (2), the slope of AB = ^ , ^ = 2, and
1—3 —1
the slope of CD = „ - = -^ '
Since either slope is the negative reciprocal of the other, the lines are
perpendicular.
EXERCISES
1. Find the slopes of the lines joining
(o) (2, 5) and (-3, -3);
(b) (3, 2) and (7, -7);
(c) (4, 3) and (-2, 5);
(d) (a, 6) and (-a, 2 6).
2. Find the inclination of the Unes joining
(a) (3, 2) and (-1, -2);
(6) (V3, O) and (0,1);
(c) (0,0) and (l, Vs);
(d) (-4, 0) and (-5, Vs);
(e) (5, 6) and (4, 7).
3. Find the slopes of the sides of the triangle whose vertices are (3, 5)
(6, 2), and (5, 7). Am. -1, 1, and -5.
4. The inclination of AB is 40°. If CD makes an angle of 20° with
AB, find the slope of CD. Arm. a/3. . .
6. Solve Ex. 7 and 8 in Art. 6 by means of formula 2. • .
6. Prove that the diagonals of the square in Ex. 9, Art. 6, cut at right
angles.
7. Prove that (6, -5), (2, -1), (-3, -4), and (-2,-5) are the
vertices of a trapezoid.
8. What is the slope of the line joining (2, 6) and (2, —4)? What
is the slope of any Une parallel to the y-axis?
9. Prove that the Une joining (5, 2) and (6, 4) is parallel to the line
joining (2, 5) and (4, 9) and perpendicular to the line joining (8, 1) and
■ (6, 2).
10. Prove that the triangle whose vertices are (0, 0), (3, 1), and (2, 4)
is a right isosceles triangle.
11. Prove by means of slopes that the figure whose vertices are
(2, 1), (1, 3), (3, 4), and (4, 2) is a rectangle.
12. Prove by means of slopes that the three points (1, 1), ( — 2, —2),
and (3, 3) lie in the same straight line.
13. Three vertices of a parallelogram are ( — 1, —2), (2, 0), and (8, 6)
joined in the order named. Find the fourth vertex. Ans. (5, 4).
POINT OF DIVISION 11
14. A line with an inclination of 60° passes through the origin. If
the ordinate of a point on the line is 6, what is the abscissa of the point?
Am. 2 Vs.
IB. A point is 4 units from the origin and the inclination of the line
joinmg it to (l, V3) is 60°. Find its coordinates. Am. (2, 2 Vs) and
(-2, -2V3).
16. A point is equidistant from the two points (2, —4) and (4, 6){
and the slope of the-|line joining it to ( — 1, 5) is —2. Find its coordi-
nates. Ans. (^,'-'#). ' '
8. Point of division. — If the point P3 is taken any-
where on the line P1P2, P3 divides the line into two segments
P1P3 and Pa-Pa- It will be understood that the segment
P1P3 starts at Pi and terminates at P3 and that the segment
P3P2 starts at P3 and terminates at P^. If both segments
extend in the same direction, the segments are said to have
Pi P3 P
2
Pi P2 P a
the same sign, if in opposite directions, opposite signs.
Thus in the first figure above, the ratio of P1P3 to P3P2 is
positive, while in the second figure the ratio is negative.
The division is called internal or external according as the
point P3 falls between Pi and Pa or on the line produced.
Formula for point of division. — The coordinates {X3, yi) of
the point P3 which divides the line joining the two points
Pi {xi, 2/1) and Pi {xi, 2/2) into segments such that the ratio
p' " = —, are given by the formulas
*8 = — — 77— — > ya — _ ,_ — w
Proof. — Let Pi and P2 be the two given points, and P3
the point which divides the line joining Pi and P2 in the
ratio of ri to r2. It is required to find the coordinates of P3.
12
CARTESIAN COORDINATES
Draw the ordinates MiPi, M2P2, and M3P3. Then since
the three parallels MiPi, MJPi, and MgPa are cut by the
P^ two transversals Miilf 2 and
P1P2, the corresponding
segments are proportional.
Therefore,
M1M3 P1P3 ^ n.
.Ml M.8. Mi
Y'
Solving for Xi, the abscissa of the point of division is
found to be
nxi + r-aXi
MzMi
P3P2
Whence
3:3
- Xi
n
X2
~ X3
Tz
X3
n + :
Similarly, by drawing the abscissas of the three points
Pi, P2, and P3, the student is asked to derive
rm + mi
y'= n + n ■
Coordinates of the middle point of a line. — The coordi-
nates (xs, 2/3) of the middle point, P3, of the line joining the
two points Pi {xi, j/i) and Pa (xa, 2/2) are given by the formulas
«^i + «a
*3 n '
ya
_ yi + ya
(4)
Proo/. — These results are derived immediately from
formula (3) by substituting n = ra.
n-LUSTRATIVE EXAMPLES
1. Find the coordinates of the point dividing the line joining ( — 2, 5)
and (3, 0) in the ratio f .
Here Xi = —2, a^ = 3, yi = 5, 2/2 = 0, n = 3, and tj = 2.
POINT OF DIVISION
13
Substituting in for-
mula (3),
3-3 + 2-(-
Xi =
yt =
-2)
2 + 3
•0 + 2-5
= 1,
2 + 3
2. Divide the line
joining the two points
(-2,5) and (3, 0) in the
ratio 2 : — 1.
Here Xi = — 2, % = 3,
~
*<;
\
N
p
,
N
?2
Y
::::::'^::::ii:::
'i;
:i::::.::s^:±
::::::i^:::^^-::
S,
\
2/1 = 5, 2/2 = 0, i-i = 2, r-2 = —1.
Substituting in formula (3),
2- 3 + (-l)-(-2)
2-1
2-0 + (-l)-5
Xi =
-5.
2-1
The figure shows that P1P3 is twice
P3P2 and opposite in sign, or PiPa :
P3P2 :: 2 : -1.
3. Prove analytically that the line
joining the middle points of two sides
of a triangle is parallel to the third side.
In examples of this class, all points chosen should be represented by
literal quantities so that the proof applies equally to all figures of the
class. The position of the origin and axes should be taken so as to
simplify the work as much as pos-
sible.
Let the base of the given triangle
be represented by a and the altitude
by 6. Taking the side a of the tri-
angle as the a>axis and one extremity
of this base as the origin, the figure
is as shown, and the vertices are
(0, 0), (a, 0), and (c, 6). Let D and
E, the middle points of OB and BA
respectively, be joined by the line DE.
B(c,6)
.A(a,0)
The coordinates of D, the
middle pomt of OB, are found by formula (4) to be f |, gj, and those of
E, the middle point of BA, to be ( — 5 — ' o ) "
14 CARTESIAN COORDINATES
The slope of DE is shown by formula (2) to be
2 2
c -\- a c
2 2
Whence DE is parallel to the ataxia which coincides with OA the base
of the triangle.
EXERCISES
1. Divide the line joining (3, —5) and (6, 2)
(o) in the ratio of | ;
—2
(6) in the ratio of -^•
Plot figure and discuss the position of points in the result.
2. Find the coordinates of the point C which divides the line joining
A ( -3, 4) and B (7, 9) in the ratio f . Check the work by showing that
the distance from A to C is | of the distance from C to B.
3. Find the middle points of the sides of the triangle ( — 1, 3),
(—3, —5) and (3, —1) and compute the lengths of the medians.
4. If the point Pj divides the line joining the points Pi and Pu in a
negative ratio numerically greater than one, will the point Ps be nearer
Pi or Pa? If the ratio is negative and numerically less than one, discuss
the position of Pi.
Find the coordinates of the point which divides the line joining
—3
(-1, 4) to (8, 1) in the ratio -^- Arts. (26, -5).
6. Prove that in the parallelogram whose vertices are (1, 2),
(—5, —3), (1, —11), and (7, —6) the diagonals bisect each other..
6. Prove that in the trapezoid whose vertices are (6, —5), (2, —1),
( — 3, —4), and ( — 2, —5), the line joining the middle points of the non-
parallel sides is parallel to the bases and equal to half their sum.
7. Find the points of trisection of the line joining (—2, —2) and
(7, 4). Ans. (1, 0) and (4, 2).
8. In what ratio does the point (3, —2) divide the line joining ( — 1, 2)
and (5, -4)? Ans. 2 : 1.
9. The middle point of a line is at the point (3, — 2) . One extremity
is ( — 1, —4), what is the other extremity? Ans. (7, 0).
10. The line joining (—4, —2) and (4, 6) is divided in the ratio --^•
Find the distance of the point of division from (2, —3). Ans. 7 Vs.
11. Prove that the lines joining the middle points of the adjacent
AREA OF A TRIANGLE
15
Bides oft he quadrilateral whose vertices are ( — 3, —2), ( — 1, 4), (3, 6),
and (5, —4) form a parallelogram.
12. One extremity of a line is at the point (—2, 3) and the Une is
divided by the point (3, —2) in the ratio |. Find the other extremity.
Am. (7, -6).
13. Find the center of gravity of the triangle whose vertices are
(-1, -2), (3, 4), and (5, -6).
Hint. — The center of gravity is the point of intersection of the
medians and was shown in geometry to be two-thirds of the distance
from any vertex to the middle of the opposite side. Ans. (J, — f).
14. The line AB is produced to C so that BC is equal to twice AB.
A is (5, —4) andBis (3, —2), what are the coordinates of C? Ans.( — 1, 2).
15. The Une joining Pi ( — 1, 3) and Pa (2, 4) crosses the 2/-axis at Pa.
Find the ratio into which Pj divides P1P2. Find the ordinate of Pj.
Ans. h; ¥■
16. Three vertices of a parallelogram are ( — 1, —2), (2, 0), and
(8, 6), joined in the order named. Find the fourth vertex by drawing
the diagonals and appljring the formulas of this article. Ans. (5, 4).
17. Prove analytically that the middle point of the hypotenuse of
any right triangle is equidistant from each vertex.
18. Prove analytically that the diagonals of any parallelogram bisect
each other.
19. Prove analytically that the line joining the middle points of the
non-parallel sides of a trapezoid is equal to half the sum of the parallel sides .
9. Area of a triangle. The area of a triangle whose vertices
are Pi {xi, yi), Pi (x^, yi), and P3 {xz, ys) is given by the formula
Area triangle P1P2P3 =
I [«i (ya - ya) + «a (ys - yi)
+ xa(yi- ya)]- (5)
Proof. — Locate the triangle
whose vertices are Pi, Pi, and
Pi, and draw the ordinates
MiPi, MiPi, and M3P3.
Then triangle P1P2P3 =
MiPiPsMi - MiPiPiMi - MiPiPMz = i [MiMi {MiPi + M3P3)
- MiMi (MiPi + MiPi) - MiMi (MiPi + ilfsPs)] (why?)
= i [{xs - xi) (2/1 + 2/3) -{Xi- xi) {yi + yi) - {xs - Xi) (yi + 2/3)].
16
CARTESIAN COORDINATES
Expanding and collecting, this reduces to
I [xi (j/2 - J/s) + Xi (2/3 - 2/1) + xs (2/1 - 2/2)].
l^J^ ■
^^3'
p^"^ ^
-x: '^^of X
• Pi
Y'
ILLUSTRATIVE EXAMPLE
Find the area of the triangle whose vertices are (1, —1), (2, 3), and
(-2, 1).
Denote (1, -1) by Pi, (2, 3) by P2, and (-2, 1) by P,.
Then from formula (5),
areaPiP^P, = J [1 (3 - 1)+ 2 (1 + D
-2(-l -3)] =7.
It will be noticed that in passing
from Pi to Pi to P3 we go in a coun-
ter-clockwise direction and that the
area hes on the left. In this case
the area is found to be positive.
If the same three points had been
lettered differently, thus. Pi (1, -1), P2 (-2, 1),
and P3 (2, 3), the formula would have given
the result in the form
Area PiPjPs = i [1 (1 - 3) -2 (3 -|- 1)
-|-2(-l-l)]= -7.
That is, if we pass through the points in a
clockwise direction, keeping the area on the
right, the formula gives a negative value to
the area.
In any example, in order to obtain a positive result, the points should
be taken in counter-clockwise order.
It is of decided advantage in remembering the formula
Area P1P2P3 = 5 fe {yi - ys) + xi {y,, - y{)
+ xz{yi -2/2)]
to notice the cyclic order of the subscripts. If the
numbers 1, 2, 3 are arranged in a circle as shown in
the figure it will be observed that the subscripts of x
in the formula follow the cychc order, that is the order
determined by following the arrow heads on the circle.
Also the three subscripts in each term follow this order, starting how-
ever with 1 in the first, 2 in the second, and 3 in the third.
l^x
^3'
p^^ J
/'^^o t X-
Pi
Y'
AREA OF A TRIANGLE 17
EXERCISES
1. Find the area of the triangle whose vertices are
(a) (-1,1), (1,2), (-1,3). Ans. 2.
(b) (0,0), (2, -1), (3, 4).
(c) (a, 0), (a, 6), (c, d).
(d) (6, 6), (-2, 3), (-5, -1).
2. Find the area of the quadrilateral whose vertices are (2, 3),
(-4, 1), (-5, -2), (3, -6). Ans. 42.
3. Prove by means of slopes that the quadrilateral whose vertices
are (2, 4), (3, 0), (5, 3), (4, 7) is a parallelogram and find its area.
4. Prove that the area of the triangle whose vertices are (2, 3),
(—4, —3), and ( — 1, 0) is zero and hence show that these points all lie
on a straight line.
5. The vertices of a triangle are (—2, —2), (4, 7), and (4, —1).
Lines are drawn from the vertex (4, —1) trisecting the opposite side.
Find the area of one of the three equivalent triangles formed. Ans. 8.
6. Are the three points (1, 3), ( — 1, —1), and (3, 7) in the same
straight line?
7. Prove that the Unes joining the middle points of the adjacent
sides of any rectangle form a rhombus whose area is one-half the area of
the rectangle.
8. In a triangle whose vertices are (1, 2), (3, —4), ( — 5, 6), Unes are
drawn joining the middle points of the sides. Prove that area of the
first triangle is four times that of the second.
9. Find the area of the triangle whose vertices are ( — 1, 5), (2, 1),
and (4, 5).
Prove the triangle isosceles, compute the altitude, and determine the
area as one-half the product of the base and altitude, thus checking the
first result.
10. Find the area of the trapezoid whose vertices are (0, 0), (a, 0),
(6, c), and (d, c). Show that this area is the product of the altitude by
one-half the sum of the parallel sides.
CHAPTER II
LOCI
10. Equation of a locus. — One of the most important
functions of analytic geometry is the appUcation of algebra
to geometry.
The two fundamental problems are
(1) To find the equation of a locus, having given certain
geometric conditions.
(2) To plot and discuss the geometric figure or locus which
corresponds to a given equation.
The first of these two problems will be considered in this
article.
The equation of a locus is an equation which is satisfied by
the coordinates of all points on the locus and not satisfied by
the coordinates of points not on the locus.
Sometimes the equation of a locus can be written imme-
diately from the above definition.
Thus, if a line is parallel to the y-axis and 2 units to the
right of it, its equation is a; = 2, for the equation is satis-
fied by the coordinates of every point on the line and by the
coordinates of no point off the Une.
EXERCISES
1. What is the equation of a line parallel to the ai-axis and 3 units
above it? Parallel to the a^axis and 5 units below it?
2. What is the equation of the x-axis? Of the 2/-axis?
3. What is the equation of a line parallel to the j/-axis and 4 units to
the left?
4. What is the equation of a line half way between the lines y = 2
and y = 81
18
EQUATION OF A LOCUS
19
6. Find the equation of the Une half way between the hnes x = —1
and a; = 6.
6. Find the equation of a Una parallel to y = —2 and 5 units above it.
7. What is the equation of the line joining (—2, 5) and (3, 5)? Is
the point (7, 5) on the line?
From the definition of the equation of a locus, it is evident
that a point whose coordinates satisfy the equation of a locus
lies on that locus, and one whose coordinates do not satisfy the
equation is not on the locus.
The steps in finding the equation of a locus are, in general,
as follows :
1st. Construct a figure in which all the given data is
located and let P (x, y) represent the coordinates of any
point on the locus.
2nd. From the figure or from given data, equate two
geometric magnitudes which are known to be equal.
3rd. Replace the geometric magnitudes by equivalent alge-
braic values expressed in terms of x, y and given constants.
Ath. Simphfy the result.
bth. Discuss why the coor-
dinates of all points on the
given locus satisfy the equation
obtained and why the coordi-
nates of all points off the locus
fail to satisfy it.
ILLUSTRATIVE EXAMPLES
1. Find the equation of the
straight Une through the points
(4, 1) and (6, 7).
tst. Plot the known points A
and B and draw the straight Une
through them. Choose P (s, y) any
point on the required locus.
2nd. Since A PB is a straight Une, it is evident that
the slope of AP = the slope gf AB,
\
B
/
J
y
1
'
/
J
J
A
/
3
1
y
X-
/
/
\
'
i
20
LOCI
3rd. The slope of AP =
V
- , formula (2).
7 — 1
The slope of AB = ^ = 3, formula (2).
Whence,
y-1
6-4
= 3.
X — 4
4th. Clearing of fractions and simplifying,
3x-y-n=0.
5th. Since the point P (x, y) was taken as any point on the desired
locus, it is evident that the first condition of the equation of a locus is
fulfilled, viz., that the equation is satisfied by the coordinates of all
points on the loousv
To prove the second condition, viz., that any point not on the locus
does not satisfy the equation, choose any point, Pi, not on the locus and
draw its ordinate crossing the given fine at Pi. Since Pi is on the given
line, its coordinates satisfy the equation, 3x — j/ — 11=0, and we have
after solving for y and substituting,
2/2 = 3x2 — 11.
If the coordinates of Pi are substituted in the same equation we obtain
2/1 = 3 xi — 11.
The second members of these two equations are equal since Xi = xs,
whUe the ordinate j/i is either greater or less than 2/2 according as Pi is
above or below the Une. Hence the equation 3x — y — ll=Ois not
satisfied by the coordinates of Pi (xi, j/i).
Since we have shown that the equation is satisfied by the coordinates
of every point on the locus and by the coordinates of no other points,
it is the desired equation of the locus.
2. Find the equation of a
line through the point (3, —2)
and perpendicular to the line
joining (4, 1) and (2, 2).
Is*. Plot the points A (4, 1)
and B (2, 2) , and draw through
them the line AB. Plot C
(3, —2) and through it draw
a line perpendicular to AB.
Choose P (x, y) any point on
this Une.
2nd. Since the lines CP and AB are perpendicular,
thedopeofCP=-^-j3j^^ij^(Art.7).
^
■
V
}
V
N,
B
f
<.
Vj
\J
-
f^
s
/
^
^
'
1
1
EQUATION OF A LOCUS
21
3rd. Slope of CP = ^-i-^, formula (2).
X -3'
Slope of AB = l^
2 — 4
Therefore ^-i-^ = 2.
X — 3
— J, formula (2).
Y
^'"~-s
z s
7 ^
^^
A ^^
L. J
jz_3,^ ^t T
,, — ^__ ^ ^^,
'=i ^^
4th. Simplifying,
y-2x + 8 = 0.
5th. The proof of this step is left to the student, being in general
similar to that given in illustrative example 1.
3. Find the equation of the locus of the point which moves so that it
is always at a distance 5 from the
point (1, 3).
Ist. Plot the point A (1, 3).
It is evident that the locus is
a circle with center A and radius
5. Let P {x, y) represent any
point on this circle.
2nd. Then AP = 5.
3rd. By formula (1),
AP = V(x - ir + iy- 3y,
whence
V{x - ly + {y~ 3r =5.
ith. Squaring, expanding, and collecting,
a?-2a; + ?/2-62/-15=0.
5th. The proof of this step is left to the student.
EXERCISES
1. Find the equation of the locus of the point for which the ordinate
is always three times the abscissa.
2. Find the equation of the line through the point (2, 3) and with
inchnation of 120° Is the point (5, 6) on the line?
3. Find the equation of the line through (1,2) and ( — 3, — 4) . Check
work by showing that the coordinates of these points satisfy the equation.
4. A point moves so that its distance from the point ( — 1, 2) is
always equal to its distance from the origin. Find its equation.
6. Find the equation of the straight line passing through the middle
point of the Une joining (2, —7) and (10, 5) and making an angle of 45°
with the X-axis.
22 LOCI
6. Find the equation of the straight line through the point ( — 1, 5)
and parallel to the hne joining (1, 3) and ( — 5, 5).
7. Find the equation of the straight hne perpendicular to the hne
joining the two points (2, 1) and (5, 4) and dividing the distance between
them in the ratio of 2 to 1.
8. Find the equation of the straight line through the point (1, 2) and
with slope |. Find the ordinate of the point on the line for which the
abscissa is 0, and thus find where the Hne crosses the y-ajda. Similarly,
fiind where the hne crosses the a^axis.
9. Find the equation of the straight hne perpendicular to the hne join-
ing the points ( — 2, 1) and (6, —3) and passing through its middle point.
10. Find the equation of the locus of a point which moves so as to
be always equidistant from the two points ( — 2, 1) and (6, —3). Prove
that this is the perpendicular bisector by showing that its equation is
the same as that in Ex. 9.
11. Find the equations of the following circles:
(a) center (0, 0), radius 4;
(b) center (3, 2), radius S;
(c) center (a, 6), radius c.
12. Find the equation of the circle whose center is (2, 3) and which
is tangent to the x-axis.
13. Find the equation of a circle whose radius is 5 and whose center
is the middle point of the line joining ( — 1, —3) and (3, 7).
14. Find the equation of the circle in the first quadrant which is
tangent to both axes and whose radius is 2.
16. Find the equation of the circle whose center is (1, 3) and whose
circumference passes through the point (—3, 0).
16. Find the equation of the circle of radius 3 which is tangent to the
y-sjds at the origin.
17. Find the equation of the circle whose diameter is the line joining
the points (.5, -7) and (3, -1).
18. Find the equation of the circle whose center is the middle point
of the hne joining ( — 1, 6) and (5, 2) and whose circumference passes
through (1, 1).
11. The locus of an equation. — In the last article,
equations of loci were derived from geometric data given.
The second problem of analytic geometry, viz., to plot and
discuss the geometric figure which corresponds to a given
equation, will now be considered.
PLOTTING THE LOCUS OF AN EQUATION
23
This second problem divides itself into two parts, plotting
the locus of an equation and discussing an equation.
12. Plotting the locus of an equation. — A pair of coordi-
nates X and y locate definitely one point in a plane. If,
however, these two coordinates must always satisfy a given
equation, then a series of points may be chosen, the coordi-
nates of each of which satisfy the given equation, for to each
value of one variable corresponds one or more values of the
other and hence an infinite number of points may be located.
Thus, if in the equation y = ^ > we give to a; a
series of values
differing by
unity, we obtain
x=-2.
2/ = 4;
x= -1,
2/ = i;
x = 0,
y = Q;
x=l,
2/=-l;
x = 2,
2/ = 0;
x = 3,
2/ = t;
4,
2/ = 4.
Plotting, the points are as shown above.
It will be noticed that these points are not located indis-
criminately over the plane, but apparently all lie on a curve
as drawn.
More points on the curve may be obtained by giving frac-
tional values to x, between those already used, and thus a
more perfect approximation to the correct curve be obtained.
This curve is called the locus of the equation.
The locus of an equation is a curve which contains all the
points whose coordinates satisfy the equation and no other points.
In examples hke the preceding, it is generally best to
solve for y in terms of x, but in particular examples, it may
be convenient or even necessary to solve for x in terms of y.
24 LOCI
If two variables are so related that when the first is
given, the value of the second is determined, then the
second is said to be a function of the first.
That variable to which values are arbitrarily assigned is
called the independent variable and the other the dependent
variable.
If the two variables are connected by an algebraic equation,
that is, by one which contains functions which are the result
of a finite number of algebraic operations, such as addition, sub-
traction, multiphcation, division, involution, and evolution,
either function is said to be an algebraic function of the other.
An illustration is given by y = x* — 8x^+1 or by
xY + 2xy^ = 7.
In many cases of great importance, the equation con-
necting the variables is not algebraic, in which case one
variable is said to be a transcendental function of the other.
Examples are, y = log x, y = e'', y tan~i x = 3.
The present chapter will be concerned with plotting and
discussing algebraic equations only. Transcendental equa-
tions will be considered in a later chapter.
The locus is sometimes evident directly from its equation.
For example, find the locus of the equation x = 2. Since
no mention is made of y, the ordinate is unrestricted. Our
problem then is to find a locus for which x is always 2, while
y may have any value whatever. Such a locus is the fine
parallel to the y-aids and 2 units to the right of it.
EXERCISES
1. What is the locus oi y = 57
2. What is the locus oi x = -8?
3. What is the locus oi y = xt
4. What is the locus oi y = — x?
In general, the locus of an equation will be determined by
the process called plotting. The steps are as follows :
PLOTTING THE LOCUS OF AN EQUATION
25
1st. Solve for one variable in terms of the other.
2nd. Assign values to the independent variable and compiUe
those of the dependent variable, expressing the results in the
form of a table.
3rd. Plot the points thus obtained and connect with a smooth
curve.
aLUSTRATIVE EXAMPLES
1. Plot the locus of the equation x -{- 2 y = i.
1st. Solving for y in terms of x,
4: — X
2nd. It is convenient when assigning values to x ana computing the
corresponding values of y, to state the result in the form of a table.
t
V
X
y
2
2
1
f
-1
J
2
1
-2
3
3
4
-3
i
4
etc.
5
-\
"-;
r
V
K
^
"N
s
^
^
N
s
■V,
^
>
''
3rd. Locating the points obtained
and connecting by a smooth curve,
the figure is approximately as shown.
It will later be demonstrated that every equation of first degree be-
tween two variables represents a straight line, a fact which corresponds
with the appearance of this figure.
' 2. Plot the locus of the equation
4 x" + 9 2/« = 36.
1st. Solving for y in terms of x,
2/ = ± f V9 - xK
2nd. It will be noted that for values of x numerically greater than 3,
y is imaginary and therefore no points of the curve can be constructed
for which x is greater than 3 or less than —3. In making our table,
values are taken between —3 and +3.
26
LOCI
-
/
^
"^
s.
Vm
r
\
V
J
\
<,
,--
->
/
■1
z
v
X
-1
-2
-3
V
1
2
3
±2
±fV2 = ±1.8
±|V5 = ±1.4
±2
±fV2=-±1.8
±f V3 = ±1.4
3r<2. Plotting the points and construc-
ting a smooth curve through them, the locus is approjdmately as drawn.
3. Plot the locus ofj/'-2a;-y = 0.
\&i. While, in general, it is better to solve for y in terms of x, in this
example it is necessary to solve for x in terms of y. This gives
„ _t -V
2nd. Assigning values to y and computing x,
V
z
V
X
1
-1
2
3
-2
- 3
3
12
-3
-12
etc.
etc.
It will be noticed that in the table above there are three values of y
which give z = 0, and therefore three points are located on the ^-axis.
In order to draw the curve more accurately in the vicinity of these
points it is advisable to give to y the fractional values — \ and -f-J. Two
additional points (j^, —J) and (— j^, 5) are thus determined.
Y
_____-=.
_^„ = = ,
iz i^ ,
_|. ^j. _|^
-~'V^
- — -====
Y
Zrd. Locating the poi^its and drawing a smooth curve through them,
the figure is as shown.
DISCUSSION OF AN EQUATION
Sometimes, as in the equation i? + y^ + 15 = 0, there are no^
values which satisfy the equation. In this case there are no real points
on the locus.
Again, an equation may be satisfied by the coordinates of one point
only, in which case there is only one real point on the locus. Such loci
are called point-loci. An illustration is the equation a? + y^ = 0, the
locus of which is the origin.
EXERCISES
Plot the locus of each of the following equations:
1. y+2x-S=Q.
2. s2 + j^ = 4.
3. a^ - j/« = 4.
i. y = aS.
6. ^ = x'.
6. a? + 4j/2 = 16.
7. 3? -\y^ = 16.
8. 2:2 + 43/' =0.
9. ia? +8x = iy -5.
10. 2x-y = 12.
11. 2/2 = 8a; + 8.
12. a;2 + 2/2 = 16.
13. y = x>>-2.
13. Discussion of an Equation. — The method of deter-
mining loci by plotting separate points is in general satis-
factory in simple examples, but in those in which the equa-
tions are somewhat complicated the work is often long and
the results more or less inaccurate. These difficulties are
lessened in many cases, by making a study of the properties
of the curve by means of the discussion of its equation.
The properties which will be discussed are as follows:
1. Intercepts.
2. Symmetry.
3. Extent.
4. Asymptotes.
Intercepts. — The intercepts of a locus are the distances
from the origin to the -points where it cuts the axea
14.
a2 = 82/.
16.
a2+2a;-l=2/.
16.
42/2 -9a? = 36.
17.
4s2=2/'.
18.
a;2 + 2/2 = 25.
19.
2/2+6a;=0.
20.
2 ^ = x* — a;.
21.
2/2 = (1 - a:) (x + 3).
22.
2S + 52/ + 2 =0.
23.
2/2 = a; (3 - a;).
24.
2/2 = 8-8a;.
26.
a;2 + 4a; + 3 + 42/ =0,
26.
2/2 = a? - 1.
28
LOCI
Thus in the figure, the a;-intercept is OA, which is the
abscissa of the point A, i.e., of the point on the locus whose
ordinate is 0. It is then evident that to find the a;-intercept,
substitute 2/ = in the equation,
and solve for x.
Likewise to find the i/-inter-
cept, substitute a; = and solve
for y.
a;2 + 4 a; _ 3 = 0.
rLLUSTRATIVE EXAMPLE
Find the x and y intercepts of the curve 2y -
1st. Let 2/ = 0, then x^ - 4 a; + 3 = 0.
Whence, x = 1 or 3. The x-intercepts are therefore 1 and 3.
2nd. Let x = 0, then 2 y — Z = 0, ov y = \, the 2/-intercept.
Symmetry. — Two 'points are symmetrical with respect to
a line if that line is the perpendicular bisector of the line joining
the two points.
If A and A' are symmetrical with respect to the a;-axis,
then if the coordinates of A are (x, y)
the coordinates of A' are {x, —y).
Similarly, the point symmetrical to A
with respect to the y-axis is (— x, y).
A curve is symmetrical with respect to
a line if the curve is made up of pairs of points symmetrical
with respect to the line.
If a locus is sjmametrical with respect
to the X-axis, there i s a point {x,—y)on the
cm-ve corresponding to every point (x, y)
on the cinrve. The coordinates (x, —y)
must therefore satisfy the equation of
the curve, i.e., y can be replaced by — y
and the equation remain imchanged.
The equation of the cmve here sketched is j/^ = 4 x.
Replacing y hj —y in this equation, the result is (—yY =
4 X, which is the same as the given equation, y^ = ix.
Iv
DISCUSSION OF AN EQUATION
29
By a similar discussion, it may be shown that whenever x
can be replaced by — x without causing any change in the equa-
tion, then the locus is symmetrical with respect to the 2/-axis.
In the equation a;^ + j/^ = 9, a; can
be replaced by —x, and y by —y,
therefore the curve is symmetrical x" '-■-SI
with respect to both axes.
The method of replacing x hj —x
and yhy—y applies equally well in testing for symmetry in
either algebraic or transcendental equations. It is evident,
however, that in case of algebraic equations, if there are no
odd powers of y, then y can be replaced by —y, and the locus
is symmetrical with respect to the a;-axis, while if there are
no odd powers of x, then x can be replaced by —x, and the
locus is symmetrical with respect to the y-axis.
Two points are symmetrical with respect to the origin if the
origin bisects the line joining the two points.
If A and A' are symmet-
rical with respect to the
origin, then when the co-
ordinates of A are {x, y)
the coordinates of A' are
{-X, -y).
A curve is symmetrical
with respect to the origin if the curve is made up of pairs of
points symmetrical with respect to the origin.
From this definition, it is readily seen that the curve is
symmetrical with respect to the origin if the equation re-
mains imchanged yhen x and y are replaced by —a; and
—y respectively.
An algebraic equation always represents a locus symmet-
rical with respect to the origin if each term is of odd degree
or if each term is of even degree. A constant term is con-
sidered of even degree.
30
LOCI
Discuss each of the following equations for intercepts and
symmetry and plot the loci:
1. y-x = 3. 3. a;!' = 4 7/ + 4.
2. 2/" = 4 a;. 4. y = x*.
Extent. — In order to find how far the curve extends
left and right from the origin, the equation is solved for y
in terms of x and the values of x which make y real are then
determined. If y is equal to an integral expression in x,
or if the radicals involved are all of odd index, y is real for
all values of x and the curve extends indefinitely left and
right from the origin.
For example, the loci oiy = 2x + l and y = 5 + "^5 x'^ + x
extend indefinitely left and right from the origin.
When y involves a radical with even index, values of x
which make the quantity under the radical negative must
be determined, as foi* these values y is imaginary, and conse-
quently there are no points corresponding to such abscissas.
Similarly, the extent of the curve above and below the
a;-axis may be determined by solving for x in terms of y.
The method of determin-
ing the extent of a curve is
made clear in the following.
n.LnSTRATIVE EXAMPLES
1. Discuss for extent
2/» + 2j/+3z = 3.
Solving for y,
y= -l±V4-3a;.
It is often helpful to make
the coefficient of x in factors of
first degree either +1 or —1,
thus
y= -l±V3(f-a;).
It can now be readily seen that if a; is greater than i, yw imaginary
and therefore there are no points on the curve to the right of the Une
>
*•
IS
f
\
1
/
y
•
DISCUSSION OF AN EQUATION
31
a; = f . Since y is real for all values of x less than f , therefore the curve
extends indefinitely to the left of that line.
Solving for z in terms of y,
2/g + 2;/-3
X ^
Since x is real for all values of y, therefore the curve extends in-
definitely above and below the x-axis.
Plotting a few points the curve is found as shown.
2. Discuss for extent 2/* + 4 a;^ - 16 a; + 12 = 0.
Solving for y, y = ±2 V— a? + 4a; — 3.
Factoring the expression under the radical,
2/ = ±2 V(a;-l).(3-a;).
It is evident that the first factor is negative when x is less than 1, and
positive when x is more than 1, also that the second factor is negative
when X is more than 3 and positive when x is less than 3.
The product then is positive when x is greater than 1 and less than 3
and therefore y is real for 1 = a; = 3. The product is negative when x
is less than 1 or more than 3 and y is then imaginary.
The whole curve therefore lies between the Unes a; = 1 and a; = 3.
Solving for x in terms of y,
4± V4 - 2/2
Since x is real when —2 = y=2 and is imaginary for all other values,
therefore the whole curve lies between the lines y = —2 and y = +2.
Y "■■■
l.\
X- Q X
)^2
v'
In plotting points it is only necessary to use values of x from 1 to 3.
The figure is foimd to be as shown.
3. Discuss for extent y' + iy - 2a^ + 6x + 1 =0.
Solving for 2/, 2/ = -2 ± V2 a;^ - 6 x + 3.
32
LOCI
If the same plan were followed as in Ex. 2, the expression 2 x* — 6 a;
+ 3 should now be resolved into factors. These factors are not evi-
dent. From a theorem in algebra, aa? + 'bx -\- c = a(x — Xi) [x — xn),
in which Xi and % are the roots of the equation ax^ + bx + c = 0.
To apply this method here, solve 2x^ — Qx + B=0, obtaining
X = ?-^^ = 2.3 + or .6 +.
Whence, the factors are 2 (x — 2.3+) (x — .6+), an expression
which can readily be seen to be negative for every value of x between
.6+ and 2.3+ and positive for all other values. Hence y is imagi-
nary when .6 + < X < 2.3 + and real when x = .6+ or x = 2.3+ .
The fact observed here is universally true, viz., that a quadratic ex-
pression has the same sign for every value between its roots, and the
opposite for all other values.
This fact is of importance as by its use the process of determining the
sign of a quadratic expression may be shortened.
In order, then, to determine the sign of a quadratic expression, find its
roots, then substitute in it some value between the roots and determine
the sign. The expression has this sign for all values between the roots
and the opposite sign for all other values.
Thus, in the above problem, substitute in2x2 — 6x + 3 any number
between the roots .6+ and 2.3+ such as x = 1. The sign is found to be — .
Hence the expression is negative for all values of x between the roots.
Solving for x,
„_ 3±V2j/'+8y + ll
^~ 2
The roots oi2y^ + 8y +
11 = are found to be im-
aginary. From a principle
in algebra it is known that
when the roots of the quad-
ratic equation ax' + bx + c
= are imaginary, the ex-
pression ax^ + 6x + c is posi-
tive for all values of x (if o
is a positive niunber).
Hence x is real for every
value of y.
That 2y' + Sy + His always positive can also be shown as follows:
22f' + 82/ + ll =2{y' + 4:y+^) = 2 [(2/ + 2)« + f ],
an expression which is always positive;
~s
V
Y
/
t —
\
/
/
V
\°
/
^
V
r
1
)
'
J
V
\
/
\
./
Y
>
\
DISCUSSION OF AN EQUATION
33
Discuss each of the following equations for intercepts, symmetry, and
extent and plot the loci:
1. x^+'f = 36.
2. 4a? + 92/2 =36.
3. 4a? -92/2 = 36.
4. a;2-4a; + 42/-8=0.
6. a?-2a; + 42/2-82/ + l=0.
6. x2 + 2/i' + 2a; + 22/-l=0.
Asymptotes. — If, as a point generating a curve recedes
indefinitely, the curve approaches coincidence with a fixed
straight line, the line is
called an asymptote to
the curve.
At this time, only
those asymptotes which
are parallel to the axes
will be considered.
If OA = a, then the
asymptotes in the ad-
joining figure are a; = 0,
X = a, and y = 0.
It will be noticed that
as X approaches either zero or a, y increases indefinitely;
also, that as y approaches zero, x increases indefinitely.
This fact leads readily to the method of finding vertical and
horizontal asymptotes, viz., solve for one variable in terms of
the other and determine those values of the second variable
for which the first is infinite.
Thus, find the vertical and horizontal asymptotes of
xy + 2x — y = 0.
Solving for y, 2 -^
y = T^x
As X approaches 1, y approaches infinity, therefore a; = 1
is a vertical asymptote.
Solving for x,
X =
y
y + 2
34
LOCI
As y approaches —2,x approaches infinity, therefore
2/ = —2 is an asymptote.
In plotting the locus, care must be taken in determining
points near the asymp-
totes. Thus one or
more points should be
plotted between a; =
and x=l, also between
X = 1 and x = 2. It
will be observed that
in the locus of an alge-
braic equation, there
can be no asymptotes
parallel to the axes,
imless when one variable is expressed in terms of the other
the result is a fraction with a variable denominator.
The process of determining intercepts, symmetry, extent,
and asymptotes involves the following:
Steps in discussion of an equation
\st. Let 2/ = 0, solve for x, thus finding the x-dntercept.
Let X = 0, solve for y, thus finding the y^ntercept,
2nd. In an algebraic equation observe:
If no odd powers of y are present the locus is symmetrical
with respect to the x-axis.
If no odd powers of x are present the locus is symmetriOil with
respect to the y-axis.
If every term is of odd degree or if every term is of even de-
gree the locus is symmetrical with respect to the origin.
3rd. Solve for y in terms of x and find what values of x
make y imaginary. Points hamng these values as absdssas
are excluded from the locus. Find what values of x make y
real. Points having these values as abscissas are on the
locus.
DISCUSSION OF AN EQUATION 35
Similarly, solve for x in terms of y and determine the valves
of y which make x real or imaginary.
Ath. Determine asymptotes parallel to the axes by finding
those finite values of either variable which make the other in-
finitely great.
ILLUSTRATIVE EXAMPLES
1. Discuss the equation j/^ = (a; + 2) (x + 1) (a; — 2) and plot the
locus.
1st. Let 2/ = 0, then x = —2, — 1, and 2, the intercepts on the avaxis.
Let a; = 0, ^ is imaginary and hence the curve does not cut the y-axis.
2nd. No odd powers of y are present, hence the curve is S3nnmetrical
with respect to the a;-axis.
Odd powers of x are present, hence the curve is not symmetrical with
respect to the y-axis.
The terms are partly of odd degree and partly of even degree, there-
fore the curve is not symmetrical with respect to the origin.
3rd. Solving for y in terms of x,
2/ = ± V(a; + 2) (a; + 1) [x - 2).
Placing the quantity under the radical equal to zero, the roots are
foimd to be —2, —1, and +2. For values of a; < —2, the quantity
under the radical is negative and hence y is imaginary. Points to the
left of the line x = —2 are therefore excluded. For values of x more
than — 2 but less than — 1, the quantity under the radical is positive and
hence y is real. Part of the curve then lies between the Unes a; = — 2
and a; = — 1. For values of x more than —1 but less than +2, the
quantity under the radical is negative and y is imaginary. Points be-
tween the Unes x = —1 and x = +2 are therefore excluded. For
values of x greater than +2, y is real and the curve extends indefinitely
to the right of the hne x = +2.
In attempting to' solve for x in terms of y, it is observed that the
equation is of third degree in x. It is not usually convenient to solve an
equation of third or higher degree. By remembering, however, that
every equation of odd degree has at least one real root, it is seen that this
curve extends indefinitely both above and below the a;-axis.
4JA. There are no asjrmptotes parallel to the ^-axis, since the variable
does not appear in the denominator of the equation in the third step.
In plotting the locus, it is noticed that the intercepts have already de-
termined three points of the curve on the a>axis. From the third step
36
LOCI
of the discussion, it is observed that it is only necessary to assign values
to X between —2 and —1, and values greater than +2.
Thus,
X
V
-2
-f
±0.93+
-1
+-2
1-3
±4.4+
1-4
±7.7+
4
(
r
/
/
/
— X
r\
X—
u
\
\
\
■
'
V
In every problem of this article, care should be taken to see that the
locus agrees fuUy with the discussion.
2. Discuss the equation (x — 2) j/* = a; and plot the locus.
1st. Let y = 0, then x = 0, the intercept on the a>axis.
Let a; = 0, then ^ = 0, the intercept on the y-snaa.
2nd. No odd powers of y are present, hence the curve is symmetrical
with respect to the avaxis.
Odd powers of x are present, hence the curve is not symmetrical with
respect to the 2/-axis.
Terms are partly of odd and partly of even degree, hence the curve is
not symmetrical with respect to the origin.
3rd. Solving for y,
= -v/A-
It is seen that the numerator is positive when x is positive and nega-
tive when X is negative, also that the denominator is positive when x is
greater than 2 and negative when x is less than 2, and hence y is imaginary
for values of x between and 2 and real for all other values. Therefore
no part of the curve lies between the lines a; = and x = 2.
Solving for x,
x = ^^.
2/'-l
DISCUSSION OF AN EQUATION
Whence x is real for every value of y.
37
4ih. In the equation 2/ = ± V 5, ii x = 2, y = infinity, there-
1 X — ^
fore c = 2 is an asymptote.
2 y'
In the equation x = , _ . , H y = ±1, a; = infinity, therefore
2/ = — 1 and y = +1 are asymptotes.
Plotting the locus:
t
v
Z
V
2
±00
1
±V5
-1
±V|
3
±V3
-2
±vi
4
ta —
±V2
-3
ip
\
\
^
V
/
t'
/
■ !
EXERCISES
Discuss and plot the following:
1. a^ = 4 2/.
2. a;2 + j/! = 6.
3. y = 2a?.
4. (a: -1)2/ =2.
6. 25 a;2 + 9 2/" = 225.
6. 9 a? -162/' = 144.
7. a;2/ = 16.
8. xy^ = 9.
9. a? = 4 2/'.
10. x'y — y = X.
11. 2/2 = 3x-9.
12. y = {x-2){,x- 3).
13. 2/2 = a;3 + a?.
14. y^ = x^ -3?.
16. y = 3? — X.
16. a:'2/ - 2/ + 6 = 0.
38 LOCI
17.
^r^ —'f=X.
18.
xy — x^—y =0.
19.
y{x - 2Y = 1.
20.
t = x(x - 3) (a - 5),
21.
3? = yiM- 1) (2/ - 2)
22.
3? + 2x + lQy-S =
23.
(y - 1)» = (x- 2y.
24.
3?+xy + y'=3.
26.
y(Sx-2) =2x + 4.
26.
y(,x - 1) (X - 4) = 1.
27.
3/ + ya? = X.
28.
a?-4«2+4x = 0.
29.
a? + 2/2-4x =0.
30.
y^ = 12(4 - x).
31.
j/».= x(x-l) (a;-2)(x-3),
32.
x^ + Z = &xy.
33.
a? + 2x + y^+iy-20 = 0.
34.
!fi+2y'-^x+'ty-W = 0.
36.
y{x - 1) (X - 3) = X + 1.
36.
3^2/ - 2/ + 1 = 0.
37.
2/'(4 + x») = l.
38.
2, = (X - 1) (X - 2) (X - 3).
39.
x2_42/'-2x + 83^-7=0.
40. ai!'-a^ + 2x-9 =0.
In some equations, the constants are represented by
letters instead of by figures. There will be a different locus
for each value given the constant, but it will be seen that
these loci have properties common to all.
Discuss the equation ay^ = x{x — 2a)^ and plot the locus.
\st. Let a; = 0, then y = 0, y intercept.
Let y =.0, then a; = and 2 a, x intercepts.
2nd. No odd powers of y are present, therefore the locus
is symmetrical with respect to the a;-axis.
Odd powers of x are present, therefore the locus is not
symmetrical with respect to the 2/-axis.
Terms are partly of odd and partly of even degree there-
fore the locus is not symmetrical with respect to the origin.
3rd. Solving for y,
y = ±{x
-2a)s/l-
y is imaginary when x has a sign opposite to a, and real
when X has the same sign as a, therefore the curve is entirely
to the right of the y-axis when a is positive and entirely to
the left when a is negative.
It is not convenient to solve the third degree equation for
X in terms of y, but as every equation of odd degree has at
DISCUSSION OF AN EQUATION
39
least one real root, the curve extends indefinitely above
and below the a;-axis.
4:th. There are no asymptotes parallel to the axes.
In computing the coordinates of the points, it is best to
assign to x values which are multiples of a, and to assign to
a any convenient length when plotting. The figure for a
positive and equal to 2 is shown.
X
V
a
±a
2a
3a
±aV3
4a
±4 a
-X
?::
5;
EXERCISES
1. Discuss the following equations and plot the loci:
(a) 2/^ = 4aa;.
(6) 4a;2+93/' = 36aK
(c) ay' = X*.
(d) 2/2 = 4o(o -x).
(e) 2/2 = a?(a2 - x").
(/) 2/ + (x-3o)' = 0.
ig) a^ + 2/2 = 4 a2.
(h) a? -2/2 = 4o2.
(i) ^ = (a;-2o)(a;-3o).
0') aV = o^a^ — x".
(k) y^a? - x2) = o2x2.
© aY =a;2(2a - x).
(to) o'x = 2/^(2/ — 2 o).
(n) 2/2(2 a - x) = x«.
(o) 2/(x2 + 4a2) = 8 a'.
(p) 9 02/2 = (x - 2 a) (x-5a)-.
(g) 27a2/2 = 4(x-2a)'.
(r) x2y2(a;z _ 0,2) = ajo.
2. Show that the following equations either represent point-loci or
have no loci.
(a)x2 + l=0. (e) (X -2)2 + (2/ -3)2 + 1 =
(6) 2,2 + 4 = 0. (f) (a: + 1)2 + (2/ + 3)2 = 0.
(c) x2 + 82/2 + 2=0. (ff) x2+2x+2/'+6 2/+16=0.
(d) a2 + 42/2 = 0. (^) (X- 1)2 +2/2 + 4=0.
40
LOCI
3. Find the equations of the following loci, and discuss and plot them.
(o) A point moves so as to be always equidistant from the a;-axis and
the point (0, 3).
(6) A point moves so that the square of its distance from the origin
is four times its ordinate.
(c) A point moves so that its distance from the y-axis is equal to its
distance from the point ( — 2, —4).
(d) A point moves so that its distance from the line x = 2 is equal to
its distance from the point (4, 1).
(e) A point moves so that its distance from the point ( — 1, —3) is
twice its distance from the point (2, 1).
(/) A point moves so that the smn of its distances from (4, 0) and
(—4, 0) is equal to 10.
(g) A point moves so that the difference of its distances from (0, 5)
and (0, —5) is equal to 6.
14. Points of intersection of two curves. — If two equa-
tions are given, the loci will, in general, be two distinct
curves. These may
or may not inter-
sect. If the ciu-ves
intersect, they have
one or more points
in common. It fol-
lows from the defini-
tion of the locus of
an equation that a
point hes on two
curves if and only if
its coordinates sat-
isfy each equation.
In order then to
find the coordinates of the points of intersection, it is neces-
sary to find those values of x and y which satisfy the two
equations, that is, the two equations must be solved simul-
taneously.
If there are no real roots, the curves do not intersect.
POINTS OF INTERSECTION OF TWO CURVES 41
For example, find the intersection of the curves x^ —
4 a; + j/2 = and x^ - y^ = 6.
Solving simultaneously, a; = —1 or 3. When a; = 3, !/ =
±V3, when x = — 1, y is imaginary. Thus, there are but
two real intersections (S, +V3) and (3, — Vs). See figure
on opposite page. r
, " EXERCISES
Find the points of intersection and plot the following curves:
1. a;2 + y2 = 100 and 2 s - 2/ = 4.
2. 3/2 = 4 a; and x^ + j/^ ,= 5.
3. 4x+2/-5 = and 7a;-32/-4 = 0.
4. a; + 2/ = 6 and '^ = %x.
6. xy = 12 and x* + j/' = 25.
6. 2/ = x* and y = x.
7. 4 x" - 2/2 = 7 and 3 X + 2 y = 12.
8. x2 = y + 2 and 2 X + 3 2/ = 10.
9. x2 + 2/2 = 25 and 3 x^ - 2 2/2 = 30.
10. y^ = x^ — x^ and 2 2/ = x.
11. 2/^ = 4 ax and x + 2/ = 3 a.
12. 62-C2 + a22/2 = 0252 and 6x + 02/ = 06.
13. Find the distance between the points of intersection of x2 = 4 2/ +4
and X = 2/ + 1- •A'^- 4-v'2.
14. Do the curves 0? -\- y"^ = ^ and ^2 = a; — 4 intersect?
15. Find the points of intersection of the loci x2 + 1/2 = 9 and
2/ = X + 6. For what values of 6 are these intersections real and dis-
tinct? imaginary? coincident?
16. Find the area of the triangle whose sides have the equations
22/ - 3x + l = 0, 42/ + 3x + 11 =0, and2/ + 3x - 4 = 0. Am. 9.
17. Find the area of the polygon whose sides have the equations
X = —4, y — 3, y = 2x + l, and y = —2.
18. Show that the three loci x' + y^ = 25, y — x + 1 = 0, and
2/ — 2 X = 2 pass through a common point.
19. Find the slopes of the sides of the triangle formed by the hnes,
x = 8, x + y — 3, X — 2y = 6.
20. Prove that the quadrilateral whose sides have the equations
2/ = 4, 2/ = —2, X — 2 2/ = 6, and x — 2y = —6 is a parallelogram.
21. The equations of the sides of a triangle are 2 x + 4 2/ = 2,
X — 3 2/ =p 6, and 12 y + x = Q. Find the lengths of the medians.
42
LOCI
15. Locus by factoring. — IJ an equation whose second
member is zero has for its first member the product of variable
factors, then the locus of the equation is found by setting each
factor equal to zero and plotting the resvdt.
Proof. — Let u and v represent any two functions of
X and y. The given
equation can be writ-
ten
uv = 0, (1)
and the equations
formed by setting the
factors separately
equal to zero are
M = (2)
and V = 0. (3)
Assume the loci of (2) and (3) to be the figures as shown
above.
To prove the proposition it will be necessary to show:
1st. that the coordinates of any point on the locus of
equation (2) or (3) satisfy equation (1) ;
2nd. that the coordinates of no other points satisfy
equation (1).
Let Pi {xij yi) represent any point on the locus of (2),
then the coordinates (xi, yi) must satisfy the equation u = 0.
If the same coordinates are substituted in the equation
uv = 0, the equation will be satisfied, since one of the
factors is zero and consequently the product is zero.
Similarly, the coordinates of P^ (a^, j/a) any point on the
locus of equation (3) can be shown to satisfy equation (1).
Let Pi {xs, Vi) represent any point not on either locus. The
coordinates of this point will not satisfy equation (1) since
neither factor is zero and consequently the product is not
zero.
LOCUS BY FACTORING
43
It has therefore been shown that uv = has for its locus
the combined loci of w = 0, and d = 0, since the locus of an
equation is the curve which contains all the points which
satisfy the equation and no other points.
ILLUSTBATIVE EXAMPLES
1, Plot the locus of the equa-
tion 2x!'-3xy + x + y^-y = 0.
Grouping the second degree
terms together, the equation can
readily be factored thus:
(2x' - 3xy + t/') + {x - y) = 0,
{x-y)(,2x-y) + (x-y) =0,
(,x-y)(,2x-y'+l) =0.
Plotting the loci represented
by the equations x — y = and
2x — y + 1 =0, the locus is found
to be as shown.
2. Plot the locus of the equa-
tion2 a^ -3xy-2y^ + 5y-2
= 0.
Since the factors of this equa-
tion are not readily found, the
principle is used that ax^ +bx + c
= a{x — Xi) {x — X2) where Xi and
xt are the roots of the equation
aa^ + 6a; -|- c =5 0.
Solving for x,
3j/± V252/2 -402/ + 16
^= 4
V I 4.
X' ^f. X
r/ \f'
N
,
\
^
\
/
\
^
^
\
)'y>
\
\
'
1
V
= 2 w — 1 or
-y + 2
2
The equation therefore may be written
2{x-2y^-\)(x~ -^V"^) = "'
or (a:-22/ + l)(2x + 2/-2) =0.
Plotting separately the loci represented by a; — 22/ + 1 =0 and
2x + y — 2 =0, the figure is found as shown.
44
LOCI
EXERCISES
Plot the loci represented by the following equations:
1. y-' + Qy = Q.
2. a;2 - 9 2/2 = 0.
3. a? + 2 2:2/ = 0.
4. 4 2;2 - lls!/ - 3 2/^ = 0.
6. a;2 - 9 2/^^ + 2 a; - 6 2/ = 0.
6. a;'' - S2/ - 2 2/2 + 3 2/ - 1 = 0.
7. (a;' - 2 2/) (2/ + ? - 1) = 0.
8. 2 a;2 - 7 X2/ + 3 2/^" + 5 2/ - 2 = 0.,
9. (a;2 + 2/'' - 9) (a;2 + 4 2/2 _ 9) = q.
10. Write the single equation which represents:
(a) the two coordinate axes;
(6) the two hnes parallel to the a;-axis and at distances 2 and 4
units respectively above it;
(c) the two hnes which bisect the angles between the axes.
11. Show that the locus of a;^ — 7 a; + 12 = is a pair of parallel
lines.
12. Plot the locus of (2/^ - 2 !/ - 8) (a;' - 2 a; - 3) = 0, and show
that the lines enclose a parallelogram.
16. Loci through intersections of two given loci. — If an
equation whose second member is zero is mvUiplied by any
constant and added to another equation whose second member
is zero, the resulting equation represents a- locus through all
points common to the two given loci and through no other points
on either locus.
Proof. — Let
M = (1)
and v = (2)
represent the
equations of two
given loci, also let
Pi (xi, yi) repre-
sent any point
common to the two given loci and Pz (x2, 2/2) any point on
one of the loci but not on the other.
LOCI THROUGH INTERSECTIONS OP LOCI
45
To prove that
u -\- kv = 0,
(3)
in which k represents any constant, positive or negative, is
satisfied by the coordinates of Pi but not by those of Pj.
Since Pi is a point on the locus of equation (1), its coordi-
nates must satisfy the equation u = 0. Since Pi is a point
on the locus of equation (2), its coordinates must satisfy the
equation v = 0. Wherefore the equation m + fo = is
satisfied by the coordinates of Pi.
The coordinates of P2 will cause one term oi u + kv to
equal zero, but not the other, therefore u + kv = is not
satisfied by the coordinates of P2.
Hence u + kv = represents a locus through the points
common to w = and v = and through no other points
on either locus.
ILLUSTRATIVE EXAMPLE
Find the equation of a system of loci through all the intersections of
the two loci whose equations are x' + y'' = 18 and ix' — y^ = 27.
Find the particular curve of the system which passes through the point
(6, 0). Check the accuracy of the result by plotting the curves.
Multiplying the first equation
by k and adding to the second,
(fc + 4)x2 + (S; - i)v' = 18fc + 27.
Since the point (6, 0) is on the
locus, its coordinates must satisfy
the above equation, whence
36 i; + 144 = 18 fc + 27.
Solving, k = — Ss*.
Substituting this value instead
of k in the equation of the system,
we get
a;2 + 3 2/2 = 36.
The three curves are shown in the figure above.
"
~
—
V
—
—
—
■
7
>
V
\
__
]r
,'_
^
L;;
/
\
/
\
-)
1
~
'■
■
*v
/
V
*Sl
]p
5 —
^
■^ '
~~
~
/
—
\
/
\
/
,
,,
Z
w
u
_
_
46 LOCI
EXAMPLES
Find the equation of a system of loci through the intersections of the
following loci:
1. x + 2y + l=0 and 3x - 4y -8 =0.
Ans. (it + 3)a; + (2 A; - 4)2/ + fc - 8 = 0.
2. a? + 3/2 = 9 and x — iy = 8.
3. 2 x'' - 2/2 = 7 and a? + 3 2/! = 10.
4. Write the equation of a system of loci through the intersections
ot y^ + ix — and y — 2x = 0.
Test the accuracy of the work by finding the coordinates of the points
of intersection of the given curves and substituting in the resulting
equation.
6. Write the equation of a system of loci through the intersections
of the cirrves whose equations are a;^ _ j^ — g = o and x + y = 6. So
determine k that the resulting curve shall pass through the origin. Fac-
tor the resulting equation and plot.
6. Write the equation of a system of loci through the intersec-
tions of the loci whose equations are a? — 4 y = and y — x = 0.
Give k such a value that the resulting equation shall not contain y.
Ans. a;2 — 4 a; = 0.
7. Write the equation of a system of loci through the points of
intersection of a^ -|- 2/" — 4 = and x^ + y^ — ^x = 0, and by giving k
the value —1, determine the equation of first degree, the locus of which
passes through the common points of the two loci. Plot the loci of the
three equations.
8. Find the equation of first degree which represents a locus through
the intersections ofa;2_2a;-l-^ = and 3? + y' = 1.
CHAPTER III
THE STRAIGHT LINE
17. This chapter will be concerned with a study of the
equations and properties of straight Unes. Later chapters
will consider other well-known curves.
It was observed in plane geometry that a straight Hne
was fully determined if two conditions regarding the line
were known; for example, two points on the line or one
point and the direction.
Similarly, it is found that the equation of a straight line
can always be found if the two conditions which fix the line
are given.
18. First standard equation of a line. In terms of point
and slope. — The equation of a straight line passing through
a given point Pi (xi, y-i) and having a given slope m is
y -yi = m{x- «i). ' (6)
Proof. — Con-
struct the given hne
AB whose slope is
m and which passes
through the point
Pi {xi, yi). Let the
point P {x, y) rep-
resent any point on
the line.
The slope of the
hne PPi = m, by
hypothesis. From formula (2), the slope of PPi
47
p
/"
■^1
/
p
— v
/
y
<^
'
' X — Xx
48 THE STRAIGHT LINE
Hence, — = m ot y — yi'= m (x — Xi).
X — Xx
This is the equation of the line with slope m and passing
through the point Pi {xi, yi), since it fulfils the two require-
ments of the definition. For, P {x, y) was taken as any
point on the line, therefore the equation is satisfied by the
coordinates of every point on the hne. Moreover, that the
equation is not satisfied by the coordinates of any point
not on the Une, can be shown in a manner identical to that
given in the first illustration of Art. 10. This step is so
similar in all examples that the student will not be required
to give it, unless called for, but he should never lose sight
of the fact that this is one of the essential conditions in the
determining of the equation of a locus.
19. Second standard equation of a line. In terms of
slope and y-intercept. — The equation of a straight line of
slope m and y-intercept b is
y == mx + b. (7)
Proof. — Since the ^/-intercept determines the point whose
coordinates are (0, 6), this is a particular case of equation (6).
Substituting in that equation xi = and t/i = b, the equa-
tion becomes
y = mx + b.
This equation can also be derived from a figure in a manner
similar to that used in deriving equation (6).
In deriving many equations, the student may either locate
his given data in a figure and derive the equation according
to the method outlined in Art. 10, or he may substitute the
data in any standard equation previously derived.
Since m and b may have any values, positive, negative or
zero, the equation y = mx + b represents any Une which
cuts the y-axis, that is, the locus of this equation includes
SECOND STANDARD EQUATION OF A LINE 49
all straight lines except those parallel to the y-axis. The
equation of such a line has been shown in Art. 10 to be of
the form x = a, in which a represents the constant distance
of the line from the y-aids. The two equations y = mx + b
and X = a represent all straight lines.
rU-USTRATIVE EXAMPLE
Find the equation of a line through the point (2, 1) and perpendicular
to the Une joining ( — 3, 1) and (1, 5).
The slope of the hne joining the two points is 1 by formula (2), there-
fore the slope of the required line is — 1 . Substituting m = — 1 , a;i = 2,
and 2/1 = 1 in equation (6), the equation of the line through the point
(2, 1) and perpendicular to the line joining ( — 3, 1) and (1, 5) is
y — 1 = —1 (x — 2) or x + y = 3.
EXERCISES
1. Find the equations of the lines:
(a) through (-2,-1), inclination 60°. Am. y= Vs x+2 Vs -1.
(6) through (-3, -2), slope 2. Ans. y - 2 x = A.
(c) through ( — 2, 5), inchnation 90°.
(d) through (2, —5), inclination 135°.
(e) through ( — 1, 1), parallel to the hne joining (2, 3) and (5, 2).
(/) through (2, 6), and perpendicular to the hne joining (5, 5) and
(-1,3).
{g) through (4, 2), with equal and positive intercepts on the axes.
(h) a>-intercept 5 and slope —3.
(i) ^/-intercept —2 and slope —4.
2. Find the equation of the line with Siope —2 through the inter-
section of the lines 2y + x - S = and x — Zy + 2=0. Ans.
2x + y = 3.
3. In the equation y = mx + b, what is the relation between the
lines if b remains constant and m changes? If m remains constant and 6
changes?
4. What is the sign of m if both intercepts on the axes are positive?
If both negative? If of opposite signs?
6. Find the equation of a Une perpendicvdar to the Une joining
( _ 1^ _2) and (3, 6), through its middle point. Ans. 2y + x-5=0.
50 THE STRAIGHT LINE
6. Find the equations of the perpendicular bisectors of the sides of
the triangle whose vertices are (—2, —1), (4, 1), and (0, —3). Prove
that these bisectors meet in a point.
7. Find the equation of the line parallel to the 2/-axis through the
middle point of the line joining (2, 3) and (4, —3). Ans. a; = 3.
8. The vertices of a triangle are (5, -3), (3, 7), and (-3, 1). Find
the equations of the line through the vertices and parallel to the sides.
9. An isosceles right triangle has its hypotenuse along the x-axis and
its vertex at the point (2, 3). Find the equations of its sides.
10. The vertices of a triangle are (7, 1), (5, —3), and (—3, 5). Find
the equations of the perpendiculars dropped from the vertices on the
opposite sides. Prove these lines meet in a point.
11. Two lines are drawn through (2, 4) with inclinations 30° and 60°.
Find the equations of the two lines which bisect the angles between the
two given lines.
12. If tan e = 3, find the equations of the lines through the origin
whose inclinations are
(o) e - 45°;
(6) e + 45°;
(c) B + 30°.
20. Third standard equation of a line. In terms of two
given points. — The equation of a straight line passi,ng
throiigh two given points Pi {xi, j/i) and Pa (a^, Vi) is
y-yi = 72^'(*-*i). (8)
Proof. — The slope of the hne through the two given
points Pi {xi, j/i) and P2 (^2, 2/2) is, by formula (2), m = — — ;
X2 — Xx
also it is given that the hne passes through the point
Pi {x\, j/i). Therefore, applying equation (6), the result be-
comes
As an exercise, the student is asked to derive the equation
by the method outUned in Art. 10.
FOURTH STANDARD EQUATION OF A LINE 51
21. Fourth standard equation of a line. In terms of the
intercepts. — The equation of a straight line whose x^ntercept
is a and whose y^ntercept is b is
- + r = l. (9)
a b ^ '
Proof. — The intercepts determine two points (a, 0) and
(0, b) on the line. Substituting these results in (8), the
equation is
. b-0 , • ,
Simphfying,
a b
The student is asked to derive this equation by applying
the method of Art. 10.
EXERCISES
1. Write the equations of the Unes through the following pairs of
points:
(a) (3, 2) and (4, 5). ! (c) (1, 6) and (-2, 4).
(6) (2, -3) and (-3, -2). (d) (a, 2 o) and (3 a, - a).
2. Write the equations of the lines which make the following inter-
cepts on the X and y axes respectively:
(o) 1 and 5. (c) —4 and —4.
(6) Sand —3. (d) —a and +a.
3. Write the equation of the Une through the points (5, —1) and
(—4, —2), and check the result by showing that the coordinates of the
given points satisfy the equation.
4. Is the point (5, —6) on the straight line joining (2, 4) and (—3,
-2)7
6. Find the equations of the sides of the triangle whose vertices are
(3, -1), (-4, 2), and(-l, -1).
6. Find the equation of the line whose y-iilteroept is —5 and which
passes through the intersection of the two lines 2x +y + 5 = and
ix-y + 7=Q.
52 THE STRAIGHT LINE
7. The vertices of atriangle are (I, 3), (4, —3), and (—3, —2). Find
the equations of the lines from ( — 3, —2) trisecting the opposite 'side.
8. Determine whether the following sets of points lie on a straight
line:
(a) (2, 3), (-1, -2), and (3, 2).
(b) (2, 4), (1, 2), and (-2, -4).
9. Find the equations of the medians of the triangle whose vertices
are (4, 1), (2, -3), and (-1, 5).
10. Prove that the medians in example 9 meet at a point f of the dis-
tance from any vertex to the middle of the opposite side.
11. Find the equations of the hnes joining the middle points of the
sides of the triangle whose vertices are (—2, —3), (4, 1), and (2, —5).
12. Find the equation of a line whose a;-intercept is 4 and which
passes through the intersection of the Unes x + y = Q and 3 x — 2y = 8.
22. Locus of equation of first degree. — It was shown in
Art. 19 that any straight Une can be represented by either
y = mx + b, or by X = a, both of which are equations of
first degree. It will now be shown that the converse is
true, namely:
Every equation of first degree represents a straight line.
Proof. — If A, B, and C may have any values, positive,
negative, or zero, then the equation Ax + Bi/ + C = in-
cludes all equations of first degree. If B is not zero, the
equation may be divided by B, and after transposing and
A C
solving for y, the result isy = — ^^ ~ n'
B D
This equation is of the form y = mx + h, in which m =
—A/B and h = —C/B. Therefore the equation Ax + By
-f C = represents a straight line of slope —A/B and
j/-intercept —C/B.
If B = 0, the equation becomes ^a; 4- C = 0, and may
be written x = —C/A, a straight line parallel to the y-axis
at a distance —C/A from it.
Hence all equations of first degree represent straight hnes.
The method just outlined of changing a general equation
to a standard form is one of great practical use in analytic
PLOTTING STRAIGHT LINES 53
geometry, not only for the straight line, but also for all the
other curves which follow. After an equation has been put
into one of the standard forms, it is only necessary to com-
pare the constants in order to write down many facts of
importance regarding the locus which any given equation
represents.
Thus, given the equation 2 x + 3 y = 6.
Solving for y,
2/= -ix + 2.
This is in the form y = mx + b, in which m = — f = the
slope of the line and 6 = 2 = the t/-intercept.
Again, dividing the given equation by 6,
^ + ^=1.
3^2
This is in the form - + ? = 1, in which o = 3 = x-inter-
a
cept and 6 = 2 = ^/-intercept.
23. Plotting straight lines. — Siace every equation of first
degree has been shown to be a straight hne, therefore in
plotting the locus of a first degree equation it is sufficient to
locate two points and then draw the indefinite straight line
through them. The most convenient points are usually
those determined by the intercepts on the axes.
If the intercepts are both zero, the line passes through the
origin and it is necessary to locate another point on the hne.
EXERCISES
1. Find the slopes of the following lines:
(a) 2 a; - 6 2/ = 6. (c) 7 x + 4 j/ = 8.
(b) x + Sy -5 =0. (d) Sy -x = 12.
2. Find the slopes of the following lines and determine which of them
are parallel and which perpendicular to each other. Plot the loci.
(a) 3x + y -7 =0. (c) 3y -x = 2.
(b) 6x + 2y -1 =0. (,d) 2x-6y = i.
54
THE STRAIGHT LINE
3. Find the equation of the line through the point ( — 1, 5) and per-
pendicular to the line 2y — 3x = 7. Ans. 2 x + 3 y = 13.
4. Findtheslopesof thetwoUnes As + Bj/ + C = Oand A'a;+B'2/
+ C = and show that if the lines are parallel,
A' B''
6. Prove that if the lines in Ex. 4 are perpendicular, then AA' =
-BB'.
6. Find the equation of a Une through (xi, yi) parallel to Ax + By +
C = 0. Ans. Ax + By - (Axi + Byi) = 0.
7. The equations of two sides of a parallelogram are y — x = 2 and
2 a: + 2/ = 4. What are the equations of the other sides if they intersect
at the point (0, —4)? Atis. y + 4 = a; and 2/-|-2a; + 4=0.
24. Nonnal equation of a straight line. — As has been
previously stated, whenever two conditions which determine
a straight Une are known, the equation of the Une can be
found.
In the case now to be considered, the Une is determined
by its distance from the origin and the incUnation of the
Une perpendicular to the given Une through the origin.
It wiU be recalled that incUnation is always taken to be
less than 180° and consequently in Figs. Ill and IV, the
NORMAL EQUATION OF A STRAIGHT LINE 55
line OC must be produced through the origin before the
inchnation can be determined.
The inchnation a in Figs. I and III, in which the hne AB
crosses the first and third quadrants respectively, is seen to
be acute, while in Figs. II and IV, in which the hne crosses
the second and fourth quadrants, it is obtuse.
In each figure, OC = p and is positive when above the
cc-axis and negative when below it.
The equation of the line which is determined by the con-
ditions of this article is called the normal equation of a
straight line.
The equation of a straight line in terms of p, the length of
the perpendicular from the origin to the line, and a, the inclina-
tion of that perpendicular, is
X cos a + y sin a = p. (10)
Proof. — In each of the above figures, if a and b represent
the intercepts on the a;-axis and y-asas respectively, then
from the triangle AOC, p = a cos a, and
from the triangle BOC, p = 6 sin /3 = 6 sin a,
since j8 is either equal to a or to 180° — a (why?) and there-
fore sin j3 = sin a.
Computing the intercepts a and h,a= and h = -—■ —
^ ° ^ cos a sm a.
Substituting these in the intercept form of the equation,
-4-^=1
the equation becomes,
V V
cos a sm a
Simplifying,
X cos a + ysma = p.
56
THE STRAIGHT LINE
EXERCISES
1. Write the equations of the lines, having given:
(a) p = 3, a = 120°. (c) p = -i, a = 30°.
(6) p = 5, a = 180°. (d) p =0,a = 60°.
2. What system of Unes is given by x cos a + ysiaa — p = 0, when
a is constant and p varies? when p is constant and a varies?
3. Given p = 5 and <l) the incUnation of the hne = 120°. Compute
a and write the equation of the hne.
4. Write the equations of the hnes, having given:
(o) ,f = 135°, p = 5. (6) <^ = 135°, p = -5.
5. Draw each of the following lines, find p and a and write their
equations :
(a) intercepts each equal 3.
(6) <^ = 120°, ^/-intercept = 3.
(c) intercepts each = —3.
(d) a;-intercept = 3, ?/-intercept = —3.
(e) <l> — 30°, s-intercept = — 5.
6. For what values of p and a will x cos a -{■ y sin a — p = be
jiarallel to the a^axis? to the i/-axis? pass through the origin?
7. Write the equation of the line
through the point (3, 0) if a = 60°.
8. For the equation y — x = 4:, find
slope, incUnation, a and p.
9. Derive equation (10) by the method
of Art. 10, using the figure here given.
Hint. — p = OK + KC.
10. Derive equation (10) by computing
m and 6 in terms of p and a and substitut-
ing in equation (7).
25. Reduction to normal form. — It is required to reduce
the general equation of a straight Hne, Ax -{- By -{• C = 0,
to the normal form x cos a + y sin a = p.
Equating the a;-intercepts in each,
P
cos a
-c
(1)
REDUCTION TO NORMAL FORM 57
Equating the ^/-intercepts in each,
sin a B '^^^
Dividing (1) by (2),
tan a = -r-
A
A B
Whence, cos a = ± . and sin a = ± ■
/ aLL\X Dill U. 3Z •
This is readily seen by drawing a right
triangle with leg B opposite angle a and A
adja cent to it. The hypotenuse is then
— C C
From (1), p = —V- cos a ~
■4 VA^ + B^
Since a is always less than 180°, sin a is always + and
therefore the sign of the radical is always the same as the
sign of B.
Substituting these values of sin a, cos a, and p in the
normal form, the general equation becomes,
^ -g ^ -c
Va^ + 52 Va^ + B^ Va^ + 52
These results can be summarized as follows:
To reduce an equation in the form Ax + By + C = to
the normal form, divide the equation by ± Va^ + B^, in which
the sign of the radical is the same as the sign of B. If B is
missing, choose the sign of the radical the same as A.
Example. — Reduce 4a; — Sj/ — 15 = 0to normal form
and decide from the signs of sin a, cos a, and p, which quad-
rant is crossed by the Hne.
A = 4:,B= -3,C= - 15, ±Va^ + B^ = ± V16 + 9 = ±5.
Dividing by — 5,
58 THE STRAIGHT LINE
Comparing with
X cos a -\- y sin a — p = 0,
it is seen that cos a is negative and sin a positive, therefore
a is obtuse; also p is negative and thus the perpendicular
falls below the a;-axis. The line then crosses the fourth
quadrant. Check by plotting the line.
1.
and I
EXERCISES
Reduce the following equations to normal form and determine p
(a) 3x - Ay = 25.
(6) 3x + y - 10 =
(c) 2/ + 2 = 0.
(d) 3 a; - 4 2/ = 0.
(e) 3x + Ay = 25.
0. (/) ^ - 3 s + 4 = 0.
(g) y-2=0.
ih) X -2 =0.
2. A line passes through ( — 2, — 1) and is perpendicular to 2 a; + ^ +
3=0. Find its equation and distance from the origin.
3. A line passes through (—4, —5) and has its intercepts equal and
both negative. Find its equation and distance from the origin.
26. Perpendicular distance from a line to a point. — The
solution of a particular case of this problem will be illus-
trated in the fol-
lowing example:
Find the dis-
tance from the
hne 3x — 4y +
15 = to the
pointPi(-4,3).
Let Li in the
figure represent
the given line
and Pi the given
point.
Through Pi, draw L2 parallel to Li and RPi perpendicular
to Li. Then RPi is the required distance since it is meas-
y
y
/
/
F
y
/
/
/
P.
/
\
/
/
/
\
\
r
/
/
/
y
\
s
,
^
/
,/
/
\
'l^
/
L<
f
DISTANCE FROM A LINE TO A POINT
59
ured jrom the line to the point. Draw the perpendiculars
OB = ipi to Li and OA = pi to Li. Then from the figure,
RPi = OB-OA = p2- pi.
The slope of the given hne Li is f , whence the equation
of the line Li through Pi and parallel to Li is by standard
equation (6),
i/-3 = f(a; + 4) or 3a; - 4z/ + 24 = 0.
Reducing to normal form, the equations of Li and La become
respectively,
-f:c + |2/-3 = and - ix + iy - -%^ = 0.
Whence pi = 3 and Pi = V, and therefore RPi = ^j*- _3 = |.
It is observed that this resu t is positive. This checks with
the figure as RPi has the same direction as OA which is posi-
tive. The distance from a hne to a point is always positive
if the point is above the hne, and negative if below the line.
The point and
the line may he on
opposite sides of
the origin as in the
accompanying fig-
vire. The same
process as that used
in the preceding ex-
ample will lead to
the correct result,
but care must be
taken to give the
correct signs to the
perpendiculars.
Thus, find the distance of the point Pi (-1, -3) from
the line Li of the preceding example.
Make the construction. as before. The distance required
is i?Pi = AB = 0B-0A = p2- Pi-
/•
^
/
/
y
^
K
/'
F
/
/
\
\,
/
y
V
/"
^.
\
/
/
\
V,
\
/
L
1
\
\
^
\
/
/
Pi
y
/
,1
L
l_
60
THE STRAIGHT LINE
The equation of La is 3 a; — 4 y — 9 = 0, or in normal
form,
-fx + |j/ + f = 0.
Whence p2 = — f, Pi = 3, as in Ex. 1.
RPi = V2-Vi= -1-3= -^.
The minus sign indicates that Pi is below the line Li.
The general formula for the distance from a Hne to a point
will be determined in a manner similar to that used in the
examples above.
The distance d, from the line Ax + By + C = to the
point Pi (xi, 2/i), is given by the formula
± V^2 + B" '
(11)
the sign of the radical being taken the same as that of B.
Proof. Let Li rep-
resent the given line
and Pi the given point.
Through Pi draw Li
parallel to Li and let
pi and Vi represent
the perpendicular dis-
tances from to L\
and Li respectively.
Then d — f^ — pi.
Slope of Li = slope
-A
of Li =
B
Whence the equation of Lj is
-A
y-yi = -^i^-xi)
or
Ax + By- {Axi + Byi) = 0.
BISECTORS OF ANGLES BETWEEN TWO LINES 61
Reducing equations of Li and In to normal form,
-C
and
Pi =
P2 =
±VA2 + 52-
± VA2 + B^'
where the sign of the radical in each case is the same as that
of JS.
Hence, d = 4E1+M±^,
± Va^ + £2
the sign of the radical being the same as the sign of B.
ILLUSTRATIVE EXAMPLE
Find the distance from the line 5 x
— 12 2/ = 25tothepoint (-1, 3).
Substituting in the formula,
-5 - 36 - 25 ^ 66
-13 13
d
This positive value of d checks with
the figure since it is measured upward
from the line.
Y
y
P.
^L
5
-X- ot ^^^^
L ^^
- u^
EXERCISES
1. Find the distance from the line Zx + iy = 5 to the point
(-1, -1).
2. Find the distance from the line 5 a; + 12 y = 13 to the point of
intersection of the lines y — x + 1 =0 and 2y — x = 1.
3. Find the altitudes of the triangle whose vertices are (1, 1), (—3, 4),
and (-3, -2).
jil. Bisectors of the angles between two lines. — Since
the bisector of an angle is the locus of a point which moves
so as to be numerically equidistant from the sides, the
equation may always be readily found as in the following
example.
62
THE STRAIGHT LINE
Find the equations of the bisectors of the angles between
the lines 3 a; — 4 y = lO and 4 x + 3 y = 7.
Draw the given
lines Li and La.
There are two bi-
sectors Ls and Li.
Let Pi (xi, j/i) be
any point on the bi-
x-l sector Li.
Then the perpen-
diculars AiPi and
BiPi are equal in
length. They are
each positive, being measured upwards from Li and L2 re-
spectively.
Then
AiPi = BJ>i.
3 xi - 4 1/1 - 10
•
<
s.
L
3
\
/
s,
/
,
/'
''
/'
\
iyl
s
t^'
y
-X
I
.1^
y
X-
— ■
■-
—
1/
\
^
'\
>
3
'
_
— 1
r~
/
/
3<
L/
/
^^
AxPi
and BiPv
Therefore,
-5
4x1 + 32/1-7
-, formula (11),
3xi-4j/i- 10_4xi + 3t/i- 7
Since Pi (xi, yi) was taken as any point on the hne L3,
the subscripts may be dropped and the equation of L3 is
7x-2/- 17 = 0.
Similarly, let P2 (xa, j/2) be any point on L4. Then A2P2 =
—BnPi, since A2P2 is negative and B2P2 is positive.
Then
3 X2 - 4 j/2 - 10 ^ 4 X2 + 3 ^2 - 7
-5 ~ 5 ■
Hence the equation of L4 is x + 7 y + 3 = 0.
BISECTORS OF ANGLES BETWEEN TWO LINES 63
EXERCISES
1. Find the equations of the bisectors of the angles between the lines
4 a; + 3 2/ = 6 and ix — Zy = 6 and show that they are perpendicular.
2. A line is drawn through (0, 0) perpendicular to 3 a; + 4 j/ = 6.
Find the equations of the bisectors of the angles between these two Mnes.
MISCELLANEOUS EXERCISES
1. How far from the origin is the line through (1, 6) parallel to
2/ + 4 a; = 7? Ans. 10/VI7.
2. Show that a is the same for all parallel hnes. Find the equation
of a hne parallel to 3 a; + 4 y = 25 and nearer the origin by two units.
Ans. 3 a; + 4 2/ = 15.
3. Find the equations of the lines halfway between the parallels:
{ay ix-Sy = 15, {b) 5 x + 12y = IS,
ix-Sy = -15. 5a; + 12 2/ + 39 =0.
i. Find the equation of a Une parallel to 12 x — 5 ?/ + 13 =0:
(o) at a distance of 3 from it;
(6) at a distance of —3 from it.
6. Find the equation of a line with slope 2 at a distance of 5 units
from the origin.
6. Find the distance from the line to the point in the following
examples constructing a figure for each:
(o) 4a; -32/4-6 = to (2,1). (c) 5a; - 122/ + 6 = to (3, 4).
(6) 3a; + 42/-5=0to(-l, -5). (d) 6a; + 2^ + 7 = Oto(-1, 5).
7. Find the area of the triangle whose vertices are the points (3,-2),
(4, 3), and (—2, 1) by finding the lengths of a side and the corresponding
altitude.
8. Find the altitudes of the triangle formed by the Unes y + x =■ 3,
y — 5x = 9, and y = —1.
9. Find the point which is equidistant from the points (1, 3) and
(5, 5) and is at a distance of 2 from the line 3x + 4:y — 10=0.
10. Find the equations of the bisectors of the angles of the following
triangles and prove that these bisectors meet in a point, the equations of
the sides being:
(o) 3 a; - 4 2/ = 12, 4 a; + 3 ^ = 12, 3 a; + 4 2/ + 12 = 0.
(6) 5 a; - 12 2/ = 24, 12 a; + 5 2/ = 24, 5 2/ - 12 a; = 20.
(c) 2/ = 4, a; = —4, 3x — iy = i.
64
THE STRAIGHT LINE
11. A triangle has sides 4 a; + 3 i/ = 24, 3 y — 4 a; = 24, and y =
—4. Prove that
(a) the triangle is isosceles;
(6) the bisector of the exterior angle formed by the first two sides
is parallel to the third side.
12. Given the triangle whose sides are 4a; + 3j/ = 24, 4a; — 3t/ +
12 = and 2/ + 4 = 0. Prove that the bisector of the angle formed by
the first two lines divides the opposite side into segments which are pro-
portional to the sides adjacent to the angle.
13. Find the locus of all points which are twice as far from 3 x — 4 j/
+ 12 = as from bx -\2y = Z0.
14. Find the distances between the parallel lines
(a) 3 a; + 2 2/ = 13 and 3 3; + 2 2/ + 26 = 0.
(6) a; + 2 2/ = 5 and a; + 2 2/ + 10 = 0.
28. The angle which a line makes with another line. —
In Art. 7, the ange which one straight line makes with
another was defined as the angle less than 180° measured
coimter-clockwise from the second to the first.
Thus, in both figures 1 and 2, 6 is the angle which L\
makes with L^.
U
Pig.l
Fig.S
// mi and rrh are the slopes of two lines and B is the angle
which the first line makes with the second,
tane =
nil — wtg
(12)
1 + mimz
Proof. — Let <^i and 02 represent the inclinations of the
two lines Li and L2 respectively, then tan <i>i = mi and
tan <j)i = rn^
THE ANGLE A LINE MAKES WITH ANOTHER 65
In Fig. 1, <t>x = (h + 6. (Why?)
Whence B = 4>i — <i>i,
therefore tan B = tan (<^i - <l>,) = tan ^i - tan <^-, .
1 + tan 01 tan 02
tan 01 = mi and tan 02 = mz; •
But
therefore
tan 9 =
lUi —
mi — nh
minii
1 + mim2
In Fig. 2, 02 = 01 + (180° - d).
Whence d = 180° + (0i - 02),
and therefore
tan d = tan [180° + (0i - 02)] = tan (0i - 02) = ^
The student should not fail to fix in mind that the angle
6 is always measured from the second hne to the first.
ILLUSTRATIVE EXAMPLES
1. Find the angle which the line y — 3x + 2=Q makes with the
line 2y — X = 0.
Reducing each equation to slope form, mi = 3, ma = J.
3 — -
Substituting in formula (12), tan 8 =
Therefore 8 = 45°.
2. Find the equation of the
line through ( — 1, —2) making
the angle tan"' | with the line
2x + y-3 =0.
The facts given are sufficient
to determine the slope of the line.
In the formula
l+f
= 1.
tan e =
mi — Ml
1 +mimi
if any two quantities are known
the third may be foimd. Here
tan9 = §. Since the angle (tan"' i)
is to be measured from the
given hne to the required Une,
and the slope of the given Une
is —2, therefore vh = —2.
Y
^
^
V
s 5
X' "s o\ X
^ ^ ^
N V
s 3
^^ 5
^^3
^V
3^
Sk '
± tr
66
THE STRAIGHT LINE
Substituting in formula (12),
1 ^ ra, + 2
2
whence mi =
1 -2mi'
The equation of Li may then be written by substituting in standard
equation (6) and is •
2/ + 2 = -I (X + 1)
or 3 a + 4 2/ + 11 = 0.
3. Compute the angles of the triangle formed by the intersection of
the lines whose equations are x — 5 y = 10, 2 x + 3 y — 12, and
11 a; + 10 2/ + 33 =0.
Since the angles
must always be
taken in a covmter-
clockwise direction,
the angle A is meas-
ured from Li to Li,
B from Z/8 to L^ and
C from Li to La.
The slopes of
la, Li, and La are
respectively ^, -|,
and — iJ.
Substituting in
the formula
i*.B w
5,B
^^
S"^^
\^
\^^
N^ ^v.
S -i^^
\ ^^
^^ ^!S
^^ -P^
s ^^
V V
\ -^ ^s M
V X^ ^i^__„^=:=y
^.., |^2j^„. ,,.X
— ■^5^ s
<s
•^^ ^s
s r
i\i- t
tane =
■Ml — mi
1 + mimi'
being careful to use for mj the slope of the line from which the angle is
measured, the results are:
tan A =
tanB =
= 1.
i+1
3 ^ TS _ 1
1+M "
_1JL
tan C = — ^
1-i
Therefore A = 45°.
Therefore B = tan-i i.
Therefore C = tan"' (-|).
EXERCISES
1. Find the tangent of the angle which the first line makes with the
second in the following:
SYSTEMS OP STRAIGHT LINES 67
(o) 3a; -2/ + 4 = and 2x + 4:y + 5=0.
(b)y-2x = 3 and y+3x = 5. Am. -1.
(c) y — 2 = and x — y = 1.
id) x + 2y = 3 and y -2x = 4t.
(e) ax + by = ah and ax — by = ah.
2. Find the equation of the Une passing through the point ( — 2, — 4)
and making an angle tan-^ J with the Une y — 2x = 7. Am. 11 x +
2 2/ + 30 = 0.
3. Find the equation of the hne through the origin making an angle
of 120° with the Une y = x/Vs. Prove that the s-axis bisects the
angle between these Unes.
4. Find equation of the line through ( — 2, —1) making an angle of
135° with the Une x — y = 2. Atis. y = —1.
5. A right isosceles triangle has the ejctremities of the hypote-
nuse at the points (1, 2) and (—3, 4) and the vertex of the right angle
below the line joining the points. Find the equations of the three
sides.
6. Find the interior angles of the triangle whose sides are the lines
x + y = 2, y — x='2, and y — 2 x = i.
7. Given the triangle formed by the three Unes x — 2y = 2, x + 3 y
+ 3=0, and y = 2. Prove that the exterior angle formed by the'first
two Unes is equal to the sum of the two opposite interior angles.
8. The vertex of an isosceles triangle is (2, 3) and the base is
along the Une x + y =0. Given that the vertex angle is 120°, find the
equations of the other two sides.
9. Prove that the Une through the origin which makes the angle
(tan"' I) with the Une 2y = x + 3 bisects the angle between the co-
ordinate axes.
10. Two opposite vertices of a rhombus are (2, 1) and ( — 2, —3).
Find the equations of the sides if the interior angles at these vertices
are 60°.
11. Two opposite vertices of a square are (2, 3) and ( — 1, — 3) . Find
the equations of the sides.
12. The base of an equilateral triangle Ues in the Une y — 3x = 6.
The opposite vertex is at (4, 3). Find the equations of the other two
sides.
29. Systems of straight lines. — Each of the standard
equations of a straight Une contains certain arbitrary con-
stants. An arbitrary constant is represented by a letter
68 THE STRAIGHT LINE
to which different values may be assigned. For example,
in the equation y = mx -\- b, m and b are arbitrary con-
stants.
Often a fixed value is given to one of these constants
while the other is left arbitrary. Thus, in the equation
2/ = 2 a; + 6, 6 is an arbitrary constant and by assigning
different values to b, an infinite number of Unes is obtained
each having slope 2. Such a set of lines is called a system
of lines, and the arbitrary constant is called the parameter
of the system. Other illustrations are y = mx + 2 which
represents a system of straight lines whose ^/-intercept is 2,
and a; cos a + 2/ sin a = 5 which represents a system of
hnes each five imits from the origin.
The equation of a straight fine can always be foimd at
once if the two facts determining the line give the values
of the arbitrary constants in one of the standard equations.
Thus, if slope = 2 and 2/-intercept = 3, the equation of
the fine, by substituting in equation (7), is y = 2 a; + 3.
When the two conditions determining the fine do not give
the values of the arbitrary constants in any one standard
equation, either of two methods may be used.
First, from the data given, find the values of the arbi-
trary constants in some standard equation, then substitute
these constants in that equation; or second, write the equa-
tion of the system of Unes satisfying one of the conditions
given. This equation will contain one parameter. Deter-
mine the value of this parameter from the other condition.
n,LTJSTRATIVE EXAMPLE
Find the equation of the straight Une whose slope is — | and which is
3 units from the origin.
It is given that the Une L is 3 units from the origin, with inclination
<^ = tan~' ( — f ) . The problem will be first worked by finding the con-
stants for substitution in equation (10).
Let a represent the inclination of the perpendicular from the origin
SYSTEMS OF STRAIGHT LINES 69
to the line. Then <^ = a + 90° and tan <t> = tan (a + 90°) = — cot a
= — J. Therefore tan a = f, sin a = 4, and cos a = |.
Substituting in equation (10),
the equation of L is
ix+iy-S =0
or 3 a; + 4 2/ = 15.
The problem might also be ,
worked by finding the coordinates
of the point A and substituting in
equation (6).
In the triangle ARO, OA =
3 CSC (180° - 0) = 3 esc = 5.
Substituting in (6), the equation of L is y — = — | (x — 5) or
3 a; + 4 2/ = 15.
The second method of solving problems of this class is to first write
the equation of the system of lines satisfying one of the given conditions.
Thus, in the equation
y=-lx+h, (1)
which represents the system of Unes with slope — J, the parameter b must
be so determined that the line shall be 3 vmits from the origin. Reducing
equation (1) to normal form, fx+fj/ — |6 =0.
Whence, 4 6 = 3, or 6 = J^, and the equation of L is found to be
3x + 47/ = 15.
Another apphcation of the second method is to use
a; cos a + y sin a — 3 =0, (2)
which represents the system of Unes 3 units from the origin. Here a
must be so determined that the slope of the line shall be — f. Reduc-
3
ine (2) to slope form, y = —cot a x -\-—. — , whence, —cot a = —\,
° ^ sm a
sin a = I, and cos a = |. Substituting in (2), the required equation is
found to be 3 a; + 4 2/ = 15.
EXERCISES
1. Write the equations of the systems of straight lines which satisfy
the following conditions:
(o) distance from the origin = 5. Ans. x cos a + ysiaa = 5.
(6) x-intercept = 3.
(c) slope = 5.
(d) passes through (1, 4).
70 THE STRAIGHT LINE
(e) inclination of perpendicular from origin to line is 60°.
(/) slope of perpendicular from origin to line is |.
(jg) sum of intercepts on axes = 6.
2. Write the equation of the system of lines 3 units from the origin
and so determine a that a Une of the system shall pass through the point
(2, 3).
3. Find the equation of the line with slope 2 and which in addition
satisfies the following condition:
(a) distance from the origin = 5.
(b) a;-intercept = 5.
(c) simi of intercepts = 6.
(d) distance from the origin = —3.
4. Find the equation of the straight hne through ( — 4, —2) and sat-
isfying in addition the following condition:
(a) distance from origin = t/XQ.
(6) parallel to 2 x — 5 2/ = 6.
(c) sum of intercepts = 3.
(d) inclination of perpendicular from origin to Une = 45°.
(e) portion of hne intercepted by axes is bisected by given point.
6. Find the equation of the straight line 4 units from the origin and
satisfying in addition the following condition:
(a) perpendicular to the line 2x — y = Z.
(6) through the point (2, 4).
(c) ^/-intercept = 5.
(d) product of intercepts = 32.
6. A Une through the point (3, 1) intersects the x- and ^-axes at A and
B respectively. The Une AB is divided by the point in the ratio \.
Find its equation.
7. Find the equations of the two hnes through (1, 4) and making the
product of the intercepts 18.
8. Find the equation of the Une through (—3, —4) and making the
j/-intercept twice the avintercept.
9. Find the equations of the two Unes in which the incUnation of the
perpendicular from the origin on the Une is 45° and the product of the
intercepts 8.
MISCELLANEOUS EXAMPLES ON CHAPTER IH
1. The equations of two sides of a paraUelogram are 2x — y = Z
and 3x + 2y = 1. Find the equations of the other two sides if they
intersect at (2, 5).
SYSTEMS OF STRAIGHT LINES 71
2. Find the equations of the Unes through the point ( — 1, —2) tri-
secting that portion of the line 2y + & = x which is intercepted be-
tween the axes.
3. A tangent to a circle with center (— 3, 5)is3a; — 42/ — 6 = 0.
Find the length of the radius. Ans. 7.
4. One side of an equilateral triangle has its extremities at (3, —4)
and (3, 2). Find the equations of the other sides.
6. The line joining A (1, 3) and B (3, 0) is cut by the line y — x+ 8
= 0. In what ratio does the point of intersection divide AB1 Ans.
(-2/1).
6. Find the center and radius of the circle circumscribing the tri-
angle whose vertices are (0, 2), (3, 3), and (6, 6). Ans. Center ( — 1, 10),
T = V65.
7. Find the center and radius of the circle inscribed in the triangle
the equations of whose sides are 3a; + 4^ = 6, 4^ — 3a; = 6, and
y = -2. Ans. Center (0, -|), r = V-
8. An isosceles right triangle is constructed with its hypotenuse
along 4 a; — 2 2/ = 3 and the vertex of its right angle at ( — 1, 3). Find
the equations of the equal sides and the coordinates of the other vertices.
9. Find the equations of the following loci. Prove that they are
straight Unes and construct the lines.
(o) A point moves so as to be always equidistant from the points
(-1, 2) and (3, 4). Ans. 2x + y = 5.
(6) A point moves so that the sum of its distances from j/ — 2 =
and 5x + 12y — 26 =Ois equal to 7.
(c) A point moves so that its distance from the hne 3 x + 4y —
6 = is one-half its distance from the line 5x — 12y +
13 =0.
(d) A point moves so that the square of its distance from (—2, 3)
minus the square of its distance from (1, 4) is equal to 10.
10. A point moves so that five times its distance from the x-axis is
three times its distance from the origin. Find the equation of the locus
and prove it represents a pair of straight lines.
11. The base of an isosceles triangle is the line joining (—3, 2) and
(4, 3) . Its vertex is on the line y = —2. Find the equations of its sides.
12. Show that 6a^ + 5xy — 6y'— x + 5y — 1 =0 represents a
pair of perpendicular lines.
13. The sides of a quadrilateral are given by the equations x' + ixy
-|-4 2/''-l-3a;-|-6y = and y' + y — 6 = 0. Prove that the figure
is a parallelogram.
72 THE STRAIGHT LINT!
14. Prove analytically that the perpendiculars drawn from the ver-
tices of any triangle to the opposite sides meet in a point.
15. Find what relation must hold among the coeflficients in the
general equation of a line Ax + By + C = Oin order that
(a) the x-intercept shall = 3.
(6) the given line shall be perpendicular to 2x -\- Sy = 5.
(c) the slope shall = 5.
(d) the perpendicular from the origin to the line shall = 5.
(e) the line shall be parallel to the z-axis.
(f) the line shall pass through the point (3, 5).
16. Write the equation of the set of lines through the point of inter-
section of the two hnes 3 x + 2y + 8 = and x — 3y = I and so de-
termine the parameter of the system that the line shall pass through the
point (1, 2).
17. Prove that the two lines whose equations are xy-\-2x — iy —
8 = are the bisectors of the angles between lines whose equations are
a?-2/2-8a;-4j/ + 12 = 0.
18. Find the equation of the line perpendicular to the line 2x -{-Sy
— 12 = and bisecting the portion of the line intercepted by the
.^
CHAPTER IV
POLAR COORDINATES
30. Definition. — A second method of locating a point
in a plane is by means of polar coordinates. These often
lead to simpler results than those obtained by rectangular
coordinates. A comparison of the two systems of coordi-
nates is shown by the following illustra-
tion. If in a country where roads follow
section Unes, the question were asked w-
how to reach R from 0, the answer
would be of the form, go 4 miles east and 3 miles north.
If the same question were asked in an open country, the
direction would probably be pointed out and the questioner
told to go 5 miles. in that direction. The first is an illus-
tration of rectangular and the second of polar coordinates.
In order to locate a point in any system of coordinates,
two fixed things are necessary. In rectangular coordinates
these are two intersecting perpen-
dicular lines. In the polar system
a fixed directed straight fine called
the polar axis or initial line and a
fixed point on that Une called the
pole or origin are given.
In the figure, OX is the polar
axis or initial Kne and the pole
p or the origin.
The line OP from the pole to the point is called the radius
vector and is represented by p. The angle which OP makes
with the polar axis is called the vectorial angle and is repre-
sented by d. In the figure, OP = p = radius vector, XOP =
73
74
POLAR COORDINATES
B = vectorial angle. These two quantities are called the
polar coordinates of the point and the point is represented
by P (p, 6). The radius vector is positive when measured
on the terminal line of the angle and negative when meas-
ured on that Hne produced through the origin. The vecto-
rial angle is positive when measured counter-clockwise and
negative when measured in clockwise direction. As in trig-
onometry the angle d may have an unlimited nimiber of
values differing by 2 7r, since it is any angle whose initial
line is OX and whose terminal line is OP
The position of a point in a plane is definitely deter-
mined if its polar coordinates are given. The same point
may, however, be expressed in many different ways. Thus,
in the first figure above, if the least value oi d = 30° and
p = 5, then P may be written (5, 30°) (5,-330°), (-5, 210°),
(-5, -150°), (5, 390°), etc.
The steps in plotting a point P in polar coordinates are as
follows:
From the polar axis OX construct an angle equal to 6.
If p is positive, lay off OP = p on the terminal Une of the
angle. If p is negative,
produce the terminal line
through and lay off
on it OP equal to the
numerical value of p.
Thus, locate the point
P(-5, 150°). The an-
gle XOR = 150° is first
constructed in a positive
direction from OX.
Since p is negative, the
terminal line of the angle is produced through to P making
OP 5 imits in length. P then represents the point (—5,
150°). Show that (—5, —210°) represents the same point.
DEFINITION 75
EXERCISES
1. Plot the points (-3,30°), (3, -150°), (-5, 180°), (-2, Itt),
(-3, -Itt), (-1, 330°).
2. Write three other pairs of coordinates of each of the points
(-3, 20°), (2, |,r), (-4, 240°), (3, 330°).
3. A side of a square is 3 inches. A diagonal is taken as the polar
axis and one extremity of that diagonal as pole. Find the coordinates
of the vertices.
4. Each side of a rhombus is 4 inches. One side is on the polar axis
and a vertex is at the pole. Find the coordinates of the vertices if the
angle at the pole is 60°.
B. Prove that the three points (0, 0), (3, 30°), and (3, -30°) are the
vertices of an equilateral triangle.
6. Show that (2, 30°) and (2, —30°) are symmetrical with respect to
the polar axis, that (2, 30°) and ( — 2, 30°) are symmetrical with respect
to the pole and that (2, 30°) and (2, 150°) are symmetrical with respect
to a perpendicular to the polar axis through the pole.
7. What point is sjrmmetrical to (4, —30°)
(a) with respect to the polar axis?
(6) with respect to the pole?
(c) with respect to the perpendicular to the polar axis through the
pole?
8. What point is symmetrical to (p, 9)
(a) with respect to the polar axis?
(6) with respect to the pole?
(c) with respect to the perpendicular to the polar axis through the
pole?
9. Where do the points lie
(o) for which 6 = 45°? (c) for which p = 5?
(6) for which 9=0? (d) for which p = 0?
10. Find the distance between the points (2, 30°) and (-3, 150°).
Hint. — Use law of cosines in trigonometry. Arts. V7.
11. If 9 is a positive angle less than 360°, in how many ways can the
following points be expressed:
(a) (3, 30)?
(6) (-3, 240°)?
(c) the pole?
76
POLAR COORDINATES
31. The equation of a locus: polar coordinates. — The
definition of the equation of a locus in polar coordinates is
the same as that given in Art. 10, and the steps in finding
the equation are identical to those stated in that article
except that p and d are used instead of x and y.
Thus, find the equation of a Une such that the perpen-
dicular from the pole upon it is p and the angle which the
perpendicular makes with
the polar axis is a.
1st. Given the Une L such
that OR = p and XOR = a.
Let P (p, ff) represent any
point on the fine.
2nd. Cos ROP = ;;rn,from
OP'
trigonometry.
Zrd. Cos {d - a) = p/p,
Aih. Clearing of fractions, the required equation is
p cos {6 — a) = p.
EXERCISES
1. Prove that the equation of a line
(a) perpendicular to the polar axis and at a distance of four units
to the right of the pole is p cos 6 = 4.
■ (6) parallel to the polar axis and two units above it is p sin 9 = 2.
2. Prove that the equation of a line through the pole with inclination
7r/6 is 9 = ir/6.
3. Prove that the equation of the circle with center at the pole and
radius 5 is p = 5.
4. Prove that the equation of the circle which passes through the
pole and has its center on the polar axis o units to the right of the origin
is p = 2 a cos 8.
6. Prove that the equation of the circle which passes through the
pole and has its center on the perpendicular to the polar axis through
the pole and 6 units above it is p = 2 & sin 9.
THE LOCUS OF AN EQUATION 77
32. The locus of an equation : polar coordinates. — It is
required to find a locus which contains all the points whose
coordinates (p, 6) satisfy the equation and which contains no
other points.
As in the case of rectangular coordinates, this can always
be done by assigning values to one variable and finding the
values of the other, then plotting the points and connecting
by a smooth curve. It was found in that case, however,
that the work was greatly facilitated by combining with
the plotting a certain amount of discussion. The same is ,
true in the case of polar coordinates.
The points in discussion which are particularly helpful
are:
1. Intercepts on the polar axis.
2. Ssnnmetry.
3. Extent.
Intercepts. — Placing 9 = and solving for p, points
are found at which the curve intersects the polar axis.
Other intersections may be found by letting B = 180°, 360°,
etc., and finding the corresponding values of p.
The coordinates of the pole are p = 0, 5 = any angle.
Even though the pole is on the curve not all such values
satisfy the equation. Placing p = 0, and solving for B the
particular angles are determined at which the curve passes
through the origin.
Thus, in the equation p^ = d?' cos 2 5, if = 0, p = ±a.
Two points on the polar axis are thus located. If = 180°,
360°, etc., no new points are found on the initial line. Plac-
ing p = 0, is found to be 45°, 135°, 225°, and 315°, which
shows the pole is on the locus.
Symmetry. — The tests for symmetry ordinarily used,
correspond closely to those of rectangular coordinates. It
can be shown that in polar coordinates a curve is sym-
metrical with respect to
78 POLAK COORDINATES
(a) the polar axis if d can be replaced hy —6 without
changing the equation. Why?
(6) the perpendicular to the polar axis through the pole
if 6 can be replaced hy tt — 6 without changing the
equation. Why?
(c) the pole if p can be replaced by — p without changing
the equation. Why?
In general, the test for symmetry with respect to the polar
axis will be the only one used. This is of particular prac-
tical importance, since any part of the curve determined by
giving d values from 0° to 180° can be reproduced from 0°
to — 180° by the principle of symmetry. Points should be
plotted until it is certain that any further points found are
the same as those obtained by symmetry.
While the above tests (a), (6), and (c) are universally
true, their converse does not necessarily hold. A curve
may, for example, be symmetrical with respect to the polar
axis even though the equation is changed when 6 is replaced
by —d. This point is discussed in Art. 34.
The equation p^ = a^ cos 2 6 stands all the tests of sym-
metry mentioned in this article and hence the locus is sym-
metrical with respect to the perpendicular to the polar axis
through the pole, to the pole and also to the polar axis.
Extent. — Under this head will be considered:
values of which make p imaginary;
values of 6 which make p infinite;
values of 6 which make p a maximum or miTiiTmiTn nu-
merical value.
In those problems in which p enters the equation in even
degree, it is possible that certain values of 9 may make p
imaginary. Such values of d are excluded.
In some examples, there are values of 6 which make p
infinite. Such values are important as they show that the
Qurve extends to infinity in that direction. In such cases
THE LOCUS OF AN EQUATION
79
it is well to determine values of p corresponding to values of
d a little less and a little greater than those which render
p = 00 , as important changes often take place in the vicinity
of such points.
Other important values of d are those which give to p
its maximum or minimum numerical values.
Consider again p^ = o^ cos 2 9 or p = ±a Vcos 2 d.
It is seen that values of 8 between 45° and 135°, also be-
tween 225° and 315°, make p imaginary, and therefore these
values of ^ are excluded. There are no values of 6 which
make p = oo , hence this curve has no infinite branch.
The greatest value of p corresponds to 6 = 0° or 180° for
which cos 26 = 1 and p = ±a.
It has already been shown that the curve passes through
the pole, hence the least numerical value of p is 0.
Taking into account symmetry and excluded values of 9,
the curve can be completely drawn by assigning to 6 values
from 0° to 45°. This curve is called the lemniscate.
e
2e
Cos29
p
Degrees
Degrees
1
±a
15
30
.86
±.93
30
60
.5
±.7a
45
90
When 6 in the equation has no coefiicient, it is usually
sufiicient in plotting to take values of 9 differing by 30°. If
9 has an integer coefficient as in this problem, smaller inter-
vals should be used, and when 9 has a fractional coefficient,
80
POLAR COORDINATES
it is often sufficient to take much larger intervals between
the values of 6.
In plotting curves, the student is advised to use polar
coordinate paper. Such paper is usually accompanied by
tables which faciHtate the calculation.
ILLUSTRATIVE EXAMPLES
1. Plot and discuss p =
1—2 cos 6
Intercepts. — li 9 = 0°, p = -6. If 9 = 180°, p = 2. 6 = 360°,
540°, etc., give no additional values to p. No values of 8 make p = 0.
Hence the curve crosses the polar axis in two points only, one point 6
units to the left, and the other 2 units to the left, of the pole.
Symmetry. — The equation is unchanged if 8 is replaced by —9, hence
the curve is symmetrical with respect to the polar axis.
Extent. — There are no excluded values of 8, since the value of p con-
tains no radical.
When 1 — 2 cos 9 = or cos 9 = T, p = 00 , therefore the curve has
infinite branches corresponding to 9 = 60° and 300°.
p wiU have the least value when 1 — 2 cos 9 is greatest, which will be
when cos 9 = — 1. Then p = 2 is the minimum numerical value.
A table of values is here given in which the natural values of cos 9 are
used. The figure proves to be an hyperbola.
e
Cos«
p
Degrees
1
- 6
30
.866
- 8.2
45
.707
-14.5
60
.5
00
75
.259
+12.4
90
6
120
- .5
3
135
- .707
2.5
150
- .866
2.2
180
-1
2
THE LOCUS OF AN EQUATION
81
If values of 8 greater than 180° were used, the same points would be
obtained as those determined by applying the principle of symmetry
with respect to the initial hne.
2. Plot and discuss the locus of p = a cos' ^•
Intercepts. — If 9 = 0, p = o; if fl = 180°, p = a/8; if 9 = 360°,
p = -a/8; if 9 = 640°, p = - a. If p = 0, 9 = 270°.
Symmetry. — Since 9 can be replaced by —9 without changing the
equation, the curve is symmetrical with respect to the polar axis.
Extent. — There are no excluded values of 9. p is never infinite. It
is greatest when 9 = 0°, for which value p = a.
Making a table of values and plotting, the figure is found to be as
shown below.
It should be noticed that in this curve it is not sufficient to plot from
0° to 180° but that points up to 9 = 270° are necessary before the appli-
cation of symmetry can be applied to complete the figure.
e
8
3
Cos?
p
Degrees
Degrees
1
a
45
15
.97
.91a
90
30
.87
.65 a
135
45
.71
.35 a
180
60
.5
.13a
225
75
.26
.02 a
270
90
EXERCISES
Discuss and plot the loci of the following:
1.
P = 5.
2.
p = 5 cos 9.
3.
p = 4 sin 6.
4.
p cos 9 = 4.
6.
p sin 9 = 4.
10.
•p2 cos 2 9 =
o2
11. The parabola p
1 + sin 9
6.
7.
8.
9 = 10°.
p = cos (9 + 45°).
p2 = a' sin 2 9.
8
'' l+2cos9'
82
POLAR COORDINATES
12. The cardioid p = a (1 + cos 6).
13. The elUpse p = ■= 1 ■
^ 5 - 3 cos 9
14. The hmason p = 4 (1 — 2 cos 9).
IB. p = 4(2 - cosfl).
16. p = a(l +sin9).
17. p = osin'^-
18. p2 sin 2 9 = o2.
2*
a
20. p = a sin" 5-
19. p = a csc^
33. Equations of the form p = a sin fe9 and p = a cos fe6,
where k is any integer, are of frequent occurrence. A
sketch of these curves sufficiently correct for many pur-
poses can be constructed by making use of the following
discussion.
Draw a radial hne corresponding to each value of 9 which
makes p = 0, also a radial Une corresponding to each value
of B which makes p a numerical maximum. Discuss the
changes which take place in p as 5 increases through each
interval determined by these radial lines.
Thus, plot the locus of p = a sin 2 6.
lip = 0,6 = 0°, 90°, 180°, 270°.
p has a numerical maximum of ±a when 6 = 45°, 135°,
225°, 315°. Therefore radial hnes are drawn at intervals
of 45° beginning with the polar axis.
Quad-
Aaein-
P vanes
rant oc-
creasea from
from
cupied
by curve
Degrees
Oto 45
Oto a
1st
45 to 90
a to
1st
90 to 135
Oto -a
4th
135 to 180
—a to
4th
180 to 225
Oto a
3rd
225 to 270
a to
3rd
270 to 316
Oto -a
2nd
315 to 360
-oto
2nd
MULTIPLE REPRESENTATION OF POINTS 83
The plan here used will often be of advantage in other
examples and should be kept in mind for use whenever
practicable.
EXERCISES
Construct the following loci:
1.
P = o cos 2 e.
6.
p = 4 sin 5 8.
2.
p = 4 sin 3 e.
7.
P = a cos e.
3.
p = a cos 5 e.
8.
p = 6 cos 4 e.
4.
p = 8 sin 4 e.
9.
p = 4 sin 6 9.
6.
p = 4 cos 3 8.
10.
P = a sin' e.
34, Difficulties arising from the multiple representa-
tion of points in the polar system. — The fact that the
same point may be expressed by more than one pair of co-
ordinates often leads to confusion and sometimes to error
unless great care is taken. In the rectangular system, where
each point has one pair of coordinates, and each pair of
coordinates corresponds to a single point, it is always safe
to conclude that if the coordinates of a point fail to satisfy
an equation then the point is not on the locus. This is not
always the case in the polar system, for it often happens
that if one pair of coordinates fails to satisfy an equation,
another pair representing the same point may show the
point to be on the locus.
Thus, in the equation p = a sin 2 5, if the point be taken
whose coordinates are (a/2, —15°) the equation is not sat-
isfied; but the same point when considered as determined
by (—a/2, 165°) is found to be on the curve.
Care must be taken to hold this multiple representation
of points in mind when considering the question of symmetry.
If the curve is symmetrical to the polar axis, then corre-
sponding to every point (p, 6) on the curve there must be a
point (p, —5) also on the curve. It has been shown, however,
that any point as (p, —B) may be on the curve even though
its coordinates, in that particular form, do not satisfy the
84
POLAR COORDINATES
equation of the curve, and thus the locus is sometimes sym-
metrical with respect to the polar axis even though the
equation is changed by the substitution oi —6 for 6. This
is shown in the case of the curve p = asin2B, drawn in Art.
33, which is found to be symmetrical with respect to the
polar axis, even though the usual test for symmetry fails.
Another case where confusion sometimes arises is that of
excluded values. It often happens that certain values of
6 make p imaginary and therefore these values of 6 are
excluded. It may be, however, that if the set of points
corresponding to these values of 6 were expressed by other
pairs of coordinates, these coordinates would satisfy the
equation, showing that the curve is foimd in that area from
which a too hasty conclusion would have excluded it. Thus,
in p'' = 4 sin 6, p is imaginary for values of d between 180°
and 360°. This might seem to indicate that there is no
part of the curve be-
low the polar axis. In
plotting points, how-
ever, it is found that
for every value of 6
in the first and second
quadrants, p has two
values, one positive
and the other nega-
tive, showing that the
curve is found in
each of the four quad-
rants.
p = ±2. The coordinates (-2, 90°)
Another pair of coordinates for the
These do not satisfy the equation.
Thus, when = 90°,
satisfy the equation,
same point is (2, 270°).
SPIRALS
85
EXERCISES
1. Show that the point ( — i, 150°) is on the curve p = cos 2 9
although its coordinates do not satisfy the equation. How may the
given point be written in order that its coordinates shall satisfy the
equation?
2. Determine whether the point (1, 210°) is on the curve whose
equation is p = 2 cos 4 6.
3. Discuss and plot p* = 4 cos 0.
4. Discuss and plot p^ = cos 3 6.
6. Discuss and plot p' = 1 — 2 sin e.
6. Discuss and plot p = sin 4 0.
35. Spirals. — A spiral is a curve traced by a point
which, while it revolves about the pole, continually ap-
proaches or recedes from this point.
There are five principal spirals as follows:
The spiral of Archimedes, p = ad.
The reciprocal or hyperbolic spiral, p = a/d.
The parabohc spiral, p^ = ad.
The lituus or trumpet, p^ = a/d.
The logarithmic spiral p = e"^ (e = 2.718+).
Plot the locus p = ad (where a is positive).
It is seen that when = 0, p = 0, and as B increases
without limit, p also increases without hmit. The curve
e
p
7r/2 = 1.67
1.57o
T = 3.14
3.14a
37r/2 = 4.71
4.71
27r = 6.28
6.28 a
57r/2 = 7.85
7.85a
3ir = 9.42
9.42a
0,2517
86
POLAR COORDINATES
thus starts at the pole and winds around the pole indefi-
nitely, receding from it with each revolution.
In plotting these curves, B is expressed in circular measure.
It is usually sufiicient to determine only such points as
correspond to values of B differing by 7r/2 radians. In
some examples it is more convenient to take the interval
between the successive values of fl to be 1 radian.
The curve sketched in the figure with the heavy line cor-
responds to positive values of B and that with the dotted
line to negative values of B. These two spirals constitute
the complete locus of the equation.
Plot the Iqcus p = 6°®.
Some definite value must be assigned to a. Suppose a = 1,
the equation becomes p = e*. Assigning to 6 values differing
by 1 radian, the following table is computed.
e
p
e
p
1
2
3
4
e»= 1
e = 2.72
e^ = 7.39
e3 = 20.1
e* = 54.6
-1
-2
-3
-4
e» = 1
e-i= .37
e-'= .14
e-»= .05
e-'= .02
It is seen that as 8 increases from radians to 4 radians,
p increases from 1 to 54.6, also that as B increases indefinitely,
p also increases indefinitely.
As B decreases from to —4 radians, p decreases from 1
INTERSECTIONS OF CURVES 87
to .02, and as 6 decreases indefinitely, p approaches as a
limit. Hence the curve winds around the pole indefinitely,
coming closer and closer to it with each revolution, but not
reaching it until an infinite number of revolutions in clock-
wise direction have been made.
EXERCISES
1. Plot the spiral p = a/$.
2. Plot the spiral p« = aS.
3. Plot the spiral " = a/0.
36. Intersections of curves. — As in rectangular coordi-
nates, if two equations are solved simultaneously, points
are found whose coordinates satisfy both equations and
hence such points are the intersections of the two loci. In
polar coordinates, this process does not always give all the
common points, for since the coordinates of a point may
be written in a number of different ways, it may happen
that one equation is satisfied by one pair of coordinates of
the point of intersection, and the other equation by a differ-
ent pair of coordinates of the same point.
To make sure that all intersections are obtained, the
curves should always be drawn. These will show any addi-
tional common points. Care must always be taken to make
sure whether the pole is on both curves.
ILLUSTRATIVE EXAMPLE
Find the points of intersection of the two curves p = — 1 — cos 9
and p = 1 + cos 9.
Equating the two values of p,
— 1 — cos9 = 1 + cos 9.
Hence 2 cos 9 = -2, cose = -1, 9 = 180°.
Substituting in either equation, p is found to be 0. The pole then
is a common point.
88
POLAR COORDINATES
Plotting the loci, taking account of symmetry, the figure is as shown
below.
p = -1 - COB e
p = 1 + cos 9
e
p
e
p
Degrees
45
90
135
180
-2
-1.7
-1
- .3
Degrees
45
90
135
180
2
1.7
1
.3
The point marked A in the figure has coordinates (1, 90°) for the
right hand curve and ( — 1, —90°) for the left hand curve.
The point marked B in the figure has coordinates (1, —90°) for
the right hand curve and ( — 1, 90°) for the left hand curve.
The curves then intersect in three points.
EXERCISES
Find the points of intersection of the following pairs of curves and
plot the loci.
1. p = o. 4. p = V2,
p =acos0. p = 2sin9.
2. I? =0? sin 29, B. p = sin 2 B,
p = o sin fl. p = sin 8.
3. p = 1 + cos9, 6. p = 1 + sine,
2p = 8ec4fl. p(2-sine)=2.
7. p = 2 sin 3 B,
p = 2smB. Ans. (0, 0), (V2, 45°), (VI, 135°).
8. p2 = 2 a' cos 2 B,
p = a. Am. (a, ±30°), (a, ±150°).
9. p2 = a? cos B,
p = a.
INTERSECTION OF CURVES 89
10. p (3 - 2 cos e) = 1,
p = 1 - cos 9. Am. (I, 60°), (I, 300°).
11. p = 6 — cos 6,
p(l -2cosfl) = 6.
12. p =2 -2 sine,
p = 2 cos 2 9.
13. p =aB,
p = a/d.
14. p = 2 a sin 9 tan 9,
p = a sin 9.
16. Show that 9 = 60° and p = a. intersect in two points.
MISCELLANEOUS EXAMPLES
Discuss and plot the following:
1. p = 2 sec 9.
2. p = o tan^ 9 sec 9 (semi-cubical parabola).
3. p = a? sin 4 9.
4. p = 2 a sin 9 tan 9 (cissoid).
5. p = a sec^ 9.
6. p = 1 + sin 2 e.
7. p =a(sin29 + cos2e).
_ 2 o sec 9
'' ~ 1+ tans'
{Hint. Change to sine and cosine when calculating for 9 = 90°.)
3 o tan 9 sec 9 ,. ,. , _ . ,
9. p = , , , — 7-— (folium of Descartes).
1 + tan' 9
10. p2 =a?0. -cos 9).
11. p = 2 sin fl + cos 9.
12. p" = cos 4 9.
13. p2 cos 9 = a^ sin 3 9.
14. p = 4 sin 5 9.
16. p« = "^^
a^ sm2 9 + 62 ^os* g
9 , . 6
16. p = cos s + sin s*
,
\
1
/
/
/
/
J
0'
/
1
CHAPTER V
TRANSFORMATION OF COORDINATES
37. If a point is referred to a given system of axes, its
coordinates are fixed. If the axes are changed, the coordi-
nates of the point are also
changed. Thus, if the
point P when referred to
OX and OY is (5, 5), it is
seen that if referred to the
parallel system, O'X' and
O'Y' through 0' (3, 1) the
coordinates of P are
changed to (2, 4). Simi-
larly, the equation of the
line O'P when referred to
OX and OY is y = 2 x — 5) and when referred to O'X' and
O'Y' is y = 2x.
This example illustrates that an equation of a locus is
sometimes simphfied by a change of axes and it is therefore
often desirable to find the equation of the curve in a new
system. To do this it is necessary to determine the laws
which connect the coordinates of a point in the given system
with the coordinates of the same point in the new system.
Transformation of coordinates is the operation of chang-
ing the axes. There are two principal transformations
in rectangular coordinates. When the new axes are re-
spectively parallel to the old through a new origin the
transformation is called translation of axes. When the
origin is unchanged but the axes are each rotated through
a given angle, the transformation is called rotation of axes.
90
TRANSLATION OF AXES
91
38. Translation of axes. — Ij x and y are the coordi-
nates of any point before translation to a new origin (h, h) and
x' and y' the coordinates of the same -point after translation,
then
X = x' + h,
y = y' + k.
(13)
Y
Y'
■
0'
M'
v'
— X
N
- M
to meet the a;-axis in N.
Proof. — Let OX and 07 be the given set of axes. Through
0' having coordinates (h, k)
when referred to the given
axes, draw a new set O'X'
and O'Y' parallel respec-
tively to OX and OY. Let
P be any point in the
plane. Its coordinates in
the given system are rep-
resented by X and y and
in the new system by x'
and y'. Draw the ordi-
nate MM'P and extend O'Y'
Then
X = OM, x' = O'M', h = ON,
y = MP, y' = M'P, k = NO'.
From the figure it is seen that
OM = 0N + NM = 0N + O'M'
and MP = MM' + M'P = NO' + M'P,
whence x = x' + h
y = y' + k.
In recaUing all formulas of transformation it is well to
hold the figure in mind as an aid to the memory.
To transform an equation referred to a given systern of
axes to another system parallel to the first through the
point (h, k), replace x in the given equation hy x' + h and y by
92 TRANSFORMATION OF COORDINATES
y' + k and simplify the result. This gives the new equation
of the given locus in which x' and y' are the variable coordi-
nates in the new system. It is customary to drop the primes
when the work of transformation is finished.
EXERCISES
1. What are the new coordinates of the points (3, —3), (—4, 2),
(0, —2), (4, 0) referred to parallel axes through (1, 2)?
2. Transform the equation 3x + 2y = 12 when the axes are
translated to a new origin (—2, —3). Construct the two sets of
axes and plot the locus of each equation, showing that they represent
the same Une. Ans. Sx -\-2y = 24.
3. Transform the equations y — x = 3 and 3y + 2x = i when
the axesiare translated to a new origin at their point of intersection.
4. Transform the following equations to a new set of axes parallel
to the old, the new origin as indicated. In each case draw both sets
of axes and the curve.
(a) a;2-2x-2/2 + 42/ = 4, (1, -2).
(6) 4a;2 -8a; + 9^-362/ + 4 = 0, (1,2). Ans. ii? +9y''=Z6.
(c) x' -2hx + y^ -2ky + h' +¥ =0, (h, k). Ans. 3?]+ y^ = 0.
(d) y + 2 = (,x + iy, (-1, 2). Ans:y = 3^.
(e) y' + iy^ix-iy, (1, -2).
G. The equation of a curve after translation to a new origin ( — 1, 2)
is a^ + 2/' = 9. What was the original equation?
Ans. (x + 1)« + (y - 2)^ = 9,;
39. Rotation of axes. — If x and y are the coordinates
of any point before rotation through an angle 0, and x' and y'
the coordinates after rotation, then
X = x' cos 6 — y' sin 9, . .
y = «' sine +y' cose. ^^^
Proof. — Let OX and 07 be the given set of axes, and let
OX' and OY' be the positions of the axes after they have
been rotated about the origin through an angle 6. Take
P any point in the plane whose coordinates in the given
ROTATION OF AXES
93
Draw
system are x and y, and in the new system x' and y'.
the ordinates MP and M'P.
Then OM = x, MP = y,
OM' = x' and M'P = y'.
Draw through M' the lines
RM' and NM' parallel to
the x and y axes respec-
tively. The angle RPM'
is equal to 8. (Why?)
It is seen from the figure
that
x = OM = 0N- MN = 0N - RM' = a;'cos0 - y'sinO,
y = MP = MR + RP = NM' + RP = a;' sine + y'cosd.
EXERCISES
1. Find the coordinates of the points (3, 1), (—5, 0), and (0, —2),
after the axes have been rotated through 45°, also through 90°.
2. Transform the equation x^ + y^ = 16 when the axes are rotated
through 60°. Ans. o? + y' = 16.
3. Show that the equation x^ + y^ = a? will be unchanged after
rotation of the axes through any angle $.
4. Transform the following equations when the axes are rotated
through the angle given. Construct both sets of axes and the curve.
(o) xy = 4, 7r/4.
(b) y'=4:x, 7r/2.
(c) x^ +2xy + y^ - X -y =0, ir/4.
id) 3a;2 - 4x2/ + 8x - 5 = 0, tan-' 2.
(e) x/a + y/b = 1, tan"' a/6.
(/) 3y' +Sxy - Sx^ = 0, tan"' i
5. The equation of a locus after the axes have been rotated through
—45° is 2/ — a; = 1; what was the equation before rotation?
6. Through what angle must the axes be rotated in order that the
new X-axis shall pass through (3, 4)?
7. Transform the equation xy— y+2x — 6=0 to new axes
whose origin referred to given axes is (1, —2) and which make an angle
of 45° with those axes. Ans. a? — y^ = S.
Hint. — First translate to the new origin, then rotate the axes.
94 TRANSFORMATION OF COORDINATES
8. . Three sides of a triangle are x — y = i, x + y = 6, and y + 2x
= 20. If the first two lines are chosen as axes, what will be the equa-
tion of the third? Ans. 3x-y = 9V2.
40. Degree of equation not changed by translation and
rotation. — Since in each of the formulas of transforma-
tion the values of x and y are of first degree in x' and y',
therefore the transformed equation will never be of higher
degree than the given equation. That it cannot be of lower
degree is shown by the fact that if this were the case, a
transformation back to the original system of axes would
have to raise the degree in order to give the original equa^
tion. This has been shown to be impossible.
41. Simplifications by transformation. — One of the
principal advantages obtained from transformation is the
simpKfication of equations. Some of these simplifications
are best accompHshed by translation, others by rotation.
By translation to a proper new origin it is often possible
to remove the first degree terms, to make the constant
term disappear, or to ehminate one first degree term and the
constant term.
The methods by which these results are usually accom-
pUshed are illustrated in the following examples:
1. Simplify the equation x^ — 2x + y^ — 6y= 15 by
translation to a new origin.
Substituting x = x' + h and y = y' + k in the equation
x^-2x + y^-6y = 15 (1)
and collecting terms, the equation becomes
x'^ + y'^+i2h-2)x'+{2k- 6) y'
+ h^ + k^ - 2h - 6k - 15 = 0. (2)
It is readily seen that it is possible to so choose h and k
that the coefficients of x' and y' shall be 0. Thus
2h-2 = 0, h = l, (3)
2 fc - 6 = 0, fc = 3. (4)
SIMPLIFICATIONS BY TRANSFORMATION 95
Substituting these values back for h and fc, the equation
becomes
a;'2 + 2/'2 = 25. (5)
One advantage gained by the transformation is that the
new equation shows that the locus is symmetrical with re-
spect to both axes since there are no odd powers oi x ov y
in the equation.
Another method of accomplishing the same result is to
complete the squares of all x terms and of aU y terms, thus
(a;='-2a;+l) + (2/^-6y + 9) = 15+1+9, (2)
or {x - \Y +{v- 3)2 = 25. (3)
It is readily seen that if the axes are translated to a new
origin at (1, 3) the equation will have no first degree terms.
Although often desirable, it is not always possible to re-
move the first degree terms. This is illustrated in the
second example.
2. Simplify by translation 2/^ + 42/ — 8a; — 4 = 0.
Substituting x = x' + h and y = y' + k, in the equation
2/2 + 42/-8x-4 = (1)
and collecting terms, the equation becomes
y"' + y' {2k + 4) - 8x' + ¥ + 4:k - 8h - 4: = 0. (2)
It is evident that the coefiicient of x' cannot be made
equal to zero. The quantities h and fc may, however, be
determined so that the coefficient of y and the constant
term shall be zero.
Thus 2 fc + 4 = (3)
and ¥ + 4:k-8h-4: = 0. (4)
Whence k = -2, h = - 1.
The equation then reads y'^ = 8 x'. (5)
This locus is symmetrical with respect to the new a;-axis
and passes through the new origin.
96 TRANSFORMATION OF COORDINATES
This problem can also be solved in a manner similar to
the second method used for the first example.
By rotation through a proper angle it is possible to re-
move the xy-term. from an equation of second degree as is
shown in the following example.
3. Remove the xy-term. from Sx' + lOxy + 3y^ = S.
Substituting x = x' cos 6 — y' sin 9 and y == x' sinO -\-
y' cos 6 in the equation
3a;2+10a;2/ + 32/2 = 8, (1)
x"^ (3 cos^ 0+10 sin cos 6 + 3 sin" d)
+ x'y' (10 cos^ e - 10 sin2 6)
+ 2/'2 (3 sin^ - 10 sm cos e + 3 cos^ 6) = 8. (2)
The coefficients of all the terms can be changed to func-
tions of 2 6. The equation then becomes
x'^ (3+5 sin 2 e) + x'y' (10 cos 2 (9) + y'"^ (3-5 sin 2 6)=%. (3)
Since the new equation is to contain no x'y' term, there-
fore 10 cos 2 = 0, whence cos 2 9 = 0, 2 9 = 90°, and 6 = 45°.
Substituting B = 45° in equation (3),
8x'2-2 2/'2 = 8. (4)
EXERCISES
1. Simplify the following equations by translation of axes. Plot
both pairs of axes and the curve.
(o) a;2 + 4a; + 9^2 - 18 2/ + 4 =0. Ans. I'+O^ =9.
(6) a? + 2a; -gj/! -363/ = 44. Ans. 3? -Qy^ = 9.
(c) a;2 -6a; + ^2 + 62/ = 7.
(d) 3/2 -82/ + 6a; -2 =0.
(e) a;2+4a; = 2y + 6.
2. By rotating the axes, remove the xy-ierai from the following.
Plot both pairs of axes and the ctlrve.
(a) x^ + 2xy + \^ = 9.
(b) xy = 4.
(c) 5a^ + 6a;2/ + 52/' = 8^
(d) a? + 2xy + 3/^ + 4 v'2 (a; - 2/) = 0.
TRANSFORMATION 97
3. In the following, remove the xy-teim by rotation of axes. Con-
struct the two sets of axes and the curve in each example.
(a) 3? — xy + y' + 5x~y = l.
(6) 2xy-2V2y = 4.
4. To what new origin must the axes be translated in order that
the two lines 2a; — y — 3=0 and x + 2y + 1 =0, when referred
to the new system, shall have no constant term? Find the equations
referred to the new axes.
6. Through what angle must the axes be rotated in order that the
new equation of the Une x — y = 4 shall have no a;-term? Check
from the figure. Arts. 45°.
6. Transform the equation x — y = 6to the form y = 0.
Hint. — First translate the axes to a new origin located anywhere
on the given line and then rotate the axes.
42. Transformation from rectangular to polar coordi-
nates and vice versa.
If X and y are the coordinates of a point in a rectangular
system and p and 6 the coordinates of the same point in a polar
system, the origin and the x-axis coinciding respectively with
the pole and the polar axis, then
« = pcose, ,jg.
y = p sin 6.
Proof. — Let OX and OY represent the rectangular axes,
then and OX are the pole and initial Hne respectively.
Let P represent any point whose
coordinates in the rectangular
system are x and y and in the
polar system p and 6. Draw
MP perpendicular to OX.
Then x = OM, y = MP, p = OP,
e = angle MOP.
It is readily seen from trigonometry that x = pcosd
and 2/ = p sin d.
98
TRANSFORMATION OF COORDINATES
It is seen from the following figures that if P is located
in any other quadrant than the first, the proof is identical
with that given above.
7/ p and 6 are the coordinates of a point in a polar system
and X and y the coordinates of the same point in a rectangular
system, the pole and polar axis coinciding respectively with
the origin and x-axis, then
p= iV^a + y",
e = tan~^y/*.
(16)
Proof. — These results can be read directly from the
figures. It is also seen that
cos fl = - = ,
P zhVx^ + y*
, sin e = - =
y
P ± Va;2 + j/2
It is particularly helpful in this set of formulas, as has been
suggested before in this chapter, that the student keep the
figures in mind when recalling formulas of transformation.
TBANSFORMATION 99
EXERCISES
1. Find the polar coordinates of the points (0, 3), ( — 3, 3), ( — 3, — 4) ,
(5, -12).
2. Find the rectangular coordinates of the points (3, 7r/4), (4, v),
(5, -x/6), (2, 5x/4).
3. Transform the following equations from rectangular to polar
coordinates. Plot each curve.
(a) X = a. Ans. pcosS = a. (/) xy = 4.
(6) 2/ = 6. (g) x^+y' + 4:x = 0.
(c) y = X. (h) y'(2a -x) = x'.
(d) 3?+^ = a". Ans. p' = a?, (i) (x^ + j/^)" = a? (x^ - y^).
(e) a;2 _ 2/2 = 6. 0') a;^ + j/^ + 2 ox = o Va;^ + y''.
4. Transform the following equations from polar to rectangular
coordinates. Plot each curve.
(o) e = 45°.
(6) p cos e = 2.
(c) /) = 2 a cos 9.
ffinJ. — In (c) and similar examples it is sometimes best to multi-
ply by p before transforming.
(d) p2sin2e = 4. (h) p = a (cos29 + sine).
(e) f^ = a*cos2fl. (i) p = 2atanflsine.
(/) p = ocosfl +6sine. (j) p = 2 + 3cos9.
(g) p = a (1 — cose). (A;) p = o (1 + cos29).
6. Translate axes to new origin and then transform to polar coordi-
nates:
(a) x' + j/' + ix + 8y - 20 = 0, new origin ( -2, -4).
(6) «» - 2/« -H 2a; -1- 62/ = 24, new origin (-1, 3).
CHAPTER VI
THE CIRCLE
43. A circle is a locus traced by a point which is every-
where equidistant from a fixed point, called its center.
The distance of any point from the center is called the
radius.
A circle, therefore, is determined, and its equation can be
written if its center and radius are known.
First standard equation of a circle. Center and radius
known. — The equation of a circle whose center is C {h, k) and
whose radius is r is
{x-hY+{y- kY = r". (17)
Proof. — Let P (x, y) represent any point on the circle.
By the definition of a circle, PC = r. From the formula
for the distance between two points,
formula (1),
PC= V{x-hy+iy-ky,
whence V{x — hy + (y — ky = r.
Squaring, (x - hy + (y - ky = r\
Second standard equation of circle. Center at origin,
radius r. — The equation of a circle whose center is at the
origin and whose radium is r is
x^ + y> = r>. (18)
Proof. — Substituting h = and A; = 0, in equation (17),
it reduces to equation (18).
100
^
P
^p-
/
\ ^
J. /N
1
GENERAL FORM OF EQUATION OF CIRCLE 101
EXERCISES
1. Write the equations of the circles whose centers and radii are
as follows:
(a) C (1, 4), radius 5. (d) C (-4, 0), radius 2.
(6) C (0, 0), radius 2. (e) C (-1, -2), radius 7.
(c) C(-3, 4), radiusS. (/) C (5, -1), radius 3.
2. Write the equations of the circles, having given:
(o) Center at the intersection of the Unes 2a; — y — 3=0 and
x + 3y — 5=0, and radius 5.
(6) Center at origin and passing through the point (5, 6).
(c) Line joining (1, 5) and (—3, 1) as diameter.
(d) Center at (5, 6) and tangent to a;-axis.
44. General form of equation of circle. — Equation (17)
when expanded becomes
x^ + y^ - 2hx-2ky + h^ + ¥ -r^ = 0. (1)
It is thus seen that the equation of a circle is of second
degree. If the constants are collected, equation (17) is
seen to be of the form
x^ + y^ + Dx + Ey + F = 0. (2)
It will be shown that every equation of this form repre-
sents a circle.
Completing the squares of the a;-terms arid of the ^/-terms,
equation (2) becomes
[^+2)+[y+2) = — T — ' (^^
from which it is seen by comparison with (x — hY +
{y — hy = r^ that equation (2) represents a circle whose
center is at (— D/2, —E/2) and whose radius is
I VD2 + jB2-4F.
If D^ + B" — 4 F < 0, the radius is imaginary and no
circle is possible. If D^ + -E^ — 4 F = 0, the equation rep-
102 THE CIRCLE
resents only one point, the center. The foregoing may be
summarized as follows:
The equation x" + y'^ + Dx + Ey + F = (19)
represents a circle whose cente r is { — D/2, —E/2) and whose
radius is J VW+W^^TF, providing D^ + E^ - iF > 0.
It should be noticed that equation (19) is not the most
general form of the equation of second degree, this being
Ax^ + Bxy + Cy'' + Dx + Ey + F = 0.
If, in this equation, B = and C = A, it is possible to
divide through by A and thus reduce it to the form of the
general equation of the circle x^ + y^ + Dx + Ey + F = 0.
Whence:
The general eqiudion of second degree Ax^ + Bxy + Cy^ +
Dx + Ey + F = represents a circle if B = and A = C.
In plotting loci of equations of second degree, the student
should look for the presence of the conditions which make a
second degree equation a circle, as when these exist, he can
save himself all the work of discussion and of plotting
points, since a circle can be readily drawn as soon as its
center and radius are known.
In determining center and radius, he can either complete
the square of the a;-terms and of the ^/-terms or can make
use of the facts learned in connection with the general
equation.
Thus, plot the locus of 2 a;^ + 2 j/^ - 18 a; + 16 2/ + 60 = 0.
Since the coefficients of x^ and y^ are equal, this can be put
in- the form of the general equation of a circle by dividing
by 2, giving
a;2 + 2/''-9a; + 8y + 30 = 0. (2)
Completing the squares,
(a;-|)='+(2/ + 4)2=(f)^ (3)
RADICAL AXIS
103
Whence by comparison with standard equation (17), the
center is (|, -4) and radius f .
Or, comparing (2)
with the general equa-
tion (19), ^:_^
h = -D/2 = I,
k= -E /2 = -4,
J" = i Vsi + 64-120
= f,
whence the circle is as
shown. y'
45. Radical axis. — In Art. 16, it was learned that if
an equation is multiplied by any constant and added to any
other equation, the result represents a locus through the
points of intersection of the two given loci. If the equation
of a circle is multiplied by k and added to the equation of
another circle the resulting equation represents a system of
circles, since for every value of the constant multiplier k
the coefficients of x^ and y^ are the same.
If the equations of the two circles are put into general
form (19), the terms of second degree will be eliminated
if the constant multiplier is —1, or if the equations of
the two circles are subtracted. This result being of first
degree represents a straight line. When the circles inter-
sect, this line is their common chord. When the circles
touch at one point only, it is their common tangent.
Whether the circles have any common points or not this
line is called the radical axis.
This radical axis is the locus of points from which tangents
to the two circles are of equal length as will be proved in
Ex. 16 of the list which follows.
In finding the intersection of two circles, it is best to first
find the radical axis and then find the intersection of this
with either of the given circles.
104 THE CIRCLE
Exercise. — Find the intersections of the circles:
(a) a;" + j/2 - 6a; + 4 = and x* + 2/2 - 4a; - 41/ = 0.
lb) x' + y^-y = &nd 2x^ + 2y^ + x = 0.
46. Circle determined by three conditions. Since the
equation of the circle in either of the two forms
(x - ny + (2/ - kY = r\
or x^ + y^ + Dx + Ey + F = 0,
has three arbitrary constants, therefore three conditions are
necessary in order to determine its equation.
Sometimes it is best to use the data given to obtain three
equations in h, k, and r and sometimes in D, E, and F. From
the three equations, the three constants can be determined,
and the required equation obtained by substituting their
values back in the corresponding standard equation.
In other examples, it is better to determine more directly
the center and radius by using the given data in connection
with equations and formulas already derived. Thus, the
center is often at the intersection of two lines whose equations
can be found, and the radius the distance between two kno^yn
points. Whenever the coordinates of the center and the
radius are known or have been found, it is only necessary to
substitute in standard equation (17).
ILLUSTRATIVE EXAMPLES
1. Find the equation of the circle through the three points (4, 6),
(-2, -2), and (-4, 2).
Let the required equation be
'3? + 'f+Dx-\-Ey + F = Q, (1)
in which Z>, E, and F are unknown constants. Since each of the
points is on the circle, therefore the coordinates of the three given
points must satisfy equation (1), whence
16 + 36 + 4D + 6£? + f = 0, (2)
4 + 4-2D-2£; + F = 0, (3)
16+4-4D + 2JS + F = 0. (4)
CIRCLE DETERMINED BY THREE CONDITIONS 105
Solving (2), (3), and (4), for D, E, and F,
D = -2, E = -i, F = -20.,
Whence the equation of the circle is
a;2 -{- ^2 _ 2x - 42/ - 20 = 0. (5)
Changing to form (17), (a; — 1)* + (y —
2)2 = 25, from which it is seen that the
center is (1, 2) and the radius is 5. The
figure is as shown.
This problem may also be solved by
finding the equations of the perpendicu-
lar bisectors of the Unes joinijig two pairs of the points. The intersection
of these bisectors will be the center, and the distance from this center
to any one of the given points will be the radius of the required circle.
Substitution in standard equation (17) will give the equation of the circle.
2. Find the equation of the circle whose center hes on the Une
y — X = 1, and which is tangent to each of the lines ix — Zy = IS
and 3 a; + 4 s^ = 10.
Represent the three lines in the order given by Li, Li, and Lt.
It is seen from the figure that there are two circles which fulfil the
conditions mentioned, and from geometry it is known that the center
of each hes on one of the bi-
/'-i sectors of the angles between
1/2 and Lt. Let the bisectors
be represented by Lt and Li.
The equation of that circle
whose center C Ues on the bi-
sector L4 will first be deter-
mined. By the method of Art.
27, the equation of Lt is found
to be 7 a; + 2/ =25. The in-
tersection of this line with Li
determines the center, C (3, 4).
The radius is the distance from
either L2 or Lt to C, and by
Art. 26 this is 3. Substituting
the coordinates of the center
land the radius in standard equation (17), the equation of the circle in
the first quadrant is found to be {x — 3)^ + {y — AY = 9. The equa-
tion of the other circle can be found in a similar manner to be (a: + 2)^
+ (2/ + 1)= = 16.
106
THE CIRCLE
3. Find the equation of the circle tangent to the Une ix + 3 y
= 15 and passing through Pi (7, 4) and
P2 (1, 4).
Let C {hj k) represent the center of the
required circle and r the radius. Also let
L represent the given line. The three
given conditions lead to three equations
in h, fc, and r. Since L is tangent to the
circle, the distance from line L to C is r.
By formula of Art. 26, this distance is
4;i + 3A;-15
5 = '■• (^^
The points (7, 4) and (1, 4) are on the circle. Therefore their
coordinates must satisfy the equation
(x - hy + {y- k)' = r^.
Whence (7 - hy + (4 - kY = r' (2)
(1 - hY + (4 - ky = T^. (3)
Subtracting (2) from (3), ^ = 4. (4)
1 +3fc
Substituting in (1),
(5)
Combining (4) and (5) with either (2) or (3), fc = 8 or -'/ and r = 5
or ^. The equations of the required circles then are
(x - i.r + (2/ - 8)2 = 25,
and (x - 4)2 + (^ - Y)' = W-
EXERCISES
1. Find the coordinates of the center and the radius of each of the
following circles:
(a) a;2 + 2/2 - 4a; + 8?/ + 4 = 0.
(6) 3x2 + 32/'-6a; + 122/ = l.
(c) 1? -^'f = ^x.
(d) 2x2+22/2 + 4s + 82/ = 0.
(e) x2 + 2/2 -I- 10 ax — 24: ay = 0.
(/) x2 + 2 (o + b) X + 2/2 + 2 (o - 6) 2/ = 4ob.
(g) 2a?+22/= = 32/.
(h) x2 + 4x + 2/2 - 62/ + 13 = 0.
2. Find the equation of a circle through the three points (3, 1),
(6, 0), (-1, -7). Am. x2+2/2_6x + 82/ = 0.
CIRCLE DETERMINED BY THREE CONDITIONS 107
3. Find the equation of a circle '
■ (a) center at ( — 1, 4), tangent to 5 a; + 12 y + 9 =0.
(6) center on y-axis, passing through the points (3, —1) and (3, 7).
(c) center on s-axis, passing through (0, 0) and (1, 5).
(d) passing through (5, —5), having the same center as 2 a;^ + 2 2/*
+ 4x - 12 2/ + 3 =0.
(e) having Une joining ( — 1, 6) and (5, 2) as diameter.
(/) passing through (1, 0) and (6, 1) and having center on line
2a; + 2/ + 4 = 0.
(g) radius 4, tangent to a;-axis at (3, 0) and lying above it.
4. Find the equation of that diameter of the circle 3 a;' + 3 j/'
+ 12 a; — 12 y — 1 =0 which makes an angle of 45° with the x-axis.
6. A diameter of the circle x' + y^ + 4:X + 6y = 3 passes through
(1, —1). What is its equation and the slope of the chords it bisects?
6. Find the equation of that chord of the circle a;^ + y^ = 25 which
is bisected at (2, 3).
7. Prove that a circle can be drawn through the four points (0, 2),
(3, 3), (6, 2), and ( — 1, —5). Find its center and radius.
8. Find the equation of the circle
(o) radius 10, passing through (—2, —2) and (0, —4).
(6) in the first quadrant, of radius 3, and tangent to both axes.
(c) tangent to both axes, center on the line j/ — . 2 x = 3.
(d) passing through (1, —3) and (2, —2) and tangent to 3 x — 4 ?/
= 15.
(e) center on 2 x + y = 4 and tangent to y — 3 x = & and
3x + 2/ + 6 = 0.
(f) tangent to both axes, distance from center to origin = 4,
and lying in the fourth quadrant.
9. Find the equation of the circle inscribed in the triangle whose
sides are the Hues y — d = 0, 12 x — 5 y = 21, and 12 x + 5 y + 21
= 0. Ans. x2 + v^ - 2 2/ = 3.
10. Find the equation of the circle circumscribed about the tri-
angle whose sides are the Unes y + x = 0, 3y +x = 0, and 2y + x
-1=0.
11. Find the equation of the circle tangent to the x-axis, through
the point (4, 1) and center on the hne y = 5x.
12. Find the equation of the circle whose center is on the y-axis
and which passes through the points of intersection of the two circles
xSi + yi -5x - 7 y + Q = a,nd x^ + y' - 4:x - 4:y + i =0.
108 THE CIRCLE
13. Find the equation of the common chord of the two circles
a^+3/' + 6a;-4^ + 3 =0 and a;^ + j/^ - 2a; + 4 2/ - 5 = 0, _and
prove that it is perpendicular to the line of centers.
14. Prove that the square of the length of the tangent from Pi (xi, yi)
to the circle x^ + y'^ + Dx -irEy + F = Qi9x{? + yi^ + Dxi + Eyi + F.
Hint. — Join the center with the point Pi and with the point of
contact. These Unes with the tangent form a right triangle.
16. Prove that the point ( — 1, —1) is on the radical axis of the
two circles a;2 + 62; + 2^-42/ + 9=0 and x'^ + 'f — ^x — 2y + l
= 0, and show that the tangents from this point to the two circles are
equal.
16. Find the equation of the locus of the point which moves so
that the lengths of the tangents from this point to the two circles
3? + T/^+Dx + Ey + F = Q and x^ -^ y^ + D^x + Eiy + Fi = are
equal. Show that this locus is the radical axis of the two circles.
17. Given the three circles s' + ex + j/^ — 42/ + 9 = 0, 3? + jf
-4a;-2j/ + l=0, and x^ + y'^-Zx~y + l = Q. Taking the
circles in pairs, find the equations of the radical axes and prove that
they meet in a point.
18. Prove analytically that every angle inscribed in a semicircle
is a right angle.
Hint, — Take the extremities of the diameter as (— o, 0) and (o, 0),
thus making the equation of the circle a? + 2/^ = o*.
19. Prove analytically that if a perpendicular is drawn from a
point on a circle to a diameter, the length of the perpendicular is a
mean proportional between the segments it cuts oflf on the diameter.
20. Prove that the foUowing loci are circles and find the radius
and the coordinates of the center in each:
(a) A point moves so that the sum of the squares of its distances
from (3, 0) and ( — 1, —4) is always 40.
(6) A point moves so that its distance from (1, 3) ia twice its
distance from (—2, —3).
(c) A point moves so that the square of its distance from (2, 3)
is equal to its distance from the line 4a; — Sj/ — 15=0.
21. Prove that the following loci are circles:
(a) A point moves so that the sum of the squares of its distances
from two fixed points is constant.
Hirii. — When no mention is made of axes or coordinates, it is
always advisable to choose these in such a way as to make the work
CIRCLE DETERMINED BY THREE CONDITIONS 109
as simple as possible. Thus, in the above problem take the x-axis
through the two points with the origin halfway between them.
(6) A point moves so that the sum of the squares of its distances
from the four sides of a square is constant.
(c) A point moves so that the square of its distance from the base
of an isosceles triangle is equal to the product of its dis-
tances from the other two sides.
(d) A point moves so that the square of its distance from a fixed
point is proportional to its distance from a fixed line.
22. A point moves so that its distances from two fixed points are
in a constant ratio K. Show that this is a circle excepting when
£ = 1, in which case it is a straight line.
CHAPTER VII
THE PARABOLA
47. Conic sections. — The three curves next considered
belong to a general class called conic sections. This name
arises from the fact that each of these curves can be ob-
tained by passing a plane through a right circular cone.
Many of the properties of these curves were known by
the early Greek geometers among whom the principal in-
vestigators were Archimedes and Appolonius about 200 B.C.
The former computed the area of a parabolic segment and
of an elUpse. The latter discovered that all- three curves
can be cut from the same cone and investigated many
problems peculiar to the hyperbola.
That the knowledge of conic sections could be made of
great practical use in studying the laws of the universe was
not learned until after the passage of many centuries.
About 1600, Kepler in Germany discovered their importance
in the study of the motion of the heavenly bodies, and, about
the same time, Gahleo in Italy discovered that the path of
a projectile is a parabolic curve. The field of their useful-
ness has spread until a large group of problems in physics,
mechanics, and architecture are now known to depend upon
a knowledge of these curves for their solution.
Although these conic sections differ very much in appear-
ance, it is found that they can all be generated by the same
law, viz., A conic section is the locus traced by a point which
moves so that its distance from a fixed point bears a constant
ratio to its distance from a fixed straight line.
The fixed liae is called the directrix, the fixed point the
focus, and the fixed ratio the eccentricity, represented by e.
110
EQUATION OF THE PARABOLA
111
Equation of a conic section. — Take the directrix as the
j/-axis and the perpendicu-
lar through the focus on
-the directrix as the x-axis.
Let P {x, y) be any point
on the curve. Draw PD ,
perpendicular to YY'. Call
the distance OF = 2 p.
FP
•By definition, jyB = ^•
By formula (1), FP = ^{x-2vY + y', DP
V{x-2py + y^
X
= e.
Therefore,
Clearing of fractions and collecting,
(1 - e^) «® - ipx + 4p2 + y« =
(20)
From this equation it is seen that the curve is symmet-
rical with respect to the a^axis which is the perpendicular
from the focus on the directrix. For this reason the line
is called the principal axis of the curve.
By letting y = 0, the intercepts on the principal axis can
2P _^ 2p _
be found to be
and
1+el-e
When e = 1, the curve is called a parabola. It cuts the
principal axis in one finite paint, halfway between the focus
and the directrix.
When e < 1, the curve is called an ellipse.' It cuts the prin-
cipal axis in two points on the same side of the directrix as
the focus.
When e > 1, the curve is called an hyperbola. It cuts the
principal axis in two points on opposite sides of the directrix.
48. Equation of the parabola. — Since, in the parabola,
e = 1, the definition of this curve can be stated :
112
THE PARABOLA
A parabola is a locus traced by a point equidistant from a
fixed point called the focus and a fixed line called the directrix.
It was seen in Art. 47 that the parabola passes through
a point halfway between the focus and directrix. This
point is called the vertex. It is found that the simplest
form of the equation is obtained when this point is taken
as origin and the x-axis coincides with the principal axis.
First standard equation of the parabola. — The equation
of a parabola whose vertex is at the origin and whose axis is
the x-axis is „ .
p being the distance from the vertex to the focus.
(21)
D
A-
Y
B P/
C
//
/
-y.^•
II'
\
Proof. — Let DD' be the
directrix and F the focus.
Through F draw the a;-axis
perpendicular to the direc-
trix, meeting it in C. At
halfway between C and F
erect the y-axis. Let
P (x, y) be any point on
the curve and draw AP
perpendicular to the direc-
trix, meeting the j/-axis in
B. Let CF be represented as before by 2 p.
coordinates of F are (p, 0), also CO = OF = p.
By the definition given above,
FP = AP.
From formtila (1),
FP = Vi^
Then the
Therefore,
p)2 + y\
AP = AB + BP = p + x.
V{x — pY + y^ = p-\r X.
Clearing of fractions and simphfying,
2/2 = 4 px.
EQUATION OF THE PARABOLA
113
The equation shows, as has been previously discovered,
that the curve is symmetrical with respect to its axis and
passes through the origin, that is, through a point half-
way between the directrix and the focus. It also shows that
when p is positive, the curve extends indefinitely to the
right, while no part lies to the left of the origin. When p is
negative, the curve extends indefinitely to the left, while
no part lies to the right of the origin.
A chord through the focus of any conic section is called
a focal chord.
The latus rectum is that focal chord parallel to the
directrix.
The equation of the latus rectum is x = p. Solving this
simultaneously with the equation of the parabola 2/^ = 4 px,
the ordinates of the intersections are y = ± 2 p. Whence
the length of the latus rectum is 4 p.
It is helpful in sketching a parabola, to locate the vertex
and focus, then erect the latus rectum equal to 4 times the
distance from the vertex to the focus. The parabola passes
through the extremities of this latus rectum and the vertex.
The second standard equation of a parabola. — The
equcdion of a parabola whose vertex is at the origin and whose
axis is the y-axis is
x^ = ipy,
(22)
p being the distance from the vertex
to the focus.
Proof. — Rotating the axes
through (-90°), equation (21) be-
comes
Y'
or
[a;sin(-90°)-|-2/cos(-90°)P
= 4p[x cos (-90°) - y sin (-90°)],
a;2 = 4 py.
114
THE PARABOLA
This equation may also be obtained directly from the
figure by taking steps similar to those used in deriving
equation (21).
The third standard equation of a parabola. — The
equation of the parabola whose vertex is at the point (h, k)
and whose axis is parallel to the x-axis is
{y-kY = 4p {x - h),
(23)
V
p being the distance from the vertex to the focus.
Proof. — Let the figure be drawn as indicated with the
vertex V having coordinates {h, k) and principal axes VN
parallel to X'X. The
equation of the parab-
ola, considering V as
the origin and VN as
-N the a;-axis, is y^ = i px.
The equation with
the origin at V is
known, and the equa-
tion with the origin at
is required. The
problem then is to translate the axes to a new origin. The
coordinates of the new origin with respect to the old axes
through V are {—h, —k). Hence the equation becomes,
after translation of axes to 0,
(y — ky = ip (x — h).
The fourth standard equation of a parabola. — The
equation of a parabola whose vertex is at {h, k) and whose axis
is parallel to the y-axis is
(x -hr = ip(y- k), (24)
p being the distance from the vertex to the focus.
The proof is identical with that used in deriving (23).
EQUATION OF THE PARABOLA
115
ILLUSTRATIVE EXAMPLES
1. Find the equation of the parabola with axis parallel to the
2/-axis, vertex at ( — 1, 2), and passing through the point Pi (1, 3).
The equation of a parabola whose vertex is at ( — 1, 2) and whose
axis is parallel to the ^-axis is (a; + 1)* = 4 p (y — 2) by equation
(24). Since the point (1, 3) is on
this locus, its coordinates must
satisfy the equation, whence
(H-l)2 = 4p(3-2), or p = l.
The equation of the parabola then
is
(x + iy = 4{y- 2)
or x^ + 2x -iy +i^= 0. ^"
2. An arch is in the form of
a parabola with vertical axis. Its ^
highest point is 18 feet above the base which is 36 feet wide. Find the
length of the beam horizontally across the arch, 10 feet above the base.
Let A'BA represent the given arch. If the origin is taken at the
center of the base, the coordinates of the vertex are (0, 18), and the
equation of the parabola is
a;2 = 4p(y — 18), by stand-
ard equation (24). The
point A (18, 0) is on this
parabola and its coordi-
nates must satisfy the equa-
tion, whence (18)^ = 4 p
(0 -18), or p = -f. The
equation then becomes x^
= -18 (2/ -18).
Let D'CD represent the
position of a beam 10 feet
above the base. Then the ordinate of D is 10. , Substituting 10 for y
in the equation of the parabola.
"/
B
C \
'
A
Whence
x^ = -18(10 - 18) = 144.
X = ±12 or D'D = 24 feet.
116 THE PARABOLA
EXERCISES
1. Find the coordinates of the focus, the equation of the directrix,
and the length of the latus rectum for each of the following parabolas
and plot the curves:
(a) if- = &x. (c) 2/' = -4a;. (e) x^ = -&y.
{h)Zx^ = by. id)x + ^y^ = Q. (J) 2'jf = -bx.
2. Find the equations of the parabolas satisfying the following
conditions:
(a) vertex (0, 0), axis ^ = 0, a point on curve ( — 1, 3).
(6) vertex (-2, -2), focus (-3, -2).
(c) focus (0, 0), vertex (0, -3).
id) directrix j/ = — 2, focus (1, 4).
(e) vertex (0, 1), axis parallel to x-&Taa, and the point (1, 3) on
curve.
(J) focus (1, -2), directrix 3 a; - j/ + 6 = 0.
Hint. — Use the definition of a parabola.
3. Find the equation of the Une joining the vertex and the upper
extremity of the latus rectum of the parabola y'^ = —8 a;.
4. The equation of 'a parabola is ^ = 8x. With center at the
origin, and diameter equal to three times the distance from the vertex
to the focus, a circle is described. Prove that the common chord of
circle and parabola cuts the a>axis halfway between the vertex and the
focus.
6. Find the equation of the circle through the vertex and the
ends of the latus rectum of a;^ = 4 y.
6. Find the equations of the parabolas with the axes parallel to
the ^-axis and satisfying in addition the following conditions:
(o) vertex (2, —5) and a point on curve (6, —1).
(6) three points on curve (0, 3), (4, 3), and (—2, 6).
7. Find the equation of the focal chord of the parabola 2/^ = 6 a:
through the point on the curve whose ordinate is 4.
8. A parabola has its vertex at the origin and axis along the j/-axis.
A focal chord has one extremity at (3, —3). Find its equation and
the coSrdinates of the other extremity.
9. A trough whose cross section is a parabola with vertex down-
ward is partly filled with liquid. The width of the trough one foot
above the vertex is 4 feet and the width at the surface of the Uquid is
8 feet. Find height of hquid. Ans. 4 feet.
CONSTRUCTION OF THE PARABOLA
117
10. An arch has the form of a parabola with vertical axis. The
width of the base is 36 feet and the height above the base at a point
12 feet to the right of the center of the base is 10 feet. Find the height
of the arch at its highest point. Ans. 18 feet.
49. Construction of the parabola. — Having given the
directrix and the focus there are two principal methods of con-
structing the parabola mechanically.
First method. — Let DD' be the
given directrix and F the focus. Place
a right triangle ABC with one leg BC
on the directrix, the other leg lying
on the same side of the directrix as
the focus. Fasten one end of a string
of length CA at A and the other end
at the focus. With a pencil point
against the triangle at P, keep the
string taut and move the triangle
along the directrix. The pencil point
will describe a parabola, since CP = FP, and therefore P is
equidistant from the focus and the directrix.
Second method. — Lo-
cate the focus and direc-
trix as in the first case.
Draw OX through F per-
pendicular to the directrix,
on it lay off a number
of points, as Mi, Mi, Ms,
etc., and erect ordinates
MiKi, MiKi, M3K3 at
these points. With F as
a center and a radius equal
to the distance from the
directrix to the foot of
any ordinate as CMi, de-
118 THE PARABOLA
seribe an arc cutting the ordinate in two points as Pi and
Ri. Similarly, locate the points P2 and R2, P3 and R3, etc.
These points all he on the parabola since they are equidistant
from the focus and the directrix. Connect by a smooth
curve and the figure is approximately a parabola.
50. General equation of a parabola, axis parallel to one
of the coordinate axes. — When equation (23) is expanded,
it takes the form
2/2 - 2 % - 4 pa; + fc2 + 4 pA = 0. (1)
Similarly, equation (24) becomes
x^ - 2hx - Apy + ¥ + Apk = 0. (2)
These results show that every equation of a parabola with
axis parallel to a coordinate axis contains one and only one
term which is the square of a variable and no xy term. It
will be shown that every equation of the form
y^ + Dx + Ey + F = (3)
or x^ + Dx + Ey + F = (4)
represents a parabola.
Completing the squares and collecting, equation (3)
becomes
(»+f)■-(-f)^-^fI^o■ <^'
which is in the form of equation (23) if D is not 0.
Similarly, (4) becomes
which is in the form of equation (24) if E is not 0.
Comparing the general equations of the parabola (3)
and (4) with the general equation of second degree Ax^
+ Bxy + Cy^ + Dx + Ey + F = it is seen that:
GENERAL EQUATION OF A PARABOLA 119
The general equation of second degree represents a 'parabola
with axis parallel to a coordinate axis if B = 0, and if there is
only one second degree term, either x^ or y^, providing the first
degree term in the other variable is present.
ILLUSTRATIVE EXAMPLE
Determine the vertex, focus, latus rectum, equation of the direc-
trix, and of the axis for the parabola whose equation is x' + 6x + 8y
+ 1=0.
Completing the squares of the a>terms,
{x + 3y= -8{y-l).
This is in the form of the fourth standard equation of the parabola,
{xYh? = ip(.y-k).
Whence the vertex is at ( — 3, 1), p the distance from the vertex to
the focus is —2, and the length of the latus rectum is 8.
The facts just deter-
mined are sufficient to
roughly sketch the figure.
Since the vertex bisects
the distance from the focus
to the directrix, that hne
can now be drawn and its
equation is seen to be
y = 3. The equation of
the axis VF can likewise
be read from the figure,
and is x — —3.
~]
~~
n-
^1
V
^
s
■^
X-
L'
/
N,
1
/
F
•s
('
EXERCISES
Determine the coordinates of the vertex and focus, length of latus
rectum, and equation of the directrix and of the axis for the following
parabolas. Also sketch the figures.
1. x^ + 4:X-6y-8 = 0. i. ix' + 8x + 8y = 3.
2. y' rr^y + Sx = IB. 5. y' - 5y = x -7.
3. 3x2 + 6a; + 5j/ = 7. 6. Sy' -6y = 4:X.
CHAPTER VIII
THE ELLIPSE
51. The ellipse has been defined as that conic section
which is traced by a point which moves so that the ratio of
its distance from a fixed point, called the focus, to its distance
from a fixed line, called the directrix, is constant and less
than 1.
It was shown in Art. 47 that the ellipse cuts the principal
axis in two points, both on the same side of the directrix
as the focus. The simplest form of the equation of an
ellipse is obtained by taking the principal axis as the
X-axis and the point halfway between the two intersec-
tions as origin. This point is called the center of the
ellipse.
The first standard equation of the ellipse. — The equa-
tion of an ellipse whose major axis is on the x-axis and whose
center is at the origin is
D
y;
R-
- :e
^P
D-
/
■^r
,
A
'f^^
A'
z
■ oJ
P'
.
y'
a"" ^ h" '
(25)
in which a and b are the
semi^major and semi-
minor axes respectively.
Proof. Let the direc-
trix of the ellipse be DD'
and take the a;-axis on
the principal axis which is perpendicular to DD' through
the focus F, meeting it at Z. Let A and A' represent the
two points at which the curve cuts the principal axis.
These two points are called the vertices of the ellipse.
120
FIRST STANDARD EQUATION OF THE ELLIPSE 121
At midway between A and A' erect the y-axis. Call
the distance AO = OA' = a. Take P (x, y) any point on
the ellipse and drop PB perpendicular to the directrix, cut-
ting the 2/-axis at E.
From the definition of an ellipse,
^ = e. (1)
•
In order to compute the values of FP and BP, it is first
necessary to find the distances from the directrix to the
center and from the focus to the center. In finding these
lengths, use is made of the fact that A and A' are on the
ellipse. Applying the definition,
ZA = '' ^2)
■and
FA' .„.
^ = e. (3)
Clearing (2) and (3) of fractions and adding,
AF + FA' --e{ZA + ZA').
Substituting from the figure,
AA' = e [{ZO -a) + (ZO + a)].
Whence 2 a = e (2 ZO) and ZO = a/e.
The distance from the directrix of an ellipse to the center
is a/e. (26)
Similarly, by subtracting (2) from (3),
FA' - AF = e {ZA' - ZA).
Whence (FO + a) - (a - FO) = e (AA') and FO = ae.
The distance from the focus of an ellipse to the center
is ae. (27)
122 THE ELLIPSE
The coordinates of F are (— ae, 0), whence
FP = V(x + aey + j/S by formula (1).
BP = BE + EP = a/e + x.
Substituting in equation (1),
V{x + aey + y" ^
a/e + X
Clearing of fractions and collecting,
a;2 (1 _ e2) + 2/2 = a^ (1 - e'*), (4)
or ^+ 2tf 2N = 1- (5)
a^ a^ (1 — e^)
If a; = 0, y = ±a Vl — e^ hence the ellipse cuts the
2/-axis in two points equidistant from the center. This dis-
tance will be represented by b, The equation of the elhpse
then is
S + P = l' (6)
where b« = a^ {1 - e^). (7)
This relation also shows that ae = Va" — 6". (28)
The portion of the principal axis cut off by the ellipse
is called the major axis. It is represented by 2 a. The
portion of the perpendicular to the principal axis through
the center, cut off by the ellipse, is called the minor axis.
It is represented by 2 b.
From the form of the equation, it is readily seen that the
ellipse is symmetrical with respect to both axes.
When the equation of the ellipse is solved for y,
y = ±- Va^ - x\
SECOND FOCUS AND DIRECTRIX
123
from which it is seen that y is imaginary for values of x
numerically greater than a, and hence the curve lies entirely
between the lines x
= —a and x = -{-a.
Similarly, by solv-
ing for X in terms of
y it can be shown
that the curve lies
entirely between the
lines y = —h and
y = +h.
x^
Y'
When points are plotted and the curve drawn, it is found
to be as here shown.
52. Second focus and directrix. — It will now be proved
that an ellipse has a. second focus and directrix on the right
of the center and similarly situated with respect to the center.
In the figure locate a second focus F', making OF' = FO
= ae. Also draw a second
directrix MM' parallel to
DD' meeting the principal
axis at N and making ON
= CO = a/e. It will now be
shown that the ellipse which
has F' for focus and MM'
for directrix has the same
equation and therefore is the same ellipse as the one hav-
ing F as focus and DD' as directrix.
Let P (x, y) represent any point on the eUipse whose
focus is F' and whose directrix is MM'. Draw PK perpen-
dicular to the directrix meeting it in K and meeting the
F'P
y-Bsas in L. Then by the definition of an ellipse, -p^
but
and
= e,
F'P = ^/{x — aeY + y^, from formula (1),
PK = LK-LP = a/e - x,
124 THE ELLIPSE
, V(a; — oeY + y^
whence — ^^ — -, = e.
a/e — X
Clearing and collecting,
a;2 (1 _ e2) +y^ = d?{l- e"),
which is the same as equation (4) of the previous article, in
which F is the focus and DD' the directrix.
53. The lotus rectum of the ellipse is the chord throujh
either focus parallel to the directrix. Its length is 2 ¥/a.
Proof. — The equation of this chord is a; = ±ae.
Solving simultaneously with the elUpse -5 + rj = 1, V =
±b Vl -e" = ±¥/a, since b^ = a" (1 - e^). Therefore the
latus rectum, which is twice the ordinate at the focus, 13
equal to 2 b^/a.
54. The second standard equation of an ellipse. — The
equation of an ellipse whose major axis is on the y-axis and
whose center is at the origin is
& + P = l' (29)
where a and b are the semi-major and semi-minor axes, re-
spectively.
Proof. — Rotating the axes through 90°, equation (25)
becomes
(x cos 90° - y sin 90°)^ (x sin 90° + y cos 90°)^ _
a^ ■'" 62 ~ '■'
The third standard equation of an ellipse. — The equa-
tion of an ellipse whose major axis is parallel to the x-axis
and whose center is at the point (h, k) is
^^^" + ^^^^ = 1. (30)
FOUETH STANDARD EQUATION 125
where a and b are the semi-major and semi-minor axes re-
spectively.
Proof. — The proof is identical to that given in deriving
the third standard equation of the parabola.
The fovirth standard equation of an ellipse. — The equa-
tion of an ellipse whose major axis is parallel to the y-axis
and whose center is at the point Qi, k) is
^^ + ^^-1, (3.)
where a and b are the semi-major and semi-^minor axes re-
spectively.
Proof as above.
ILLUSTRATIVE EXAMPLES
1. An ellipse with semi-minor axis equal to 5 and passing through
the point (6, 4) has its center at the origin and its major axis on the
a;-axis. Find the equation of the ellipse, the coordinates of the foci
and the equations of the directrices.
Substituting the value of 6 = 5 in standard equation (25),
- + ^ = 1 (1)
a2 ^ 25 ^ '
This is the equation of a family of eUipses all having 5 as semi-
minor axis.
This ellipse must pass through the point (6, 4) whence the coordi-
nates of this point satisfy equation (1).
Therefore, f + M"^- ^^^
Solving, a2 = 100.
Substituting back in equation (1),
.^4-^ = 1
100 "^ 25
vs
Since 6^ = a^ (1 - e^*), therefore 25 = 100 (1 - e*) and hence « = "o"*
The distance from the center to the focus is ae = 5 v 3 and from
the center to the directrix is o/e = 20/V3,
126
THE ELLIPSE
The coordinates of the foci are, therefore, (±5 Vs, O) and the
equations of the directrices i = ±20/ Vs.
2. Find the equation of the eUipse one of whose foci is at (0, 2),
the equation of whose corresponding directrix is y = 5 and whose
eccentricity equals |.
The data given shows that the eUipse is in the fourth standard form
(y - kY {x - hy
a? ¥
1.
Locate the focus F at (0, 2) and draw
the directrix DD' 5 units above the ori-
gin meeting the y-aids at E. Then CE
= a/e, CF = ae, whence by subtraction,
--ae = FE = 3.
e
Substituting e = |, a is found to be 2.
62 = 02 (1 _ e2) = 4 (1 _ 1) = 3.
OC = OF- CF. Since OF = 2 and CF
= ae — 1, therefore OC = 1.
The coordinates of the center then are (0, 1) and the equation is
{y -lY x" _
4 "^3 •
EXERCISES
1. Determine the vertices, foci, equations of directrices, and length
of latus rectum for each of the following eUipses. Plot each curve,
(a) 9x' + 25y'^ = 225. (d) 4:x' + 9}/^ = 36.
(6) 3x2 + 42^2 = 48. (e) 4x2 + 3yi = 108.
(c) 162/' + 25x2 = 400.
(f) ±. J. w_ = 1.
^' 36 ^ 64
2. Find the equations of the following ellipses which have their
centers at (0, 0), major axis along the x-axis. Construct the figures,
(a) Semi-major axis = 6, e = |.
(6) Distance between the foci = 6, e = J.
(c) Minor axis = 12, a focus at (8, 0).
(d) Equation of a directrix is x = 6, e = |.
(e) A focus at (3, 0), the equation of the corresponding directrix,
EXERCISES 127
(f) Major axis = 16, and (4, 3) is a point on the curve.
(g) Minor axis = 4, and (3, 1) is a point on the curve.
(h) The two points (4, 2) and (V6, S) are on the curve,
(i) Latus rectum = 3, e = |.
(j) Latus rectum = 9, one vertex (8, 0).
3. Find the equations of the following eUipses, the coordinates of
foci and vertices and length of latus rectum. Draw each curve.
(o) Center ( — 1, —2), major axis = 6 and parallel to y-asis,
minor axis = 4.
(b) Center (—4, —2), major axis = 10 and parallel to a-axis,
minor axis = 8.
(c) Center (0, 5), one vertex (0, 0), e = f .
4. Find the equations of the following ellipses:
(a) Center ( — 1, —2), major axis = 12, latus rectum equal to
one haU of minor axis, principal axis parallel to x-axis.
(6) Major axis = 10, foci at ( — 1, 3) and ( — 1, —5).
(c) Minor axis = 6, foci (-3, 4) and (5, 4).
(d) Center at (2, 1), major axis = 8 and parallel to a;-axis, and
the center twice as far from the vertex as from the focus.
6. By translation of axes reduce each of the following equations
to standard forms (25) or (29). Draw both sets of axes and the curve,
(a) x'-2x + 2if-4:y + l=0.
(6) x'-&x + iy'-8y-S = Q.
x^ ifi .1
6. Prove that in the ellipse rs + ^ = 1, the line joining the posi-
tive ends of the axes is parallel to the line joining the center to the
upper end of the left hand latus rectum.
7. Find the equation of the circle whose diameter is the major-
axis of the ellipse 9 x^ + 25y' = 225 and whose center is at the center
of the eUipse. Find the coordinates of the points where the right
hand latus rectum produced, cuts the circle.
8. Find the equations of the lines through the left hand focus of
x^ tfi
^r: + TS = 1 and the extremities of the right hand latus rectum.
25 lb
Find the distances of these lines from the origin.
9. Find the equation of the locus of a point which moves so that
the sum of the distances from the two points (0, 4) and (0, —4) is
equal to 10. Prove that the locus is an ellipse.
128
THE ELLIPSE
55. Construction of an ellipse. — A proposition which
readily leads to the construction of the ellipse is as follows:
The sum of the focal distances of any point on an ellipse is
constant and equal to the major axis.
Proof. — Draw the
ellipse with foci F
and F' and direc-
trices DD' and MM'.
From P {x, y), any
point on the ellipse,
draw PK perpen-
dicular to the direc-
trices meeting them
in B and K respec-
tively.
D
M
^
'
■ ~^P
K
B
^
F
^
\
D'
^^
"~—
M'
Similarly,
From the definition of an eUipse,
F'P = e (BP) =e(^+x) = a + ex.
FP = e (PK) = e(^-x) = a-ex.
Adding (1) and (2),
FP + F'P = 2 a = major axis.
(1)
(2)
This fact leads to a second and important definition of an
elKpse:
An ellipse is the locus of a point which moves so that the sum
of its distances from two fixed points is constant.
From this definition, an elHpse can be constructed as
follows, if the foci and the length of the major axis are
given:
In a drawing board, fasten a tack at each focus F and F'.
Tie about the tacks a string equal in length to the distance
FF' -\-2a and with a pencil point hold the string taut
GENERAL EQUATION OF AN ELLIPSE
129
while describing the curve. The locus will be an ellipse
since the sum of the focal distances is always 2 a.
By use of this property, it can be shown that the foci are
at a distance a from the extremities of the minor axis.
Hence, to locate the foci, take an extremity of the minor
axis B as center and with a
radius equal to the semi-
major axis describe an arc
cutting the major axis in
two points F and F'. These
points are its foci.
66. General equation of an ellipse, axes parallel to
coordinate axes. — When equations (30) and (31) are
expanded, they become
6V - 2 ¥hx + aY - 2 a^ky + b^h^ + aV - a'b^ =
and
1/2-2 b%y + a^x^ - 2 a%x + aW + b%^ - aW = 0.
Either of these equations is of the form Ax^ + Cy^ + Dx
+ Ey + F = 0, in which A and C are positiye and different.
It will be shown that every equation of the above type
represents an elhpse.
Completing squares and collecting,
CD^ + AE'-AACF
Aix + D/2Ay + Ciy + E/2 Cy =
4. AC
After dividing by the second member, this becomes
{x + D/2Ay , {y + E/2Cy
CD^ + AE^-4:ACF ' CD^ + AE^ - 4:ACF
= 1,
4A2C
4AC2
which is of standard form (30) or (31).
Whether the major axis is parallel to the ic-axis or the
2/-axis will depend upon whether the first or second denomi-
130
THE ELLIPSE
nator is the larger. If A and C are equal, the axes of the
ellipse are equal and the figure is a circle. If the denomi-
nators are negative, the axes are imaginary and the elhpse
impossible; if zero, it is a point-elUpse.
It is seen from the foregoing that:
The general equation of second degree, Ax^ + Bxy + Cy^
+ Dx -\- Ey ■\- F = 0, represents an ellipse with axes parallel
to the coordinate axes, if B = 0, and if A and C have like signs
but different numerical values.
ILLUSTRATIVE EXAMPLE
Determine for the ellipse 9 x' + 25y^ + 18x — 50y = 191, cen-
ter, foci, vertices, semi-axes, latus rectum, and equations of directrices.
Completing the squares,
Q{x + 1)2 + 25 (2/ - 1)2 = 225.
Dividing by 225,
{X + D' (y
25 "^
= 1.
<f
^
-
^
/
»
■
s
I-
V/
1-
'
X
-
' ^
^
'
-'
'
.
On comparing with standard equation (30), it is seen that the center
is at ( — 1, 1), the semi-major axis is 5 and the semi-minor axis 3.
The principal axis is parallel to the
X-axis. Sketching in a figure and us-
ing all the data obtained, it is seen
that the vertices are at (4, 1) and
(-6, 1).
Since b' = a? {I - ^) and 6 = 3 and
a = 5, therefore e = ^. The distance
from the center to the focus is ae = 4.
The coordinates of the foci then are (3, 1) and (—5, 1).
The latus rectum = 2 V/a = ^. The distance from the center
to the directrix is a/e = ^. Hence the equations of the directrices
are x = ^- and x = —*i:
EXERCISES
1. Determine for the following ellipses, center, foci, vertices, semi-
axes, latus rectvun, and equations of directrices,
(a) 4x2 4-16a; + 32/2 -62/ = 29.
(6) 7 2/2-1-142/ -1-16x2 -64 a = 41.
GENERAL EQUATION OF AN ELLIPSE 131
(c) 4a;2- 8a; + 82/2 -642/ + 68 = 0.
(d) x^ + 4:y' + &x-8y = 87.
(e) 3a? + 6a; + 42/! + 24s/ = 69.
Cf) 9a;2 + 54x + 8j/2 -Uy = 199.
2. Find the equation of an ellipse whose foci are at (3, 0) and
(—3, 0) and the sum of whose focal radii is 10.
3. Prove that the point (4, 1) is on the ellipse to + q = !• Find
the focal distances of the point and prove that their sum is equal to
the major axis.
CHAPTER IX
THE HYPERBOLA
57. The hyperbola is that conic section traced by a point
which moves so that the ratio of its distance from a fixed
point called the focus to its distance from a fixed line called
the directrix is constant and greater than 1.
It was shown in Art. 47 that the hyperbola cuts the prin-
cipal axis in two points on opposite sides of the directrix.
The simplest form of the equation of an hyperbola is
obtained by taking the principal axis as the a;-axis and a
point halfway between the two intersections as origin.
This point is called the center of the hyperbola.
The first standard equation of an hyperbola. — The equa-
tion of an hyperbola whose transverse axis is on the x-axis
and whose center is at the origin is
E.-yl- 1
(32)
in which a and b are the
semi-transverse and semi-
conjugate axes respectively.
Proof. — Let the direc-
trix of the hyperbola be
DD' and take the x-axis
on the principal axis which
is perpendicular to DD'
through the focus F, meet-
ing it at Z. Let A and A' represent the two points at which
the hyperbola cuts the principal axis.
132
Y
D
E
A'
B
Z
A If .;
A -
y'
D
FIRST STANDARD EQUATION OP AN HYPERBOLA 133
These two points are called the vertices of the hyperbola.
At 0, midway between A and A', erect the y-axis. Call
the distance A'O = OA = a.
Take P (x, y), any point on the hyperbola, and drop PB
perpendicular to the directrix, cutting the 2/-axis at E.
From the definition of an hyperbola,
Bp = e. (1)
The values of FP and BP are found in a manner almost
identical to that used in the case of the elUpse, use being
made of the fact that the points A and A' are on the hyper-
bola, and hence
AF
and jj^ = e. (3)
Whence A'F + AF = e {A'Z + ZA)
or (a + OF) + {OF - a) = 2 ae and OF = ae.
The distance from the center of an hyperbola to the
focus is ae. (33)
Similarly, A'F - AF = e {A'Z - ZA)
or 2a = e[{a + OZ) - {a - 0Z)\ and OZ = a/e.
The distance from the center of an hyperbola to the
directrix is a/e. (34)
The coordinates of F, then, are (ae, 0),
whence FP = V{x — ae)^ + y^, by formula (1).
BP =.EP - EB = x- a/e.
Substituting in equation (1),
V{x - ae)' + y' _^
X — a/e
134 THE HYPERBOLA
Clearing of fractions and collecting,
x" (f -1) -y^ = a^ (e^ - 1).
If X = 0, 2/ = ±a Vl — e', which since e > 1, is imagi-
nary, and the curve does not cross the y-axis.
It is found convenient to make the substitution
b'' = a'' (c" - 1).
The equation of the hyperbola then becomes -j — r; = !•
The portion of the principal axis which is cut off by the
hyperbola is called the transverse axis. It is represented
by 2 a.
The segment on the perpendicular to the princip al axis
through the center such that its length is 2 b = 2 a Ve^ — 1
is called the conjugate axis.
Since b" = a' (e^ — 1), it is readily seen that
ae = Va^ + b". (35)
From the form of the equation, it is evident that the
hyperbola is symmetrical with respect to both axes.
When the equation of the hyperbola is solved for y,
y = ±- Va;" — o^ from which it is seen that y is imaginary
for all values of x numerically less than a, and hence no part
of the curve lies between the Hnes x = —a and x = a.
For aU values of x numerically grea.ter than a, y is real,
showing that the curve extends
indefinitely both right and left.
Similarly, by solving for x in
^ a
terms of y, a; = ± =- V 6* + y^,
^ from which it is seen that for
every value of y, x is real and hence the curve extends in-
definitely above and below the a;-axis.
When points are plotted and the curve drawn it is found
to be as shown.
x^
SECOND STANDARD EQUATION OF AN HYPERBOLA 135
It can be proved, as in the case of the ellipse, that the
hyperbola has a second focus at (— oe, 0) and a second
directrix whose equation is a; = — a/e.
58. The latus rectum of the hyperbola is the chord through
either focus parallel to the directrix. Its length is 2 b^/a.
Proof. — The equation of this chord is x = ±ae.
Solving simultaneously with the equation of the hyperbola
^ _ ^ = 1, y = ±6 V?^^= ±¥/a, since b^ = a^{e^ - 1).
Therefore the latus rectum, which is twice the ordinate at
the focus, is equal to 2 b^/a.
59. The second standard equation of an hyperbola. —
The, equation of the hyperbola whose transverse axis is on the
y-axis and whose center is at the origin is
where a and b are the semi-transverse and semi-conjugate
axes, respectively.
The proof is left to the student. It is identical to that
used in the case of the ellipse.
The third standard equation of an hyperbola. — The
equation of an hyperbola whose transverse axis is parallel to
the X-axis and whose center is at the point {h, k) is
(x - hr (y - fc)'^ _ ,„„.
—^5 P 1' ^37)
where a and b are the semi-transverse and semi-conjugate axes,
respectively. The proof is left to the student.
The fourth standard equation of an hyperbola. — The
equatimi of an hyperbola whose transverse axis is parallel to
the y-axis and whose center is at the point {h, k) is
(y^.ii^ = i, (38)
a o"
where a and b are the semi-transverse and semi-conjugate axes,
respectively. The proof is left to the student.
136
THE HYPERBOLA
ILLUSTKATIVE EXAMPLE
An hyperbola in which the distance between the foci is 10 passes
through the origin, has one focus at (1, 0), and its transverse axis on the
s-axis. Find its equation.
Locate in a figure the center C, the focus F, and the vertex 0. It is
seen from the data given that the equation is in the form of the third
standard equation,
(X - hy (y
—
^ =
1.
a?
62
Here CF =
5
and
OF
= 1, therefore CO
= 4
= a,
and the coordinates of C
are (-4, 0).
_ -
Since CT
s/a? + V,
then 5 = 3.
The equation
then
be-
comes
{x + 4)»
16
1?
9
= 1.
EXERCISES
1. Determine lengths of axes and latus rectum, coordinates of ver-
tices and foci and equations of directrices for each of the following hyper-
bolas. Plot each curve.
(o) 9 x! - 25 j/* = -225.
(b) 3 a;2 - 4 2/2 = 48.
(c) 9 y2 - 4 a;2 + 36 = 0.
(d) 25 x2 - 16 2/2 + 400 = 0.
(e) 4 a;2 - 3 2/2 = 108.
36
■k = ^-
2. Find the equations of the following hyperbolas having their cen-
ters at (0, 0) and their transverse axes along the a;-axis. Construct the,
curves.
(a) Transverse axis = 4 and conjugate axis equal to one-half the
distance between the foci.
(6) Transverse axis = 6 and (5, 3) is a point on the curve.
(c) Latus rectum = 10 and e = |.
(d) Transverse axis = 12 and a focus is at (8, 0).
(e) Distance between the foci = 8 and e = f .
(/) Latus rectum = 2 and a = 2 6.
(ff) e = 2 and distance from focus to nearest vertex = 1.
EXERCISES ■ 137
3. Find the equations of the following hyperbolas which have their
centers at (0, 0) and their transverse axes along the y-ajds. Construct
the figures.
(o) Latus rectum = 3 and one vertex at (0, 2).
(6) Conjugate axis = 8 and (4, 6) is a point on the curve.
(c) e = 2 and the equation of a directrix is ^ = 3.
(d) One focus at (0, 6) and the equation of the corresponding
directrix is y = ^.
(e) The two points (3, 4) and (6, 7) are on the curve.
4. Find the equation of each of the following hyperbolas, determine
the coordinates of foci and vertices and length of latus rectum :
(a) Center ( — 1, 3), transverse axis = 8 and parallel to ^-axis,
conjugate axis = 10.
(6) Center (—2, —3), transverse axis parallel to a;-axis and = 8,
conjugate axis = 12.
(c) Vertices are ( — 1, — 1) and ( — 1, 7) and e = 2.
6. Find the equations of the following hyperbolas:
(a) Center at (2, 1), transverse axis = 6 and parallel to the x-axis
and the center twice as far from the focus as from the vertex.
(6) e = 2, one focus at ( — 1, —2) and the corresponding directrix
2/ = 4.
(c) 2 o = 6 and foci at (-2, -4) and (-2, 6).
(d) Center ( — 1, —3), transverse axis = 8 and parallel to the y-
axis and the latus rectum equal to one-half of conjugate axis.
(e) Transverse axis = 4, one directrix is a; = 6 and the corre-
spondmg focus (3, —1).
(/) Vertices at (2, 5) and (2, — 1) and latus rectum = transverse
axis.
6. Prove that the ellipse ok + q ~ ■'^ ^""^ ^^^ hyperbola -q- — y = 1
have the same center and foci. Construct each curve.
7. By translation of axes reduce each of the following equations to
standard forms. Draw both sets of axes and the curves.
(o) 9 a;2 - 36 a; - 4 2/2 - 24 2/ = 36.
(6) x2 + 6 a; - ^ + 2 2/ + 12 = 0.
8. Find the equation of the locus of the point which moves so that
the difference of its distances from the two points (6, 0) and ( — 6, 0) is
equal to 8. Prove that the locus is an hyperbola.
138
THE HYPERBOLA.
9. Prove that the foci of the hyperbolas ^V9 - a;V16 = 1 and aV16
— ^2/9 = 1 are equidistant from the center.
10. By rotation of axes, remove the a;^-term from the equation xy =
18. Show that the curve is an hyperbola and construct both sets of
axes and the curve.
11. Prove that the latus rectum of an hyperbola is a third propor-
tional to the transverse and conjugate axes.
12. Find the polar equation of the hyperbola i^/a? — y'^/l^ = 1.
13. Prove that the point (4, 1) is on the hyperbola ai'/S — ^/l = 1
and that the difference of its focal distances is equal to the transverse
axis.
60. Construction of an hyperbola. — A proposition which
readily leads to the construction of the hyperbola is as
follows :
The difference of the focal distances of any point on an
hyperbola is constant and equal to the transverse axis.
Proof. — Draw the hyper-
bola with foci at F and F' and
directrices DD' and MM'.
From P {x, y), any point on the
hyperbola, draw PK perpen-
dicular to the directrices and
meeting them in B and K re-
spectively.
From the definition of an hyperbola,
FP = e (KP) = eix - a/e) = ex - a. (1)
Similarly,
F'P = e (BP) = e{x + a/e) = ex + a. (2)
Subtracting (1) from (2),
F'P — FP = 2a = transverse axis.
This fact leads to a second and important definition of an
hyperbola:
An hyperbola is the locus of a point which moves so that the
difference of its distances from two fixed points is constarU.
Y
D
K P/^
-
-
"A
7
D'
A
V
GENERAL EQUATION OF AN HYPERBOLA 139
From this definition, an hyperbola can be constructed as
follows, if the foci and length of the transverse axes are given.
In a drawing board fasten a tack at
each focus F and F'. Let a pencil be tied
to a string at P. Let one end of the
string pass beneath F and then both ends
over F' as shown. Adjust the string so
that F'P exceeds FP by 2 a. By holding
the strings together below F' and pulHng
them in or letting them out, the point P
will, if held firmly against the string, trace
an hyperbola, for at each position F'P —
FP = 2a.
By reversing the process, the other branch may be drawn.
61. General equation of an hyperbola, axes parallel to
coordinate axes. — When equations (37) and (38) are ex-
panded, they become
¥x^ - 2 ¥hx + bW - ay + 2 d'ky - a^k^ - a^¥ = 0,
by - 2 ¥ky + bW - aV + 2 a%x - a%^ - a?b^ = 0.
Either of these equations is of the form
Ax^ + Cy^ + Dx + Ey + F = Q,
in which A and C have opposite signs.
It will now be shown that every equation of the above
type represents an hyperbola.
Completing the squares and collecting,
CD'' + AE^-4.ACF
A{x + D/2 AY + C (2/ + E/2 Cy
4:AC
After dividing by the second memberj this becomes
(x + D/2Ay . (y + E/2 CY
CD^ + AE^-4:ACF ' CD^ + AE^ - 4:ACF
= 1,
442c
4AC2
140 THE HYPERBOLA
which is of standard form (37) or (38), since A and C have
unlike signs and hence the denominators have unlike signs.
Whether the transverse axis is parallel to the x-axis or to
the j/-axis will depend upon whether the first or second
denominator is positive.
It is seen from the foregoing that:
The general equation of second degree, Ax^ + Bxy + Cy^
+ Dx + Ey + F = 0, represents an hyperbola if B = and
if A and C have unlike signs*
EXERCISES
1. Determine for each of the following hyperbolas, the center, semi-
axes, foci, vertices, and latus rectmn. Construct each curve.
(a) 9a;2 - 18x - iy^ + 16y - AS = 0.
(b) 4a;2-24a;- 16 j/^ - 64 j/ + 36 =0.
(c) x' -&x-Qy' -18y + 9 = 0.
(d) 3i/' + Gy-x^+2x + n=0.
(e) 8a;2 - 8a; - 28y' - 28y = 61.
CO 25 2/2 - 4a;2 - 50 1/ - 39 = 0.
2. Find the equation of an hyperbola with e = v'2, the line x — y = i
as one directrix and the corresponding focus at ( — 1, —1).
3. By rotation of axes, reduce the equation a:^/ + 50 = to one of
the standard forms of the equation of an hyperbola. Draw both sets of
axes and the curve.
4. Find the equation of an hyperbola whose foci are (0, 8) and
(0, —8) and the difference of whose focal radii is 10.
6. Find the equation of the hyperbola whose center is at (1, 1),
whose transverse axis is parallel to the a;-axis and which passes through
(6, 5) and (-7, -7).
62. Asymptotes to the hyperbola. — The two lines rep-
resented by the equation
o2 b^
1.
have a very important relation to the hyperbola ^ — Tj
* The general equation of second degree will under these same con-
ditions sometimes take the form of the difference of two squares, in
which case it will represent a pair of straight lines. This will be the
case when Ciy^ + AE? — 4 ACF in the equation above is zero.
ASYMPTOTES TO THE HYPERBOLA
141
It will now be shown that as a point on the hyperbola
recedes indefinitely, the curve approaches coincidence with
these lines and therefore these are the asymptotes to the
curve.
Let Li and L2 be the two
lines represented by the equa-
tion x^/a? — y^/W = 0, and take
Pi (poi, j/i), any point on these
lines, such that a perpendicular
from it on the axis of x meets
the hyperbola x^/d' — y^jW = 1
in the point P% (a;i, 2/2).
Then
2/i=±-a;x,
= ±.-V^F^^a\
Subtracting,
2/1 - 2/2 = d= - (xi - ^xf-a^.
Rationalizing the numerator,
±o&
^' ^^ ^a Xi + Va;i2 - a? Xi + Vxi^" - a^'
which approaches zei-o as xi recedes to infinity.
This shows that as the curve recedes to infinity, it ap-
proaches indefinitely close to the lines
E! _ ^ - n
which are therefore the asymptotes to the hyperbola
a TO ^*
142
THE HYPERBOLA
If the distances A'O = OA = a and B'O = OB = b are
laid off on the a;-axis and y-axis respectively, and parallels
to the axes through A, B, A', and B' are drawn, the diago-
nals of the rectangle thus formed will be the asymptotes.
It is often found convenient in constructing an hyper-
bola to first con-
struct this rec-
tangle and the
asymptotes and
then draw in the
hyperbola touch-
ing the rectangle
at A and A' and
approaching the asymptotes as it recedes to infinity.
63. Conjugate hyperbolas. — If in two hyperbolas the
transverse axis in each is the conjugate axis in the other,
the hyperbolas are said to be conjugate.
Thus, in the figure, if A'A {= 2 a) is the transverse axis
and B'B (=26) the conjugate axis, then the equation of
the hyperbola is Y
1.
If another hyper-
bola is constructed in
which B'B is the
transverse axis and
A' A the conjugate axis, the equation is
In either case the equation can be written
(Distance from conjugate axis)"
(Semi-transverse axis)^
(Distance from transverse axis)"
(Semi-conjugate axis)^
= 1.
EQUILATERAL OR RECTANGULAR HYPERBOLA 143
If the eccentricity of the first hyperbola is represented
by ei and of the conjugate hyperbola by e^, then
Va^ + ¥ , Vb^ + a"
fii = and 62 = r >
hence, aei = be^ = '\/aF+¥.
This shows that the foci of the two hyperbolas are equi-
distant from the center and thus the foci of an hyperbola
and its conjugate lie on a circle about the center with a
radius equal to the diagonal of the rectangle constructed
as in the last article.
The asymptotes to the conjugate hyperbola y^/¥ — x'/a^
= 1 are found to he y = ±. (b/a) x, from which it is seen
that the hyperbola and its conjugate have the same asymp-
totes.
64. Equilateral or rectangular hj^erbola. — If 6 = a in
the equation
it becomes x^ — y^ = a^.
The equation of the conjugate hyperbola is y^ — x^ = a'.
These are evidently equal hyperbolas.
The asymptotes of these hyperbolas are y = ±x, two
lines making angles of 45° and 135° with the a;-axis and
therefore at right angles with each other.
Since the semi-axes a and b are equal, these hyperbolas
are sometimes called equilateral; since the asymptotes
are at right angles to each other, they are sometimes
called rectangular hyperbolas.
EXERCISES
1. Given the hyperbola 9x^ — y^ = 36, find the equations of the
asymptotes and of the conjugate hyperbola. Construct the hyperbolas
and the asymptotes.
144 THE HYPERBOLA
2. Find the equations of the asymptotes to the hsTperbola 9 a? —
16 1^ = 144 and the tangent of the angle between them.
3. Write the equation of an hyperbola conjugate to the hyperbola
4 X* — 9 2/^ =36 and find the lengths of its axes and latus rectum, the
coordinates of its foci and the equations of its directrices.
4. Prove that the distance from an asymptote to a focus is equal to
the semi-conjugate axis.
6. Find the equation of an hyperbola whose foci are at the
points (5, 0) and (—5, 0), the inclination of one of whose asymptotes
is 30°.
6. Find the equation of an hyperbola whose transverse axis is along
the a;-axis, which passes through the point (5, 8) and whose asymptotes
are given by the equation y^ — ^ x'.
7. Write the equation of an hyperbola conjugate to the hyperbola
x' -2x-i'f -8y = 7.
8. Show by the method of Art. 62, that the equation of the asymp-
totes to the hyperbola 4 1' — ^' = 4 is 4 a? = ^.
9. Find the equation of the rectangular hyperbola a;^ — ^ = o^ re-
ferred to its asymptotes.
10. Find the equation of the asymptotes to the hyperbola x^—^ = 9.
Prove that any line parallel to an asymptote meets the curve in only
one finite point.
11. If ei and ei are the eccentricities of two conjugate hyperbolas,
prove that — : H — j = 1.
12. In an hyperbola, if the value of e is very little more than unity,
how does the value of 6 compare with that of at Discuss the slope of
asymptotes and form of curve. As e increases, what is the effect on the
slope of the asymptotes and the form of the curve?
13. Prove that the distance of any point on the rectangular hyper-
bola a;2 — ^2 = a' from the center is a mean proportional to its distances
ffom the foci.
14. Prove that the product of the distances of any point on an hy-
perbola from its asymptotes is constant.
15. Find the coordinates of the foci of the hyperbola a? — ^ = 9.
By rotating the axes, find the coordinates of the foci and the equation of
the hyperbola when its asymptotes are taken as axes.
16. Find the equation of an hyperbola whose transverse axis is along
the X-axis, which passes through the point (5, 2) and has the same asymp-
totes as 4 x^ - 9 2/2 = 36.
EXERCISES 145
17. Prove that an asymptote and the perpendicular from the focus
upon it meet upon the corresponding directrix.
18. Prove that the directrices of an hj^ierbola and the circle whose
diameter is the line joining the foci intersect on the conjugate hyperbola.
19. Prove that the portion of the asymptotes intercepted between
the directrices is equal to 2 a.
20. Through the point Pi on the hyperbola x^/a? — ■fj}? = 1, a
straight Une is drawn parallel to the transverse axis cutting the asjrmp-
totes. Prove that the product of the distances of Pi from these points
of intersection is equal to a^.
21. If the crack of a rifle and the thud of the ball on the target are
heard at the same instant, prove that the locus of the hearer is an hyper-
bola.
CHAPTER X
TANGENTS AND NORMALS
65. A line which cuts a curve is called a secant. The
Une P1P2 in the figure is such a fine.
If one of the points of intersection, as P2, is made to
move along the curve and approach the other point Pi,
the line P1P2 will approach a limiting position PiR. This
line is called the tangent to the curve. The point Pi is
called the point of contact of the tangent line. The follow-
ing definition may then
be stated: A tangent to
a curve at a given point is
the limiting position of the
secant line connecting the
given point with a second
point on the curve, as this
second point moves along
the curve and approaches
coincidence with the given
point.
The line which is perpendicular to the tangent at the point
of contact is called the normal to the curve. The line PiK is
the normal to the curve at Pi.
The equations of the tangent and normal to any curve
at a given point on the curve. — A point on the tangent hne
is given. If the slope of the tangent can be determined, its
equation can be readily found by substitution in the stand-
ard equation,
y - yi = mix - Xi).
146
THE EQUATIONS OF THE TANGENT
147
The method of finding this slope will now be illustrated.
Let it be required to find the slope of the tangent to the
curve j/^ = 0? at the point Pi (x\, j/i) on the curve.
Let Pi (xi + /t, 2/1 + A;) represent a second point on the
curve. Then by the formula Y
2/2- yi
m =
the slope
of
Xi — X\
P1P2 is
found to
be
h
If the point P2 is now made to
approach Pi, the values of h and
fc each approach 0, and the slope
takes the indeterminate form of
r-. This difficulty arises from
having failed to make use of the
fact that the points Pi and Pj lie on the curve and thus
their coordinates must satisfy the equation of the cvirve.
Substituting these in the equation 'ip = a^,
2/1^ = a;i',
(2/1 + W = (^1 + hf.
Expanding and subtracting (1) from (2),
2 2/ifc + fc' = 3 Xi% + 3 xji" + W,
or fc (2 j/i + fc) =/i (3x1^ + 3 Xi/i + J^^),
Whence the slope of the secaftt,
h_Zxi^ + ZxJi + h^
h~
(1)
(2)
(3)
(4)
(5)
2yi + k
If now h and k are made to approach 0, the slope of the
tangent =
li^it (1) = limit (3^!±3^lM^^) =
3xi^
2j/i"
148
TANGENTS AND NORMALS
{x - Xi),
Consequently, the equation of the tangent is
3x1='
or 3xi^x-2yiy = 3xi^-2yi\
or 3 a;i'' X — 2 yiy = x-^ (since yi^ = Xi').
Answers will, in general, be simphfied by collecting all
the variable terms in the first member of the equation and
the constants in the second, and then reducing the second
member to simpler form by making use of the fact that
•Pi {xi, 2/i) is a point on the curve.
The steps taken may be summarized as follows: To find
the slope of the tangent to a given curve at a given point Pi {xi, yi),
choose a second point P2 (xi + ^, j/i + k) on the curve. Sub-
stitute the coordinates of Pi and P2 in the given equation and
subtract. Find the value of k/h, the slope of the secant. The
limiting value of this slope as h and k approach zero is the
slope of the tangent.
Where the point of contact Pi is given by numerical
coordinates, the substitution of the coordinates of the
second point Pa gives sufficient data from which to deter-
mine the value of k/h.
This is illustrated in the
following example. Find
the equation of the tan-
gent to the circle x^ -|- 2 x
+ 2/2 _ 4 y = 20 at the
point (2, 6).
Let Pi (2, 6) and Pj (2
+ h,6 + k) he two points
on the given circle.
The substitution of
Pi{2 + h,6 + k) in the given equation gives
4: + 4:h + h' + 4: + 2h + 36 + 12k + ¥ - 24: - 4.k = 20,
THE EQUATIONS OP THE TANGENT 149
from which
k''+8k= -{¥ + 6h) or k{k + 8) = - h{h + 6).
I k h + 6
Whence the slope of the secant = t = — : ■ Letting h
and k approach 0, the slope of the tangent is found to be
— f . Therefore the equation of the tangent is
2/ - 6 = -f (cc - 2).
or 4 2/ + 3 a; = 30.
Since the normal to a curve is a line perpendicular to the
tangent at the point of contact, therefore the slope of the
normal is the negative reciprocal of the slope of the tangent
at that point. The equation of the normal may then be
determined by substitution in the equation
y - yi = m{x- Xi).
Thus, the slope of the normal to the circle given in the
preceding example at the point Pi (2, 6) is +| and the
equation of the normal isy — Q = ^(x — 2) or ix — Zy.
+ 10 = 0.
EXERCISES
1. Find the equation of the tangent to each of the following curves
at the point {xi, yi) :
(o) xy = 4. Ans. Xiy + yiX = 8.
(fc) 2/' = ipx. Ans. yyi = 2p (x -\- xi).
(c) x' + y' = r^. Ans. Xix + yiy = r*.
(e) y' + ay + bx + c = 0. Ans.' (2yi+a)y+bx+ayi+'bxi+2c=0.
2. Find the equations of the tangent and normal to each of the fol-
lowing curves at the indicated point. Draw the curve, tangent, and
normal in each case.
(a) y = x' at (1, 1).
(6) y = |±|at(2,5).
150 TANGENTS AND NORMALS
(c) f = 2 a;' at (2,4).
(d) yx^ = lat (-1,1).
(e) 9 ^2 ^ 25 2/2 = 225 at the positive extremity of the right hand
latus rectum.
(/) 2/ = a? — 4a; + 8at the point whose abscissa is 2.
(S) 2/ = xMa: - 2) at the point (3, 9).
ifC) y' = a;2 at the point whose abscissa is 8.
66. The equation of the tangent to the curve represented
by the general equation of second degree. — The most gen-
eral equation of second degree is
Ax^ + Bxy + Cy^ + Dx + Ey + F = 0. (1)
In order to find the tangent at the point Pi {xi, j/i) on
this curve, the same steps are followed as in the preceding
article. Let a second point on the curve be Pa {xi + h,
yi + k). Substitute the coordinates of Pi and Ps in equa-
tion (1).
Then Ax,' + Bx,y, + Cyi' + Dx, + Eyi + F = 0, (2)
and A {xi + hy + B (xi + h) {y, + k) + C{yi + W
+ D{x, + h) + E{y^ + k) + F = 0. (3)
Expanding and subtracting (2) from (3),
2 Axji + AK- + Byji + Bxik + Bhk
+ 2 Cyik + Ck^ + Dh + Ek = 0, (4)
which, on collecting, becomes
kiBxi+2Cyi+Ck+E)= -hi2Axi+Ah+Byi+Bk+D). (5)
k 2Ax^ + Ah + Byi + Bk + D .„^
^^^"^ A = Bx, + 2Cy, + Ck + E ^^^
Letting h and k approach
*i, 1 t4.u ^ i 2Axi + Byi + D
the slope of the tangent = - ^^^ ^2Cyi + E ' ^^^
THE EQUATION OF THE TANGENT 151
The equation of the tangent is therefore
Clearing this of fractions and placing the variable terms
in the first member and the constant terms in the second,
2 Axix + Byix + Bxiy + 2 Cyiy + Ey + Dx
= 2 Ax^ + 2 Bxxy^ + 2 Cy^ + Bxx + Ey^. (9)
From equation (2), the first three terms of the second
member equal — 2 Dxi — 2 Ey^ — 2F.
Substituting this value in (9) and transposing,
2 Axix + Byix + Bxiy + 2 Cyiy
+ D{x + Xx)+E{y + y,)+2F = 0. (10)
Dividing by 2,
» O
Axix + ^ (aJiy + yix) + Cy^y + -^ (*+«i)
+ 1 (y+yi)+F=o, (39)
which is the equation of the desired tangent at Pi {x\, y\)
on the curve Ad? + Bxy + Cy^ + Dx + Ey + F = 0. This
result is very important and should be remembered.
A convenient statement is as follows: The tangent to the
curve represented by any equation of second degree is found by
, ■ « , « 7 1 ^iV + Vi^ i_ X + Xi ,
replaang x^ by Xjx, y^ by yiy, xy by ^ ' > ^ "V 2 '
ILLUSTRATIVE EXAMPLE
Find the equation of the tangent to x'+& xy+y^—2 x+i y+6=0 at
(xi, yi).
Applying the above rule, the tangent is
,., + 6 (2^4^)+ 2..V - 2 (^) + 4 (^)+ 6 = 0,
or
(.xi + 3yi - l)x + {3xi+yi+2)y = Xi-2yi- 6.
152
TANGENTS AND NORMALS
A very convenient check on the correctness of the equation of the
tangent at the point Pi {xi, yi) is to drop the subscripts in the equation
of the tangent and to notice that the result should be identical with the
equation of the given curve.
67. Lengths of tangents and normals, subtangents and
subnormals. — The tangent and normal lines are indefinite
in extent, ,but it is customary to speak of that portion of
the tangent between its intersection with the a;-axis and the
point of contact as the tangent length and that portion of
the normal between the point of contact and its intersec-
tion with the a;-axis as
the normal length. In
the figure, TPi is the tan-
gent length and PiN is
the normal length.
The projection of the
tangent length on the
X-axis is called the sub-
tangent and the projec-
tion of the normal length
on the a;-axis is called the subnormal. In the figure, TM is
the subtangent and MN the subnormal.
There is Httle occasion to use the sign of the tangent and
normal lengths and they are usually treated as positive,
but in the case of the subtangent and subnormal the signs
are important. The subtangent is always measured from
the point where the tangent crosses the a;-axis to the foot
of the ordinate of the point of contact, and the subnormal
from the foot of this ordinate to the intersection of the
normal with the rc-axis. These lengths are easily computed,
for, from the figure,
subtangent = TM = OM - OT,
subnormal = MN = ON — OM.
LENGTHS OF TANGENTS AND NORMALS 153
The abscissa OM of the point of contact is given, and
OT and ON can be found since they are the intercepts of
the tangent and normal respectively on the a;-axis.
The tangent length is the hypotenuse of the right triangle
of which the legs are the subtangent and the ordinate of
the point of contact. The normal length is the hypotenuse
of the right triangle of which the legs are the subnormal
and the ordinate of the point of contact.
ILLUSTRATIVE EXAMPLE
Find the equations of the tangent and normal tpa;' + 2a; + 32/ = 17
at the point (2, 3), and determine the lengths of the subtangent, sub-
normal, tangent, and normal. Sketch
the figure.
Rearranging terms and collecting,
the equation x^ + 2x + Zy = VJ be-
comes (s + ly = —3(2/ — 6), which
shows that the figure is as here sketched.
From the rule in paragraph 66, the
equation of the tangent to a;^ -|- 2 a; -|-
3 2/ = 17 at Pi (2, 3) is found to be
2/ -f- 2 a; = 7. The normal, which is per-
pendicular to the tangent at Pi (2, 3),
has for its equation 2y—x = ^.
OT, the intercept of the tangent on the a;-axis = 1.
ON, the intercept of the normal on the a;-axis = — 4.
OM, the abscissa of Pi = 2.
The subtangent = TJlf = OM - OT = 2 - | = -f.
The subnormal = MN = ON - OM 4 -2 = -6.
--Vtm'
\
J
,
-1^
-J
I
^'^
s§
Af^,''
T
^
1
^'i
K, ^^
V
•i:--,-i3--
T^--^-
CpLVt""'
^-r
aS
y' r
.
The tangent length = TPi
The normal length = PiiV
' + AfPi' = V| + 9 = f Vs.
--VmK' + MN^ = VT+ZG = 3 Vs.
EXERCISES
1. Determine the equations of the tangent and normal, and lengths
of subtangent and subnormal to the following curves at the point given.
Draw the figure in each case.
154 TANGENTS AND NORMALS
(a) y = 23?aX (1,2).
(6) x2 + 2/2 = 25 at (3, 4).
(c) a;2 -4 a; = 42/at (4, 0).
{d) xy = i a,t point whose abscissa is 1.
(e) 2x2-2/!.= Mat (3, -2).
X^ y2
(/) ■=■ + rs = 1 at positive end of the upper latus rectum.
( Id
(?) 2/2-6j/-8x = 31at(-3,7).
2. Find the equations of the tangents to 2/" = 4 x — 4 at the ex-
tremities of the latus rectum. Prove that they are perpendicular and
meet on the directrix.
3. Find the equations of the tangent and normal to x' + 4 x + ^' +
62/ = 12 at (1,1) and prove that the normal passes through the
center.
4. Write the equations of the tangent and normal to each of the fol-
lowing conies at the point given. Draw each figure.
(a) 4x2 - 16x + 9 v^ - 182/ = 11 at (2, 3).
(6) x2 - 4x - 22/ - 1 = at (1, -2).
(c) 3x2 + loxj/ + 32/2 = 3 at (-i^, 1).
(d) x2 _2x2/-|-^ = 4x + 4^at (4,0).
6. Prove that the tangents at the extremities of a latus rectum of
the curve 7 x2 + 16 2/2 = 112 meet on the corresponding directrix.
6. Prove that the tangents at the extremities of the latus rectum
intersect on the directrix in the case of: (a) any parabola; (6) any
eUipse; (c) any hyperbola.
7. How far from the vertex are the tangents at the extremities of the
latus rectum oi x' + iy + i = 01
8. Find the angle formed by the tangents at the extremities of a
latus rectum of the hyperbola 9 x2 — 16 2/2 = 144.
9. Prove that the normal at one extremity of the latus rectum of a
parabola is parallel to the tangent at the other extremity.
X2 V2
10. Given the ellipse ■„ + ^ = 1. Find the equations of the tan-
gents whose intercepts on the axes are numerically equal.
11. Prove that the tangents, at the extremities of the latus rectum
of a parabola are twice as far from the focus as from the vertex.
12. Prove that the tangent at any point of the parabola ^ = 4 px,
the perpendicular from the focus upon it, and the tangent at the vertex
meet in a point.
EQUATION OF TANGENT WHEN SLOPE IS GIVEN 155
13. Prove that the subtangent of the ellipse -; + ?t = 1 at (xi, vi)
a' 0'
x{^
a'
Xl
and that the subnormal is (e^ — l)a;i.
14. What is the point of contact of that tangent to the parabola
j/2 = 4 s, whose intercepts on the axes are numerically equal and (a) of
same sign; (b) of opposite sign.
16. Prove that the perpendicular from the focus of an eUipse upon
any tangent and the line joining the center to point of contact meet on
the cortesponding directrix.
68. The equation of the tangent when the slope is
given. — The process used in finding the equation of tan-
gents to curves of second degree, when the slope is given,
will be illustrated by the following example.
Let it be required to find for the circle x^ — 6x + y^
— 6 y + 10 = 0, the equations of the tangents of slope — 1.
Let y = —x + b represent any one of the system of
parallel Unes of slope —1.
It is evident from the following figure that some of these
lines such as AB cut the
curve in two distinct
points, and that if this
Hne is moved parallel to
itself, the two distinct in-
tersections will approach
each other and eventually
coincide as at Pi and P2.
In this position, the Hne
y = — a;-l-f>isa tangent
to the curve. The prob-
lem then is to so determine b that the line y = — x + b
shall meet the curve in two coincident points.
If the equation of the line
y = - x + b
and the equation of the circle
x''-6x + y^-6y + 10 =
156 TANGENTS AND NORMALS
are solved simultaneously, in order to determine the co-
ordinates of the intersection, the result, after eliminating y, is
x'-6x+ i-x + by - 6 i-x + 6) + 10 = 0,
or when expanded,
2x^ - 2bx + ¥ - 6b + 10 = 0,
from which the abscissas of the intersections of any line of
the system y= —x + b with the given circle may be
determined.
It was shown in the theory of equations, that an equation
in the form
Ax^ + Bx + C =
has equal roots if 5' — 4 AC = 0.
Applying this principle here, the equation
2x'-2bx + b^-6b + 10 =
has equal roots if
4 &2 _ 8 (&2 _ 6 6 + 10) = 0,
or if 62 _ 12 6 + 20 = 0.
This last equation is true if & = 2 or 10, whence a line of
the system y = — x + 6 meets the given circle in two coin-
cident points if 6 = 2 or 10.
Therefore, y = —x + 2
and y = —x + 10
are the equations of the desired tangents.
EXERCISES
1. For each of the following curves, find the equations and points of
contact of the tangents whose slopes are as given:
(a) a;2 + j/2 = 25, slope = - J.
(6) yi = 4x + 4, slope = 1.
(c) xy = 4, slope = — 4.
Id) x' -ix-y^ -iy = S, slope = 2.
EQUATION OF TANGENT WHEN SLOPE IS GIVEN 157
2. For each of the following curves, find the equation of the tangent
with slope to:
(o) 3/2 = 4 Tpx. Ans. y = mx + — •
(6) ^ + ^ = 1. Ans. y =mx± VaHn'^ + V.
(c) ^ - ^ = 1. Ans. y = mx± ^dhri? - }fi.
a' 0'
3. Find the equation of the tangent to 4 a;^ + 25 j/^ = 100 parallel
to3x + 102/ = 60.
4. Find the equation of the tangent to a;^ = iy + i perpendicular
to 2 ^ + a; = 7.
X^ 7/2
6. Find the equations of the tangents to the ellipse -5 + fj = 1
which are parallel to the Une joining the positive extremities of the axes.
6. Prove that the Une Sa; — 2y — ll=Ois tangent to 5 a? —20 x
- 2 2/= + 4 2/ + 15 = 0.
7. Find the equations of the tangents to9a;2 + 4j/2 + 6a; + 42/ =
parallel to3x + 2y = 7, and the equations of the normals at the points
of contact.
CHAPTER XI
POLES, POLARS, DIAMETERS, AND CONFOCAL
CONICS
69. Harmonic division. — If a line AS is divided in-
A P B Pi
ternally by the point P and externally by the point Pi in
such a way that
AP^_APi
PB PiB' ^ ^
the line is said to be divided harmonically.
Theorem. — If two points P and Pi divide a line AB har-
monically, then conversely, the points A and B divide the line
PPi harmonically.
Proof. — If the proportion (1) above is taken by alter-
nation, it becomes
AP
APi
PB
PiB
PB
BPi
ibers,
PB
BPi
AP
APi
PA
AP,
Reversing the members,
P7? PA
From the proportion -^^ = — 7-5- , it is seen that the
line PPi is divided harmonically by the points A and B.
158
POLE AND POLAR
159
EXERCISES
1. Find the coordinates of the point P which together with Pi (2, 3)
divides harmonically the Une joining A( — 1, 4) to B (8, 1). Ans.
(-10, 7).
2. Show that the points A and B in example 1 divide harmonically
the line joining Pi and P.
70. Pole and polar. — If through a fixed point Pi, out-
side, inside, or on a conic, a secant is drawn to the conic
meeting it in the points A and B, and if P is so chosen on
the secant that the points P and Pi divide the Une AB
harmonically, then the locus which contains all positions
of P as the secant revolves about Pi is called the polar of Pi
with regard to the conic, and the point Pi is called the pole
of that locus.
Equation of the polar for the ellipse -g + ^ = 1. — Given
the ellipsis -5 + ?2=l, and the fixed point Pi{xi,yi).
Through Pi draw any secant meeting the ellipse in the
points A and B, and so locate P {x, y) upon it that the line
AB is divided harmonically
by the points P and Pi.
It is required to find the
equation of the locus which
contains P in all of its
positions.
By the theorem in Art.
69, since AB is divided harmonically by P and Pi, then PPi
is divided harmonically by A and B, and hence
PB
BPy
PA
APi"
Let the segments PB and BPi be in the ratio n : rj (n
and Ti will vary as the secant is revolved).
160 POLES, POLARS, DIAMETERS, CONFOCAL CONICS
The coordinates of A and B can now be found by formulas
(3), Art. 8. They are respectively
l r\Xx - nx nyi - r2J/ \ , / nxi + r2X nyi + r-g^ N
\ ri - r-2 ' ri — Ti J \ n + r^ ' n + n J
The points A and B are on the elhpse and their coordi-
nates satisfy the equation
&V + a^ = aW. (1)
Therefore, . .„ . .,
\ n-Ti J \ ri-Ti /
\ n + ^2 / \ r-i + r-2 /
Expanding equations (2) and (3) and clearing of fractions,
b^ (ri'xi^ — 2 nnxix + TaV) + a^ (nW ~ 2 nviyiy + r2V)
= a262(r-,2-2rir2 + r22) (4)
b° (nW + 2 rir2a;ia; + r2V) + o^ (ri^j/i^ + 2 rir-iyiy + r^y^)
= aW\r^ + 2nri-^ri). (5)
In addition to the variables x and y, equations (4) and (5)
contain ri and r^ which also vary as the secant is revolved.
It is desired to find an equation, containing no variables
other than x and y, which will be true for any position of P
as the secant revolves. Therefore such equation must be
independent of n and r^.
Subtracting (4) from (5),
b^ (4 nr^x-iX) + (^ (4 nnyxy) = a^¥ (4 nri). (6)
Dividing by 4 rir^,
¥xix + a^yiy = aW, (7)
or
a" ^ y ' ^^>
which is the desired equation of the polar. Since it is of
first degree, the polar is a straight line.
POLE AND POLAR 161
It will be noticed that equation (8) is of the same form as
the equation of the tangent to the eUipse found in example
1 (d) of Art. 65. In the polar, however, Pi may be inside,
outside, or on the conic.
If Pi is outside the ellipse, P must be inside the elhpse in
order that the secant may be divided harmonically, there-
fore only that portion of line (8) which lies inside the conic
fulfils this condition. If Pi is inside the ellipse, Pi and any
point P on fine (8) will divide AB harmonically, therefore
the whole line fulfils the condition. If Pi is on the ellipse,
the polar line is the tangent, the point of contact is the pole,
and no point other than the point of contact itself fulfils the
condition.
By the same method, the equation of the polar of the
hyperbola -j — ^ = 1 is found to be
xxi yyi
^-^ = 1. (9)
and the equation of the polar of the parabola y* = 4 pa; is
yyi = 2p (« + *i). (10)
The proof of the properties which follow is based upon
the equation of the polar of the ellipse, but can be shown to
hold for the hyperbola and parabola by using equations
(9) and (10).
Important properties of poles and polars. —
1. // thR polar of the point Pi passes through the point Pa,
then the polar of the point Pi passes through the point Pi.
Proof. — The equations of the two polars for the elUpse are
and
a^ "^ b^ ^ '
^ + ^ = 1. (2)
a'
2 > }fi
162 POLES, POLARS, DIAMETERS, CONFOCAL CONICS
Since by hypothesis Pi lies on the polar of Px, the coordi-
nates of Pi satisfy (1), and, therefore,
y_m
-2 "T >,2 ~ ■*••
(3)
If, however, the coordinates of Pi are substituted in
equation (2), the equation is identical to equation (3).
Therefore the coordinates of Pi satisfy equation (2) which
shows that Pi lies on the polar of P2.
2. If the pole Pi is outside the conic, the polar is the chord
joining the points of contact of the two tangents drawn from
Pi to the conic.
It is evident from the figure that as the secant revolves
about Pi there will be two positions, PiC and PiD, in which
it will be tangent to the
conic. As the secant ro-
tates toward the position
PiC, the points A and B
approach each other and
come into coincidence at
C. But P lies between
A and B, therefore P
coincides with C and hence the point of contact C is on the
polar. Similarly, the point D is on the polar, and the polar
passes through the two points of contact of the tangents
drawn from Pi.
EXERCISES
1. By the method of this article, find the equation of the polar of
Pi (xi, Vi)
(a) with respect to the hyperbola x^/a? — j/'/ft^ = 1.
(6) with respect to the parabola y' = i px.
2. Find the polar of the point
(a) ( — 1, —3) with respect to the conic x^ -{- iy' = 16.
(b) (2, 4) with respect to the conic ip = x.
DIAMETERS 163
3. Prove that the directrix is the polar of the focus
(o) for the parabola y' = i px.
(b) for the ellipse x^a' + yyb^ = 1.
(c) for the hyperbola xya' — y'/b' = 1.
4. Find the pole of3x + 42/ = 4 with respect to the ellipse 6 x^ +
8 j/i! = 16.
Hint. — Let Pi{xi, yi) represent the pole of Sx + iy = A. The
polar of (xi, yi) is 6 xsi + 8 yyi = 16, but 3x + iy = iis also the polar
of {xi, yi). Since these equations represent the same hne, they will be
identical if their second members are made equal.
5. Find the pole oi5x + 6y = Z with respect to the conic 15 a^ —
3 2/2 = 9.
6. Prove that the line joining any point Pi (xi, yi) to the center of the
circle x' + j^ = t^ is perpendicular to the polar of the point with respect
to the circle.
7. Prove that the radius of a circle is a mean proportional between
the distance from the center to the point Pi {xi, yi) and the distance from
the center to the polar of Pi.
8. Prove that the polar of ( — 1, 2) with respect to the parabola
^ = 4 a; passes through (2, 1). Verify the first property of this article
by showing that the polar of (2, 1) passes through ( — 1, 2).
9. Find the polars of the vertices of the hyperbola x'^/a? — ■jf/V = 1
with respect to its conjugate hyperbola.
10. Find the equations of the tangents to the circle x^ + y' = 25
through the external point (10, —5).
Hint. — Find the equation of the polar of the given point, then find
the points of contact by use of the second property of this article.
71. Diameters. — The locus of the middle points of
any system of parallel chords of a conic is called a diam-
eter of the conic.
The method of finding the equation of a diameter is
illustrated by the following examples:
1. Find the equation of the diameter of the ellipse x^/a'
-f- ^2/52 = 1 which bisects all chords of slope m.
Let y = mx + k, in which m is fixed, represent a system
of parallel chords. The constant k will have different values
for different chords.
164 POLES, P0LAR8, DIAMETERS, CONFOCAL CONICS
Let Pi and P2 be the points in which any one of these
chords cuts the elUpse and let P {x, y) be the center of this
chord.
To find the abscissas of
Pi and P2, y must be elimi-
nated between the equations
xya" + 2/76^ = 1 (1)
and
y = mx + k.
(2)-
The resulting equation is
(a'm' + &') x^ + 2 a^kmx + a%^ - a?W = 0. (3)
Similarly, the ordinates of Pi and P2 may be found from
the equation
{d'm^ + ¥) 2/2-2 ¥ky + W¥ - al'VW = 0.
(4)
It was found in the study of theory of equations that the
sum of the roots of the quadratic equation Ax^ + Bx + C
= is —B/A, therefore, from equation (3),
a;i + a;2 =
and from equation (4),
2/1 + 2/2 =
But by formula (4), Art. 8,
— 2a%m
2b%
(5)
(6)
Therefore,
Xi + Xz , 2/1 + Vi
= — 2 — ^"^ y " 2
X =
y =
— a?km
Vk
(7)
(8)
DIAMETERS 165
In addition to the variables x and y, equations (7) and (8)
contain fc which also varies as P moves along the diameter.
It is desired to find an equation, containing no variables
other than x and y, which will be true for any position of P.
Therefore, such equation must be independent of k.
Dividing equation (8) by equation (7), the equation of the
diameter of the ellipse bisecting chords of slope m is
y = - -^x. (9)
2. Let the student show that the equation of the diam-
eter of the hyperbola which bisects chords of slope m is
From the form of equations (9) and (10), it is evident
that every diameter of an ellipse or hyperbola passes through
the center.
3. Find the equation of the diameter of the parabola
y2 = 4 pa; which bisects chords of slope m.
Using a process identical with that used in example 1, the
intersections Pi and Pi can be determined from the equa-
tions
mV - (4p - 2mfc) a; + fc2 =
(1)
my'' — Apy + 4:pk = 0.
(2)
4 M — 2 mfc
(3)
4p
2/1 + 2/.= ^-
(4)
2p — mk
1, X = -^ — ^ — ■ ,
' m^
(5)
2p
y = .. •
(6)
m
166 POLES, POLARS, DIAMETERS, CONFOCAL CONICS
Following the plan of example 1, the next step would
be to find from equations (5) and (6), an equation independ-
ent of k. (6) is such an equation and therefore the equation
of the diameter of the parabola which bisects chords of slope
m is
y = ^- (7)
m
Its form shows that the diameter is parallel to the axis of
the parabola. .
Properties of diameters of central conies. —
1. Any line through the center of an ellipse or hyperbola is
a diameter bisecting some set of parallel chords.
Proof. — Let y = miX represent any line through the
center of the ellipse x^/a^ + y^/b^ = 1. The equation of
the diameter of this ellipse which bisects chords of slope m
is 52
y = — J— X.
a^m
These lines are identical if
b^
mi = ^•
a^m
For a given ellipse, a and b are fixed, while m may have
any value. Therefore, it is always possible to so choose m
that mi shall equal 5— . Hence y = mix always bisects
a^m
some set of parallel chords.
A similar proof holds for the hyperbola.
2. If one diameter bisects all chords parallel to a second
diameter, then the second bisects all chords parallel to the first.
Proof. — Let y = m^x (1)
be a diameter of the ellipse x^/a^ + y'^/V^ = 1 which bisects
all chords parallel to the diameter y = mx.
CONJUGATE DIAMETERS
167
The diameter which bisects chords of slope m has for its
equation
Therefore,
V =
mi ■■
¥
a?m'
X.
(2)
(3)
The equation of the diameter which bisects all chords of
slope mi is
from which it is seen that the slope of this diameter is
— W/a?mi.
Substituting the value of mi from equation (3), this slope
becomes
and, therefore, y — mx is the equation of the diameter bisect-
ing chords of slope mi.
A similar proof holds for the hyperbola.
72. Conjugate diameters. — If each of two diameters
bisects all chords parallel to the other, the diameters are
said to be conjugate.
168 POLES, POLARS, DIAMETERS, CONPOCAL CONICS
From equation (3) above it is seen that for the ellipse
the relation which holds between the slopes of two conju-
gate diameters is
mmi= - ^- (1;
In the case of conjugate diameters of the hyperbola, the
relation between the slopes is
mmi = — g- (2)
y
The equation of the hyperbola conjugate to -5 — Ti = 1
IS
¥
1. If the diameter bisecting chords of slope m
for the second hyper-
bola is found by the
process of Art. 71, it is
seen to be « = -;— as
in equation (10) of that
article.
It is thus seen that
the same line is a di-
ameter of each hyper-
bola, and bisects a sys-
tem of chords in each.
EXERCISES
1. Find the equation of the diameter of the ellipse ix' + 9 1/' = 36,
bisecting chords of slope 2.
2. Prove that the lines 2y + Zx = and 2 2/ + a; = are conju-
gate diameters of the hyperbola 3x^ — 4 j/^ = 12.
3. Find the equation of the diameter of the parabola y' = 8x bi-
secting chords parallel to the polar of (3, 4) with respect to the parabola.
4. Prove by means of equations (1) and (2), Art. 72, that a pair of
conjugate diameters of the ellipse he in different quadrants, and that a
CONFOCAX CONICS 169
pair of conjugate diameters of the hyperbola lie in the same quadrant
and on opposite sides of the asymptotes.
6. Lines are drawn joining the extremities of the major and minor
axes of an ellipse. Prove that the diameters parallel to these are con-
jugate.
6. Find the equation of the diameter of the parabola j/' = 4:x
through the point (1, 2). Also find the equation of the chords which it
bisects.
7. Given an extremity of a diameter of the ellipse x' + 2y^ = 24: is
(4, .2). Find the extremities of the conjugate diameter.
8. Find the equation of the chord of the hyperbola x^ — 2 y^ = 4
through (3, 1) which is bisected by the diameter 2y = x.
9. Find the equation of that chord of the ellipse a;^ + 4 2/^ = 16
which is bisected at (1, 1).
10. Prove that the tangent at the extremity of a diameter of an
ellipse is parallel to the conjugate diameter.
Hint. — Let Pi (si, yi) be the extremity of a diameter. Find the
equation of the tangent and of the conjugate diameter.
11. Prove that the polar of any point Pi (xi, yC) on a diameter of
an ellipse is parallel to the conjugate diameter.
12. Write the equation of the diameter of the parabola ^' = 4 pa;
which bisects chords of slope m. Prove that the tangent at the ex-
tremity of this diameter is parallel to the chords.
13. In the rectangular hyperbola x'^ — y'^ = a?, a diameter passes
through the point Pi {xi, yi) on the hyperbola. If Pj is the point in
which the conjugate diameter cuts the conjugate hyperbola, prove
OPi = OP2.
73. Confocal conies. — Consider the equation
in which X is an arbitrary constant and a>b.
1st. If X is positive, or negative and > —6', equation (1)
represents an ellipse.
From equation (28), Art. 51, the distance from the center
to the focus is
Via" + X) - (62 + X) = Va? - ¥.
170 POLES, POLARS, DIAMETERS, CONFOCAL CONICS
Therefore the eUipses have the same foci for every value of X.
Y If X is positive, the
ellipses lie without
-' + ^' = 1, and if it
is negative, they he
^ within.
2nd. If X is nega-
tive and is such that
— b^>'K> —a', equa-
tion (1) represents an
hyperbola.
Let X = — Xi, then equation (1) may be written
y
= 1,
a" - Xi Xi - 62
and the distance from the center to the focus is
V{a^ - Xi) + (Xi - ¥) = Va' - ¥.
It is therefore seen that the equation
__£!_, _j1_^ 1
represents a set of eUipses and hyperbolas all having the
same foci.
The curves of the system represented by this equation are
called confocal conies.
X'
EXERCISES
1. Show that through the point (2, 1), two conies of the system
■+:
= 1 may be drawn. Find their equations and plot the
1 + 1= landx'-y^ =3.
2. Find the equations of the tangents at (2, 1) on the conies foimd in
exercise 1. Prove that these tangents are perpendicular.
7 + X ' 1 + \
curves. Ans.
CHAPTER XII
THE GENERAL EQUATION OF SECOND DEGREE
74. The general equation of second degree
Ax^ + Bxy + Cy® + D« + £y + F = (1)
represents a conic section whose axes are inclined at an angle
6 with the axes of coordinates, 6 being the positive acute value
determined from the equation
l>
tan 2 9 = -r r^-
A — L
Proof. — It will first be shown that it is always possible
to so rotate the axes as to cause the xy-\^vm. to vanish.
Making the substitutions from Art. 39, x = x' cos d —
y' sin 6, y = x' sin 6 + y' cos d, equation (1) becomes
A (x' cos e - y' sin ey + B {x' cos Q - y' sin %)
\x' sin + y' cos 0) + C (x' sin Q -\- y' cos ffy
+ D {x' cos % -y' sin S)-k-B {x' sin Q -\- y' cos 6)
+ F = 0, (2)
which when expanded and collected is
x'^ {A cos^ + B sin e cos + C sin^ S)
+ x'y' (-2 A sin e cos 9 - 5 sin^ Q ■\- B cos" 9
+ 2 C sin cos B) + y'^ (A sin^ - B sin S cos 5
+ C cos^ e) + a;' (D cos + S sin 0)
+ 2/'(^cose-Dsin0)+/?' = O. (3)
The x'y' term will vanish if its coefficient is 0, that is, if
- 2 A sin e cos e - 5 sin^ ff + B cos^ e + 2 C sin cos = 0, (4)
171
172 THE GENERAL EQUATION OF SECOND DEGREE
or if B (cos=i e - sin2 ff) = {A-C)2 sin cos d, (5)
or 5 cos2e = (A - C) sin29, (6)
or tan 2 = . _ ^ - (7)
Since the tangent of an angle may have any value from
— 00 to + 00 , it is therefore always possible to rotate the
axes through such an angle that the xy-ieria shaU vanish.
Moreover, since any number may be the tangent of an
angle in the first or second quadrant, there is always a value
of 2 < 180° and a corresponding value oi 6 < 90° which
satisfy equation (7). The positive acute value of d will
always be chosen.
The xy-iercn having been removed, the general equation
becomes
A V + Cy + D'x + E'y + F' = 0. (8)
An equation of this form represents
(a) a circle if A' = C (this case never arises when
5 7^ in the general equation).
(6) a parabola if A' and E' are present with C" absent
or if C and D' are present with A' absent.
(c) an elHpse if A' and C are positive and unequal.
(d) an hyperbola if A' and C have unUke signs (except
where the equation can be resolved into first
degree factors).
(e) a pair of straight Unes. When A' and C have
unhke signs, the equation can frequently be
factored and thus represents a pair of straight
lines. When A' and D' each equal 0, or C and
E' each equal 0, the equation always represents
a pair of parallel lines, distinct, coincident, or
imaginary.
THE GENERAL EQUATION OF SECOND DEGREE 173
ILLUSTRATIVE EXAMPLE
By first removing the xy-tena, determine the nature and position of
the curve whose equation is IQx' — 2ixy + 9 y' — S5x — SOy + 175
= and plot the locus.
Rotate the axes through an angle e, by substituting a; = a;' cos fl —
y' sin e and y = x' siaB + y' cos 9, in the equation
16 a;2 - 24 sv + 9 3/2 - 85 X - 30 2/ + 175 = 0. (1)
This gives
16 (x' cos e — y' sin 6)' — 24 (x' cos 6 — y' sin 6) (x' sin O+y' cos 6)
+ 9{x'sia.e+ y' cos ey - 85 (x' cos 8 - y' sax 8)
- 30 {x' sin 9 + 2/' cos e) + 175 = 0. (2)
Collecting,
a;'" (16 cos2 9-24 sin fl cos 9 + 9 sin^ 8)
+ x'y' ( -24 cos" 9 + 24 sin^ fl - 14 sin 8 cos 9)
+ 2/'2 (16 sin2 9 + 24 sin 9 cos 9 + 9 cos^ 9) - x' (85 cos 9+30 sin 9)
+ y' (85 sin 9 - 30 cos 9) + 175 = 0.
The x'y'-teira will vanish if
—24 cos2 9 + 24 sm2 9 - 14 sin 9 cos 9 = 0,
or if 24 cos 2 9 = —7 sin 2 9,
or tan 2 9 = -Y-
If tan 2 9 = - Y> then sin 2 9 = ff and cos 2 9 = —i
2 5*
3 = 3
! 5"
(3)
TTTu • ^ 4 /l - cos 2 9 t/l
Whence sm 9 = y ^ ^ V ~
„ 4 /i + cos 2 9 4 /r
cos9 = y-^ii2 — = V-
Substituting in equation (3) and
clearing of fractions,
y" + 2y' = 3x' - f, (4)
or (y' + l)"* = 3 (x' - 2). (5)
This is seen to be a parabola which
when referred to the new axes has its
vertex at (2, — 1), principal axis par-
allel to the X-axis, and distance from
vertex to focus f , Constructing both
sets of axes and the curve, the figure
is as shown,
Y T X'
i <
^ A'
3p I ji
3s Z^
s, j/ ^
^v *l/ ,''
•s ^ _,7
!^i,zz_i:___T
5 5' zL
A-'^
X,^^ 5^^
12- ^ :
2/
t
174 THE GENERAL EQUATION OF SECOND DEGREE
75. Test for distinguishing the type of conic when the
equation contains the «y-terni. — Comparing equations (3)
and (8), Art. 74, it is seen that
A' = AcosH + Bsindcosd + Csm^e, (9)
• B' = Bcos2d- {A-C)sm2e, (10)
C = A sin^ e- Bsmdcose + C cos" 6. (11)
Adding (9) and (11) and applying trigonometry,
A' + C = A + C. (12)
Subtracting (9) and (11) and applying trigonometry,
A' -C = {A-C)cos2d + Bsm2e. (13)
Squaring (10) and (13) and adding,
(A' - Cy + B"" = (A - Cy + B\ (14)
Squaring (12) and subtracting from (14),
B'^-^A'C = B^ -'LAC. (15)
If is so chosen that tan 2 = a _ p , then 5' = and
-4A'C'=- B^-4:AC.
It was shown in the previous article that if the given
conic is a parabola either A' or C = 0. Therefore B^
-AAC = 0.
If the given conic is an ellipse. A' and C have the same
sign, therefore 5^ — 4 AC < 0.
If the given conic is an hyperbola, A' and C have oppo-
site signs, therefore B^ — 4 AC > 0.
A + C and B^ — 4 AC can be shown to remain unchanged
when the equation is translated to a new origin. Since
these combinations of coefficients are unchanged by both
rotation and translation, they are called invariants.
CONIC THROUGH FIVE POINTS 175
EXERCISES
1. Simplify the following equations by removing the xy-term. Plot
both pairs of axes and the curve.
(a) 9 x2 - 24 Ki/ + 16 2/2 - 80 X - 60 2/ + 100 = 0.
(6) x^ + ixy + iy^ +x + 2y -2 =0.
(c) 7 y' - iS xy - 7 x^ + 30 X - 40 y = 0.
2. By rotation and translation, reduce the following equations to
their simplest form. Plot the three sets of axes and the curve. Check
the result by finding the nature of the conic by the method of this
article.
(o) 5x2 - 6x2/ + 52/2 - 2 V2 (x + 2/) - 6 = 0.
(6) 7x2 -48x2/ -72/2 + 701 + 102/ + 100 = 0-
(c) 2 x2 + 3 X2/ - 2 2/2 + 3 X + 2/ + 1 = 0.
(d) 4 x2 - 4 X2/ + 2/2 + 8 X - 4 2/ = 5.
(e) 16x2 - 24x2/ + 92/2 - 45x - 6O2/ - 400 = 0.
76. Conic through five points. — Since the general equa-
tion of second degree, Ax^ + Bxy + Cy^ -{■ Dx -\- Ey + F
= 0, has six constants any one of which can be divided out,
it is seen that only five of these are independent. There-
fore five conditions are sufficient to determine the equation
of a conic. In particular, this fact makes it possible to
write the equation of a conic through five given points.
ILLUSTRATIVE EXAMPLE
Write the equation of the conic through the 5 points (0, 0), (0, 1),
(1, 2), (1, -2), (5, 0).
The general equation Ax2 + Bxy + Cy' + Dx + Ey + F = 0, after
dividing by A, takes the form
x2 + bxy + cy' + dx + ey +f = 0.
Substituting the coordinates of the given points, five equations result
as follows:
/ = 0,
c+e+f = 0,
l+2b + 4c + d + 2e+/ = 0,
l-2b + ic + d-2e+f = 0,
25 + 5d+/ = 0.
176 THE GENERAL EQUATION OF SECOND DEGREE
Solving these equations simultaneously, 6 = 1, c = l, d=— 5,
e = — 1, and / = 0. Therefore the equation of the conic is
x'+xy + y' — 5x — y = 0.
If the equation of the conic through five given points is such that a?
is not present, it would not be permissible to divide the general equation
by A in the first step of the solution, and if such step is taken, the equa-
tions determined lead to contradiction. This diflBculty can be avoided
by dividing by some other constant. This is illustrated in Ex. 2 which
follows.
EXERCISES
1. Find the equations of the conies through the following points:
(a) (0, 0), (0, 1), (1, 0), (1, 1), (-1, 2).
(6) (0, 1), (2, 1), (1, 0), (2, 0), (1, f).
(c) (-4, 0), (0, -2), (1, 0), (-5, -2), (1, 5).
2. Find the equation of the conic through (0, -2), (0, 3), (-2, 0),
(1, 3), (1, -3).
Hint. — Divide the general equation by C.
3. Find the equation of the conic through the point (3, 2) and
through the points of intersection oi x' + y' = 25 and xy = 12. Plot
all loci.
Hint. — Use Art. 16.
4. Find the equation of the conic through the point ( — 1, 1) and
through the points of intersection oii3?-\-ixy + y^ — ix — 2y =
and x'' + y'-ix-3y = 0. Plot all loci.
5. What relation must hold between the coefficients of a:^ + bxy +
ci/^ + dx -\- ey + f = in order that its locus shall be tangent to the x-
axis. Find the equation of a conic passing through (3, 2), ( — 1, 2), (3, 8),
and tangent to the ataxia at (1, 0).
6. Find the equation of the conic through ( — 1, 0), (9, 0),-( — 1, 6),
and tangent to the y-axis at (0, 3). Plot all loci.
Hint. — Use oa^ + bxy + ■)/' + dx + ey+f = Oas the equation of
the conic.
7. Find the equation of the parabola through the four points (0, 0),
(1, 0), (0, 1), and (2, 1). Make use of the fact that in the parabola
^2 = 4 AC.
CHAPTER XIII
TRANSCENDENTAL AND PARAMETRIC
EQUATIONS
77. Loci of transcendental equations. — Thus far the
discussions have been largely concerned with algebraic
equations, that is, with equations involving variables
raised to constant powers. In this chapter, attention will
be given to other forms of equations such as y = a",
y = sinx, y = log x, etc. Such equations are called
transcendental equations.
Many of the steps taken in the discussion and plotting of
such curves are the same as those used in algebraic equa-
• tions, but there are some respects in which they require
different treatment. A few examples will illustrate the
method to be used.
78. The exponential curve y = a", where a is a positive
constant. In the following discussion, a will be taken as
greater than 1. A similar discussion would result from
taking o less than 1, in which case the figure would be
changed in position, but not in character.
1st. Intercepts. — Let x = 0, then y = a" = 1, whence the
curve cuts the y-axis at the point (0, 1). Let y = 0, then
a; = — 00.
2nd. Symmetry. — If y is replaced by —y, the equation
is changed, therefore the curve is not symmetrical with
respect to the a;-axis. If x is replaced by —x, the equation
is changed, therefore the curve is not symmetrical with
respect to the y-axis. If x is replaced by —x and y by —y,
the equation is changed, therefore the curve is not symmet-
rical with respect to the origin.
177
178 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
3rd. Extent. — It is seen that y is real for all values of x
from —00 to +ix>, also that as x increases from to +<x>,
y increases from 1 to + oo , and as x decreases from to — oo ,
y decreases from 1 to 0.
Solving for x in terms oi y, x = log^i/, from which it is
seen that x is real for all positive values of y. All negative
values of y are excluded since there are no logarithms of
negative numbers.
4iA. Asymptotes. — In the equation y = a", it is seen that
no finite value of x will make y infinite, therefore there are
no asymptotes parallel to the 2/-axis. In the equation
X = logo y, the only finite value of y which makes x infinite
isy = 0, therefore y = is the only horizontal asymptote.
In plotting the curve, it is necessary to assign some value to
a. The following table of values is computed for a = 3, from
which the curve y = S" is plotted.
X
V
X
V
1
1
3
-1
.33+
2
9
-2
.11+
3
27
-3
,03+
The exponential curve of most frequent occurrence is
y = e^, in which e = 2.718+. This mmiber, 2.718+, is of
great importance in all higher mathematics.
The student is asked to plot the three curves y = 2^,
y = e", and y = 3"^, using the same set of axes. Discuss
the effect of an increase in a on the form of the curve. (Use
e = 2.7.) Plot the locus y = (|)^
THE LOGARITHMIC CURVE 179
79. Relation between natural and common logarithms.
— A system of logarithms in which the base is 10 is called
a system of common logarithms.
A system of logarithms in which the base is e is called a
system of natural logarithms.
The common logarithm of a mmaber N is the exponent
of the power to which 10 must be raised to give N. If
X = common logarithm of N, then
logioiV = a; or N = IQ'. (1)
The natural logarithm of a number N is the exponent of
the power to which e must be raised to give N. U y = nat-
ural logarithm of N, then
loge N = y or N = e". (2)
Equating the values of N from equations (1) and (2),
e" = 10^ (3)
Taking the common logarithm of each member of (3),
y logio e = X logio 10 (4)
Therefore the natural logarithm of a number is equal to
the common logarithm of that number multiplied by 2.302.
For rough calculations of natural logarithms, a table of
common logarithms may be used and each logarithm mul-
tipUed by 2i
In all higher mathematics when no base is expressed, the
base e is understood.
80. The logarithmic curve, y = lege x. —
1st. Intercepts. — When x = 0, y = - oo ; when y = 0,
X = 1, therefore the curve cuts the a;-axis at the point (1, 0)
and has no finite intercept on the y-axis.
180 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
2nd. Symmetry. — As in the exponential equation, it can
be shown that the curve is not symmetrical with respect to
the a;-axis, y-axis, or origin.
3rd. Extent. — Solving for y in terms oi x, y = log« x,
from which it is seen that y is real for all positive values of
X, but that all negative values of x are excluded, since there
are no logarithms of negative numbers.
Solving for x in terms oi y, x — e", from which it is seen
that x is real for all values of y from — oo to +00 , also that
as y increases from to + 00 , a; increases from 1 to + 00 , and
as y decreases from to — 00 , a; decreases from 1 to 0.
4ih. Asymptotes. — In the equation y = loge x, it is seen
that a; = is the only finite value which makes y infinite,
while in the equation x = e", there is no finite value of' y
which makes x infinite, a; = is then the only asymptote
parallel to the axes.
Computing a table of values, and plotting the points, the
figure is as shown.
X
logioi
log,x
— 00
— 00
.5
- .301
- .693
1
2
.301
.693
3
.477
1.098
4
.602
1.386
10
1
2.302
100
2
4.604
-J- - - -
:::::::::::E:::::=
It will be observed that the logarithmic equation y = loge x
or X = e" is the same as the exponential equation y = e"" with
X and y interchanged. Hence the curve in this article might
have been constructed in a manner identical with that used
in Art. 78, except that values would have been assigned to y,
and X computed.
THE SINE CURVE 181
These curves are of very frequent occurrence in expressing
the laws of physics, and especially in problems of electro-
motive force.
EXERCISES
1. Construct the curves y — e^ and y = e~* with the same axes and
show that these curves are symmetrical with respect to the ^-axis.
2. Plot the locus oiy = loga x by changing to an exponential equa-
tion and assigning values to y to find the coordinates of points on the
curve.
3. Discuss and plot the loci of the following equations:
(a) 2/ = e^^.
(Jb) y = 3 er^.
{c)y = e '\
ie) y = 2 log. (x + 1).
X
(d) 2/ = 2 logic (a: + 1).
(g) y = e"\
(i) ^=loge(l-2a;).
(fc) y = e-^.
Qi) y = 2 logio (1 + 3 a;)
(/) y = log. (1 + e*).
{ni)y = h (e- + e-). * (n) y = | (j ^ /-).
(o) a; = He* + e""). (p) y = log. (1 - x^).
e' — 1
(g) X = log. (2/ + 2). (r) ^ = log. ^ ■^ -
81. The sine curve, y = sin*. —
ls<. Intercepts. — When x = 0, y = 0. When y = 0,
X = mn~^ = 0, TT, 2 IT, Sir, etc., — x, — 2 7r, — Stt, etc.
Therefore the curve intersects the y-axis at the origin only,
but intersects the a;-axis at an infinite number of points at
intervals of tt.
2nd. Symmetry. — When y is replaced by —y, the equation
is changed, therefore the curve is not sjnmnetrical with re-
spect to the X-axis. When x is replaced by —x, the equation
is changed, therefore the curve is not symmetrical with re-
spect to the 2/-axis. When x is replaced by —a; and y by —y,
* The locus of equation (n) is of frequent occurrence. It is called a
catenary and has the form assumed by a heavy flexible cord suspended
between two points.
182 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
the equation is unchanged, therefore the curve is symmetri-
cal with respect to the origin.
3rd. Extent. — Solving for y in terms oi x,y = sin x, from
which it is seen that y is real for every value of x from — oo to
+ 00 . Moreover, it is seen that as x increases from to § t,
y increases from to 1 ; as a; increases from ^ tt to ir, y de-
creases from 1 to 0; as a; increases from x to | x, 2/ decreases
from to —1; as x increases from ^ir to 2t, y increases
from — 1 to 0.
Since sin (x + 2 rnr) = sin x, where n is any integer, posi-
tive or negative, it follows that if the curve is plotted from
a; = to a; = 2 X, the remainder of the curve can be obtained
by moving the portion already plotted, right and left along
the a;-axis through successive multiples of 2 x.
Solving for x in terms oi y, x = sin~' y, from which it is
seen that x is real for aU values of y between — 1 and +1, but
that points whose ordinates are > 1 or < — 1 are excluded.
ith. Asymptotes. — Since no finite value of either variable
will make the other infinite, there are no asymptotes parallel
to the axes.
In plotting curves of this type, it is customary to measure
x in radian measure, using x = 3.1416 when laying off abscis-
sas on the a;-axis.
Plotting points in this way, the curve is found to be as
follows:
X
V
X
y
x = 3.14
1= -^2
.5
lir = 3.66
- .5
1 = 1.04
.86
f7r = 4.18
- .86
1 = 1.57
1.00
1^ = 4.71
-1.00
f7r = 2.09
It = 2.61
IT = 3.14
.86
.5
|ir = 5.23
Vx = 5.75
27r = 6.28
- .86
- .5
PERIODIC FUNCTIONS
Y
183
82. Periodic functions. — The sine curve, discussed in
the last article, is an illustration of a class called periodic
functions. They are marked by the characteristic that
when a definite constant is added to the variable, the function
is unchanged. In the case of the sine function just discussed,
this constant was 2 nx where n was any integer. The least
positive value of this constant is called the period of the
function. In this case the period is 2 x.
In plotting periodic functions, it is necessary to construct
only that portion of the curve belonging to one period. The
entire curve can then be sketched by use of the fact that its
values are repeated in each successive period.
83. The curve y = a sin fe«. — Since sin kx = sin {kx + 27r)
= sin fc (a; + -77 ), therefore the period is -r--
The values of sin kx vary between —1 and +1, hence the
values of y vary between —a and +a.
It is seen that the factor k divides the period and the factor
a multiphes the function.
Letting fc = 2 and a = 3, the following table is computed :
X
sin 2 1
V
X
sin 2x
V
^ir
TT
.707
2.121
-TT
- .707
-2.121
IT
1
3
^X
-1
"3
IT
707
2.121
'•gir
- .707
-2.121
IT
T
184 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
This table, together with a discussion similar to that used
in the sine curve, shows the locus to be as follows:
Exercises. — Plot y = sin^, y = sin x, y = sin 2 x, using
the same set of axes for the three curves.
Plot 2/ = I sin a;, y = sinx, y = 2 sin x, using the same
set of axes for the three curves.
84. The tangent curve, y = tan x. — Since tan x =
tan (ir + a;), therefore the period of this curve is tt.
Is*. Intercepts. — li x = 0, y = 0. li y = 0, x = 0, v,
2ir,Zir, etc., — ir, — 2 tt, —3 tt, etc. Therefore the ciu^e in-
tersects the j/-axis at the origin only, but intersects the x-
axis at an infinite number of points at intervals of it.
2nd. Symmetry. — When y is replaced by —y, the equation
is changed, therefore the curve is not symmetrical with respect
to the a;-axis. When x is replaced by —x, the equation is
changed, therefore the curve is not symmetrical with respect
to the 2/-axis. When x is replaced by —a; and yhy —y, the
equation is unchanged, therefore the curve is symmetrical
with respect to the origin.
3rd. Extent. — Solving for y in terms o{ x,y = tan x, from
which it is seen that y is real for every value of x from — oo to
THE TANGENT CURVE
185
+ 00 . Moreover, it is seen that as x increases from to ir/2,
y increases from to + oo and as x increases from ir/2 to ir, y
increases from — oo to 0. Since the period of this curve is t,
the values of the
function are re-
peated beyond this
point.
Solving for x in
terms of 2/, a; = tan~' y,
from which it is seen
that X is real for
all values of y from
— 00 to + 00 .
4ih. Asymptotes. — y = oo when x = ±ir/2, ±3 ir/2,
±5 ir/2, etc. These lines are asymptotes to the curve.
Y'
-X'
EXERCISES
1. Discuss and plot the loci of the following equations:
(o) y = cos X.
(6) y = cot X.
(c) y = sec X.
(d) y = tan -J-
, . . TX
(e) y = sin y
(/) y = tan^a;.
(g) y = sin" x.
(h) 2/2 = tan x.
(i) 2/ = cot -g
2. (o) Plot y = cosr
' a; or a; = cos y.
(6) In example (o), rotate the axes through — 7r/2 radians and show
that the equation becomes y = cos x. Plot the locus oi y = cos x
referred to the new axes and show that the same curve is obtained as
in (a).
3. Prove that the sine curve differs from the cosine curve only in
position by finding the equation oi y = sin x when the axes are trans-
lated to new origin (7r/2, 0).
4. Plot the locus oi y = esc x. Draw ordinates at a; = ir/8 and
a; = 3 7r/8, and thus compute the cosecant of ir/8 and 3 w/S. Check
results from a table of natural functions.
6. Find the equation oi y = sin (x — ir/4) when the axes are trans-
lated to new origin at (t/4, 0). Draw both sets of axes and the curve.
186 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
6. Find the period and the greatest value of the function in each of
the following equations. Plot each locus.
(6) 2/ = 4 cos 3 X.
(a) 2/ = 2 sin 2 x.
X
(d) 2/ = 4 sin
2"
(e) y = ooty
(c) 2/ = 3 tan 2 x.
(J) y = cos \x - Ij-
(i) 2/ = tan
('+9-
7.
(g) y = 5 cos -r- (/j) y = sec 2a;.
Discuss and plot the loci of the following equations:
(a) 2/ = tan-i a;. (6) 2/ = sec"' x. (c) 2/ = cot ( ^ + 1 ) *
(d) y = log sec x. (e) y = log esc a;. (f) y = xsiax.
85. Loci of parametric equations. — It is sometimes con-
venient to express the coordinates x and y in terms of a third
variable, thus a; = i', y = f.
The third variable in such cases is called the parameter and
the two equations are called the parametric equations of the
curve. It is often possible to eliminate the parameter and
thus derive the equation of the curve in rectangular coordi-
nates. Thus, in the above example, from the first equa-
tion t = '^. From the second equation, t = Vj/, whence
Vy = ■^ ov if = x^. It is not always possible, however,
to eliminate the parameter, and even when possible it is not
always desirable.
To plot the graphs of such equations, values are assigned
to the parameter and the corresponding values of x and y
computed. The points on the»eurve are then located from
the corresponding values of x and y. The parameter does
not appear in plotting the graph. Thus in the preceding
problem, if values are assigned to t, the following table of
values of x and y is obtained:
(
X
V
I
Z
y
1
1
1
-1
- 1
1
2
8
4
-2
- 8
4
3
27
9
-3
-27
9
LOCI OF PARAMETRIC EQUATIONS
187
The parameter usually represents time or some geometric
magnitude, but may be any quantity whatever.
It is often possible to obtain equations of curves by means
of a third variable which connects x and y, when it would be
extremely difficult to derive the equation directly, without
use of this variable. A nimiber of important equations wiU
be derived by this method in Art. 86.
ILLUSTRATIVE EXAMPLES
1. Plot the locus of
a; = a (0 — sin (^), y = al^ — cos ^).
«
Sin^
Cos<^
^
V
1
i= -^2
■5
.86
■.02 a
.14a
1=1.04
.86
.5
.180
.5a
^ = 1.57
1
.57
fir = 2.09
.86
- .5
1.23o
1.5a
fx = 2.61
.5
- .86
2.110
1.860
n- = 3.14
-1
3.140
2o
J7r = 3.66
- .5
- .86
4.160
1.860
|7r = 4.18
- .86
- .5
5.04O
1.5o
^T = 4.71
-1
5.710
Jx = 5.23
- .86
.5
6.09O
.60
V IT = 5.75
- .5
.86
6.25o
.14a
2x = 6.28
1
6.28 a
188 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
This is a periodic function of period 2 ir, since when the values of i^ in
the above table are increased by 2 v, x Ls increased by 2 wa, and y re-
mains unchanged. There are therefore an infinite number of arches
like the one shown below.
2. If a projectile is thrown with initial velocity V ft. per second,
at an angle <t> with the horizon, it is found that the equation of its path
is (resistance of air neglected), x = Vt cos <t>, y = Vtsin <^ — 16 P, where
t represents the time in seconds. If 7 = 640 ft. per sec. and (ji = 30°,
plot the locus traced by the projectile and show that the equation of
its path in rectangular coordinates represents a parabola.
The parametric equations are
X = 320 VSi.
2/ = 320 < - 16 <2 = 16 < (20 - 0-
(1)
(2)
Compute a table of values and locate points as in the preceding
problem.
(
X
V
6
2,768
1200
10
5,536
1600
15
8,304
1200
20
11,072
Since u = when t = 20, the greatest value of x is at the end of 20
seconds and is equal to 11,072 ft. This is called the range. The
greatest height is when t = 10, and is equal to 1600 ft.
LOCI OF PARAMETRIC EQUATIONS 189
In equation (1) above, t = ;= . Substituting. this value in (2),
320 V3
_ _s x^
^ ~ Vl ~ 19,200'
or {x - 3200 Vs)" = - 19,200 (2/ - 1600),
a parabola with vertex at (3200 Vs, I6OO), and passing through the
origin.
In finding the locus corresponding to a pair of parametric equations,
it is often convenient to eliminate the parameter before plotting.
EXERCISES
1. Plot the loci of the following equations by assigning values to t
and computing x and y:
(a) X = 4.P, y = 2t. (b) x = t - 1, y = tK
Ic) X = t - 1, y = 3fi - 6t. (d) X = 6f, y = 1 + St.
(e) 4 a; = <', 4 2/ = i^. (f) x = 5 cos t, y = Saiat.
{g) X = 2 sin t, y = 3 cos t. (h) x = sect, y = tan t.
(i) X = siat, y = sin2 t. (J) x'^ = t, y = log (t — 9).
(fc) a; = sin 0, 2/ = cos 2 <^. (Z) a; = 1 + cos <i>, ^ = 2 cos § 1^.
(to) a; = (k^, 2/ = a (1 — cos <^). (n) x = loge t, y = h P.
2. Eliminate the parameter in Ex. 1 (c), (d), (/), (g), (h), (fc), and
(/). Name the curve and reduce to a standard form in each case.
3. A projectile leaves a gun with a velocity of 800 feet per second,
the barrel of the gun being elevated at an angle tan~' J with the horizon-
tal. Find the equation of its path, using the equations in illustrative
example 2. Eliminate the parameter and show that the curve is a
parabola. Find the range and the highest point reached.
3 3
4. Prove that 2a; = 3< + 7, 3y = 3t — j, and a; = 3 sec 0, y =
t t
2 tan 6 represent the same curve. Plot the curve.
6. Plot the loci of the following equations:
(a) x = a {e + sin e), y = a (1 — cos 9).
lb) X = a (2 cos < — cos 2 t), y = a(2 sin t — Bin 2 t).
{c) X = a cos' t, y = a sin' t.
- 3a< _ 3aP
W a; - ^_^^3, 2/ - i+ja"
190 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
86. Derivation of parametric equations. — Many equa-
tions of great importance in tlieir applications, as well as
in historical interest, are most readily derived in the form of
parametric equations. In some cases this parameter can in
the end be easily eliminated, while in others the parametric
is the only practicable form of the result. A few illustrations
will show the value of the parameter in finding the equations
of curves often used in higher mathematics.
87. The Cycloid. — If a circle rolls along a straight
line, the locus traced by a fixed point in its circumference is
the cycloid.
:^r^
X
\
ii
«X J
V.
X
M
7
^
i
The equation is derived as follows:
Let the given straight fine be taken as the a;-axis, and let
P [x, y) represent any point on the rolling circle of radius a.
Take as origin the fixed point on the a;-axis from which P
started to move. From the a;-axis erect perpendiculars NC
to C, the center of the circle, and MP to P, any position of
the generating point. Let B represent the angle NCP.
Then
x = OM =^0N - MN = arc FN - PR = ad - asm 6,
y = MP = NC - RC = a - acosd.
Therefore, the parametric equations of the cycloid are
X = a{& — sine),
y = a (1 — cos 9).
THE EPICYCLOID
191
88. The epicycloid. — If a circle rolls upon a fixed circle,
a point on the circumference of the rolUng circle generates a
figure called the epicycloid.
Let a circle of radius b roll upon a fixed circle of radius a,
and let P, a point on the rolling circle, generate the epicycloid.
Take as origin the
center of the fixed cir-
cle and as a;-axis OA,
A being the point at
which P coincides with
the fixed circle. From
C the center of the
rolling circle and from
P {x, y), any position
of the generating point,
draw NC and MP per-
pendicular to OX, also
RP perpendicular to
NC. Let angle NOC
= 6 and OCP = 4>. It can be seen from the figure that
NCP = <t>- (90° -e) = <j> + e- 90°.
Therefore, sin NCP = -cos (0 + 6)
and cos NCP = sin (<^ + S).
But by hypothesis, arc AK = arc PK ot ad = 6<^, whence
ad
Therefore, sin NCP = - cos (y + (?) = - cos ^^y-^ 6
and cos NCP = sin (^ + e\ = sin ~^ 8.
Therefore,
x = OM = 0N + NM = OC cose + CPsinNCP
= (a + b) cos $ — b cos — r — 6,
192 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
y = MP = NC - RC = OC sine - CPcosNCP
= {a + b)sm6 — b sin
The equations of the epicycloid therefore are
a+b
a + b
X = {a + b) cos e — b cos
e,
y = (a + b) sin 6 — b sin — r — 0.
When a and b are equal the curve is called a cardioid.
Its equation is
X = 2a cos 9 — a cos 2 6,
y = 2 a sin 9 — a sin 2 8.
89. The h3rpocycloid. — If a circle rolls within a fixed
circle, a point on the
circumference of the
rolUng circle generates
a figure called the hy-
pocycloid.
^ From the adjoining
figure, the equation can
be derived in a manner
similar to that used in
the preceding article, or
the result may be ob-
tained from the equa-
tions of the epicycloid by substituting —b for b. In either
case the results are
X = {a — b) cos 9 + b cos [—^ — ) 9,
y = (a — b) sin 9 — b sin ( — j- — j 9.
THE PATH OF A PROJECTILE
193
90. Hypocycloid of four cusps. — A particular case of
the hypocycloid in which 6 = J a is of frequent occurrence.
Making this substitution in the equation just obtained,
X = -;
y
T a cos + T cos 3 6,
4 4 '
T a sin 5 — 7 sin 3 5.
4 4
Changing sin 3 B and cos 3 to terms of 6, these become
x = ja cos 5 + 2 (4 cos' — 3 cos 6) = a cos' 6,
1/ = J a sin — J (3 sin fl — 4 sin' 6) = a sin' fl.
From the foregoing,
Adding, the result is
x^ = a^ cos^ 9,
y^ = a^ sin^ d.
X' + y^ = a'.
91. The path of a piojectile. — The path traced by a
body which is projected at a given angle and with a given
velocity is a curve of much im-
portance.
Let 0, the initial position of
the projectile, be taken as the
origin and take OX and OF in
horizontal and vertical posi-
tions, respectively. Let Vo de-
note the initial velocity, <^ the
angle which the initial direction
of the projectile makes with the horizontal, and t the time.
If there were no force acting other than that which origi-
nally projected the body, the path would be the straight line
OR, the distance OR being Vot.
194 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
The principal force tending to deflect the body from a
straight hne is the action of gravity which tends to pull it
vertically downward. Let P represent the position of the
body after t seconds, gravity alone being considered.
The ordinate of the point at which the body is found after
t seconds is MP instead of MR, the difference being PR which
is proved in mechanics to be 16 f. Then
X = OM = OR cos (j> = Vot cos <^,
y = MP = MR - PR = Fo< sin <^ - 16 <«.
■MR
Eliminating t, the curve is found to be the parabola
16
y = X tan <|> —
Fo'^ cos'' <|)
In the preceding problem, no account has been taken of the
resistance of the air or of other forces of which careful account
is taken in figuring the paths of projectiles in mihtary prac-
tice.
Exercise. — Find the path of a body which is projected
with an initial velocity of 300 feet per second, in a direction
incHned 45° with the horizontal.
92. The witch of Agnesi. — A circle of radius a hes be-
tween two paral-
lel tangents OX
and RS. A chord
is drawn through
the point of tan-
gency to the
other tangent,
meeting the circle
at K and the tan-
gent at Q.
Perpendiculars are dropped from Q to OX and from K to
OX and MQ. The locus of the intersection P is the witch.
THE CISSOID
195
Let e = the angle XOQ. Then
X = OM = RQ = 2acote. •
y = MP = NK = OKsine = 2asiii^e
(since the angle OKR is inscribed in a semi-
circle).
Eliminating B, y =
8 a'
a;^ + 4 a^
93. The cissoid. — A circle of radius a passes through
the origin and has its center
on the line OA which is taken
as the axis of x. A chord OR
meets the tangent AN in the
point N. If the point P on OR
is so chosen that OP = RN,
the point will describe the cis-
soid. Let d = the angle AON
and draw MP and QR per-
pendicular to OX, and RS
perpendicular to AN. Then
x = OM = QA = OA
y = MP = SN = AN
OQ = 2 a - 2 a cos^ 6 = 2 asin^d.
AS = 2a tan 9 — 2asmd cos 6
= 2asmd (sec 6 — cos 8).
The parametric equations of the cissoid are
cos 6).
X = 2asin''e,
y = 2 a sin 6 (sec 9
Solving the first for sin 6, sin 6
Substituting in the second, j/' =
V 2a
7?
2a — X
Exercise. — Extend the vertical diameter CD to E mak-
ing CE = 2 CD, and join E to A, Let F be the point where
196 TRANSCENDENTAL AND PARAMETRIC EQUATIONS
AE cuts the curve, and draw the ordinate GF. By means of
the equation of the curve and similar triangles, prove
GF' = 2 0^.
This fact makes it possible to duphcate a given cube. For
if ai represents the side of a given cube, and a fourth propor-
tional 02 is found to OG, GF and Oi, then oa will be the side of
the cube of double volume.
The cissoid was invented by a Greek mathematician named
Diodes about 150 B.C. His purpose was to solve the famous,
problem of dupUcating the cube.
EXEKCISES
1. Find the parametric equation of the
circle with radius a and center at origin, in
terms of 8, the angle between the a;-axis
and the radius to any point P.
2. A radius is drawn from the center
O of two concentric circles, cutting the
inner circle at Q and the outer circle at B.
Perpendiculars are dropped from B to OX
and from Q to OY. Find the equation of
the locus traced by the intersection P in terms of the angle 9 = XOR.
Prove this locus is an eUipse.
Hint. — Draw the perpendicular NQ.
CHAPTER XIV
A'
SOLID ANALYTIC GEOMETRY
94. Rectangular coordinates in space. — If a point is
located in a plane, its distances from two fixed perpen-
dicular lines in the plane are determined. If a point is
located in space, its distances from three perpendicular
planes are determined. In each case these distances are
called the coordinates of z
the point.
Construct three mutu-
ally perpendicular planes,
intersecting in the three
perpendicular Unes XX',
YY', and ZZ'.
The three perpendicu-
lar planes are caUed the
coordinate planes ; the
lines of intersection, the
coordinate axes; and ^
the common point of intersection of the coordinate planes,
the origin.
Let P be any point in space. Through P draw planes
parallel to the coordinate planes forming with them the
rectangular parallelopiped OLMN — P. The edges, ON
= X, NM = y, MP = z, are the rectangular coordinates
of P. These edges measure the distances of P from the
yz-, XZ-, and xy-plsmes, respectively. It is often convenient
in locating a point to draw only the lines ON, NM, and
MP, taking x = ON on the x-axis, y = NM parallel to
197
198 SOLID ANALYTIC GEOMETRY
the 2/-axis and z = MP parallel to the 2-axis. Coordinates
measured to the right, forward, and upward will be con-
sidered positive and those measured to the left, backward,
and downward, negative. The eight equal parts into
which space is divided by the coordinate planes are called
octants. That octant in which all coordinates are positive
is called the first octant. There is no estabhshed order in
numbering the other octants.
The following suggestions are helpful in constructing
figures on cross-section paper. Draw the x- and z-axes at
right angles to each other and lay off units as indicated on
such paper. Draw the j/-axis making equal angles with the
other two, and lay off units equal to one-half the diagonal of
a square whose side is the unit on the x- and z-axes. This
foreshortening of y-iuiits aids in giving the figure the appear-
ance of a soUd.
EXERCISES
1. Plot the points (0, 1, 2), (2, 3, 4), (-1, 4, -3), (1, 0, -5), and
(2, -2, -3).
2. Where are the points for which a; = 0? v = 0? z = 0? What
are the equations of the coordinate planes? Where are the points for
which both x and y = 01
3. Where are the points for which a; = — 17 j/ = 2? z = o?
4. Write the coordinates of the points sjmimetrical to the following
points with respect to each of the axes and with respect to the origin:
(o) (-1,2,3). (6) (a,fc,c). (c) (-1,0,6).
6. Find the coordinates of the feet of the perpendiculars drawn from
the point (2, —1, 3) to each of the coSrdinate planes.
6. From a point (xi, yi, Zi) perpendiculars are drawn to each of the
coordinate planes. Find the feet of these perpendiculars.
95. Distance between two points. — The distance between
two points Pi {xi, j/i, Zi) and P^ (a^, y^, z^ is given by the
formula
d = V(*i - *a)8 + (yi - y^r + («i - ^s)"- (40)
DISTANCE BETWEEN TWO POINTS
199
Proof. — Let Pi and Pi represent any two given points,
and let d rep- s
resent the dis-
tance between
them. Through
Pi and Pi pass
planes parallel
to the coordi-
nate planes
forming the
rectangular , o
parallelepiped
QS of which the
required dis-
tance P1P2 is
the diagonal. ^
Since P1QP2 is a right triangle,
P1P2 = ^M' + QP^,
also since P2QR is a right triangle.
PiQ =PiR +RQ = NiN, + LM{.
Substituting this value of P2Q ,
P1P2 = ^WWl + LMt + QPi'-
In terms of the coordinates of Pi and P2,
P1P2 = rf = V(a;i - Xif + (2/1 - ViY + (zi - ZiY-
As a particular case, let p equal the distance from the origin
to any point P (x, y, 2), then p = Vx^ + y^ + z^.
EXERCISES
1. Prove that the points (2, 1, 2), (6, -1, -3), (-2, 3, 7) are the
vertices of an isosceles triangle.
200
SOLID ANALYTIC GEOMETRY
2. Prove that the points (5, 1, 5), (0, -4, 3), (7, -2, 0), and
( — 3, 3, 5) lie on a sphere whose center is (1, 2, —1).
3. Prove that the points (1, 2, 3), (-1, -2, 1), and (3, 6, 5) are on
the same straight line.
4. Prove that the points (6, 7, 3), (3, 11, 1), and (0, 3, 4) are the ver-
tices of a right triangle.
96. Point of Division. — // the point Ps divides the line
joining the two points Pi {xi, yi, Zi) and Pi {x2, 2/2, 22) into
segments such that the ratio p „ = —, the coordinates are given
jTifi ri
by the formulas
ri -\- r2 ri -f- ra rj, + rg
Proof. — Let Pi and P2 be the given points and let P3 be
the point which divides the line joining them in the ratio
n : ri. Draw the lines MiPi, M2P2, and ilf sPs perpendicular
to the xy-plane. By plane geometry,
M1M3 _ P1P3 ^ n
MzMi P3P2 ri
The hne M\Mi lies in the aiy-plane and the x and y coordi-
nates of Mi which are also the x and y coordinates of P3 are
found, as in Art. 8, to be
rxXj -\- rjXi _ rjyi + r^yi
ri + ri ' ^' n + ri
Xs =
ORTHOGONAL PROJECTIONS 201
By dropping perpendiculars from Pi, P2, and P3 on either
of the other coordinate planes, the z-co6rdinate of P3 may be
found.
EXERCISES
1. Find the coordinates of the point which divides the line joining
(-1, 4, 3) and (-5, -8, 7) in ratio 1 : 3.
2. One extremity of a Une is at ( — 3, 2, 7) and the middle point is
( — 1, 4, 2). What are the coordinates of the other extremity?
3. In what ratio does the point (2, 3, 4) divide the Une joining
(-1, 4, 5) and (8, 1, 2)?
4. Find the lengths of the medians of the triangle whose vertices are
(2, 5,6), (3, -7, 4), and (-1, 1, 2).
5. In what ratio is the Une joining (5, —1, 4) and (2, —4, —2)
divided by the xy-pla,ne1 Find the coordinates of the point of inter-
section with this plane.
6. The Une joining A (1, 2, 2) and B ( — 1, 3, 1) is produced to C so
that BC = 3 AB. Find the coordinates of C.
7. Two vertices of a triangle are (2, 3, 0) and (—2, —3, 4) and the
center of gravity is (0, 2, f). Find the third vertex.
8. Prove that the lines joining the middle points of the opposite
edges of the tetrahedron whose vertices are (0, 0, 0), (a, 0, 0), (6, c, 0),
and (d, e, f) meet in a point.
97. Orthogonal projections. — If through a point, a
plane is passed perpendicular to a given line in space, the
point in which the line pierces the plane is called the
projection of the point on the line.
If through the extremities of a directed segment of a
line, planes are drawn perpendicular to a given line in
space, the portion of this line measured from the projection
of the initial point of the segment to the projection of its
terminal point is called the projection of the segment on
the line.
Thus, in the figure of Art. 95, the projection of P2P1 on the
a;-axis is NJ^i.
The angle between two lines which dp not intersect is the
202
SOLID ANALYTIC GEOMETRY
angle between two intersecting lines respectively parallel to
the two given lines.
Thus, in the figure of Art. 95, the angle between the line
P1P2 and the axis OX is the angle RP2P1.
Theorem I. — The projection of the segment of a directed
line upon another line in space is the product of the length of the
segment by the cosine of the angle between the two lines.
Proof. — Let
P1P2 be a directed
segment making
an angle 6 with
AB, any other Une
in space. The
planes KL and RS
through Pi and P2
perpendicular to
AB determine the projection MiM^. It is desired to prove
M1M2 = P1P2 cos e.
Draw PiQ parallel to AB, piercing the plane RS in N and
join NP2. By the definition above, the angle QP1P2 = 6.
The triangle P1NP2 is right angled at N and hence, by trigo-
nometry,
M1M2 = PiN = P1P2 cos e.
Theorem II. —
The sum of the
projections on any
straight line, of the
segments of the
broken line joining
the point A to the
point B, is equal to
the projection of the segment AB on that line.
Given the broken line AEDCB and the straight fine AB
POLAR COORDINATES
203
joining the point A to the point B. Let the projections of
the points A, B, C, D, and E on RS be A', B', C, D', and E'.
It is evident that A'E' + E'D' + D'C + C'B' = A'B',
that is, the sum of the projections on RS of the segments of
the broken line AEDCB is equal to the projection of the
straight line AB.
98. Polar coordinates. — Using the same coordinate axes
and origin as in the rectangular system, the line OP from
the origin to the point P is called the radius vector and is
represented by p. The
angles which OP makes
with the axes of x, y,
and z are called the
direction angles of the
line OP and are repre-
sented by a, 0, and y,
respectively, p, a, /3,
and y are called the
polar coordinates of the point P. The cosines of these
angles are called the direction cosines of the line OP
The direction cosines of a Une are not independent but are
connected by a very important relation, viz.:
The sum of the squares of the diredion cosines of a line is
unity, or
x~
cos'' a + cos^ p + cos** 7 = 1.
Proof. — From theorem I, Art. 97, it is evident that
X = p cos a,
y = peos /3,
z = p cos y.
Also by Art. 95, x^ + y^ + z^ = p\
Squaring and adding,
p'co&^a-'f p^cos^/S + p-'cos'y = x^ + y^ + z^ = pS
(42)
whence
cos" a + cos" /3 + cos" 7 = 1.
204 SOLID ANALYTIC GEOMETRY
The direction cosines of the line joining the points Pi (xi, 2/1)
and Pi {Xi, 2/2) o,re given by the eqiiations:
cos a = ^3^, cos p = 31^, cos 7 = ^i^, (43)
in which d is the length P1P2.
This is evident from the figure of Art. 95. Thus,
PJi N2N1 Xi — X2
cos a = cos RP2P1 =
P2P1 P2P1
ILLUSTRATIVE EXAMPLES
Find the direction cosines of a line if they are proportional to 1, —2,
and 3.
T^ . . ,, ^ cos a cosjS cos 7
It IS given that — z — = -^^ = — ^ —
Then
cos' a + cos' |8 + cos' 7 _ cos' a _ cos'g _ cos' 7
l' + (-2)' + 3' 1 ~ 4 ~ 9 ■
But the nvunerator of the &st fraction equals 1, hence
cos* a = T-T,
14
cos a = ± —p=,
Vl4
2
'°'^ = ^vil'
and «°^'>' = ±VIi-
EXERCISES
1. What are the projections of the point (3, 1, —6) on each of the
axes?
2. A line makes an angle of 60° with the x-axis and of 30° with the
j/-axis. What angle does it make with the z-axis?
3. The direction cosines of a Une are equal. Find their values.
4. The direction cosines of a Une are proportional to 3, —1, and 2.
Find their values.
THE ANGLE BETWEEN TWO DIRECTED LINES 205
6. Find the direction cosines of the line joining (6, 3, —1) and
(—2, —1, 0), and the projection of the hne upon each of the axes.
6. What are the direction cosines of a line parallel to the a>axis7
to the 2/-axis? to the z-axis? of a line perpendicular to the x-axis?
7. Abrokenhne joins (3, 1, -2), (3 ,4, 6), (-1, 2, 3), and (2, -5, -7).
Find the projections on the a;-axis of the closing line and of each of the
segments. Verify theorem II, Art. 97.
8. Find the polar and rectangular coordinates of a point, given
P = 4, a = 120°, and /3 = 135°. How many solutions?
9. Where do all the hnes Ue which
(a) make an angle of 45° with the j/-axis? with the z-axis?
(b) make an angle of 45° with both the y-axis and the z-axis? Is
there any line making an angle of 45° with each of the coor-
dinate axes?
10. What are the direction cosines of a line if a = ^ = 90°? Where
are all the points for which cos -y = 0?
11. In which octant is a point found when
(a) cosa > 0, cos|8 < 0, COST > 0?
(6) cos a < 0, cos ^ < 0, cos 7 < 0?
(c) cos a > 0, cos jS > 0, cos 7 < 0?
Name the octant by indicating the signs of the axes.
12. If the projections of P1P2 on the axes are respectively 4, 2, and
— 4 and the coordinates of Pi are (6, —3, 2), find the coordinates of Pa.
13. Given p = 30°, 2/ = 2, z = — 3. Find the coordinates of the
point in both rectangular and polar coordinates. How many solutions?
14. Prove by means of direction cosines that the points (1, 2, 3),
(—1, —2, 1), and (3, 6, 5) are on the same straight line.
99. The angle between ^
two directed lines. — The
cosine of the angle between
two directed lines is equal to
the sum of the products of
the corresponding diredion
cosines.
Let ai, Pu 7i and a^, ^i,
72 represent the direction
angles of two given lines
206 SOLID ANALYTIC GEOMETRY
and let OPi and OP2 be two lines through the origin parallel
to those lines. Also let d equal the angle P1OP2. Construct
ON, NM, and MPi, the coordinates of Pi.
If the broken line ONMPi and the straight Hne OPi are
projected on OPi, from theorem II, Art. 97,
Proj. OPi = Proj. ON + Proj. NM + Proj. MPi,
or by theorem I, Art. 97,
OPi cos d = 0N cos a2 + NM cos jSz + MPi cos 72.
But ON = OPi cos ai, NM = OPi cos ;3i,
MPi = OPi cos 71.
Substituting and dividing by OPi,
cos 9 = cos ai cos 02 + cos Pi cos % + cos yi cos 73. (44)
It is evident that if two lines are parallel and extend in the
same direction, on = a^, /3i = 182, 71 = 72; if parallel and ex-
tending in opposite directions, ai = w — an, j8i = tt — (82,
7l = IT — 72.
If two lines are perpendicular, cos d = and, therefore,
cos ai cos as + cos |8i cos 182 + cos 7i cos 72 = 0.
EXERCISES
1. Find the angle between two lines whose direction cosines are
f, — f, f and f, I, f, respectively?
2. Find the angle between two lines whose direction cosines are pro-
portioiial to 1, 2, 5 and —1, 3, —2, respectively.
3. Prove that the lines whose direction cosines are i, i, f ; — |, — f ,
f ; and f, — f, — i are mutually perpendicular.
4. Find the direction cosines of the line joining ( — 1, 2, 4) and
(6, 5, -3).
6. Find the length of the projection of the line joining (1, 1, 2) and
(2, —1, 4) upon the line joining (2, 1, —2) and (4, —5, 1).
100. The equation of a locus. — If a point moves in
space according to some law, it traces a locus which will, in
CYLINDRICAL SURFACE 207
general, be a surface. To find the equation of this locus, the
same steps will be followed as in plane analytic geometry,
viz., the discovery of some law which applies to the moving
point in all of its positions and the translation of this law
into an algebraic equation between the coordinates of the
point.
Thus, to find the equation of a sphere of radius 5 and center
at the origin, it is seen that if P {x, y, z) represents any point
on the surface of the sphere, OP = 5. Whence x' + y^ + ^
= 25 is the equation of the surface of the sphere.
Again, the equation of a plane parallel to the j/z-plane and
three units to the right of it is a; = 3, since every point in the
given plane is at a distance 3 from the j/z-plane.
EXERCISES
1. Find the equation of the plane which is
(a) parallel to the yz-plane and 4 units to the left of it.
(6) parallel to the xz-plane and 3 units in front of it.
2. Find the equation of the locus of a point which is equidistant from
the points (1, 0, -2) and (2, -3, 0).
3. Find the equation of the locus of a point which is equidistant from
the xy- and j/z-planes.
4. Find the equation of the locus of a point
(a) whose distance from the x-axis is equal to 5.
(6) whose distance from the y-a,ids is equal to its distance from the
xz-plane.
(c) whose distance from the a;-axis is equal to its distance from the
z-axis.
6. Find the equation of the locus of a point whose distance from the
point (3, 1, 0) is equal to its distance from the 2/-axis.
101. Cylindrical surface with elements parallel to one
of the coordinate axes. — The method of finding the equation
of such a surface is illustrated by the following example:
Find the equation of the cylindrical surface whose directing
208
SOLID ANALYTIC GEOMETRY
curve in the xy-Tplane is a;^ — 4 a; + 4 j/'' = 0, and whose axis
is parallel to the z-axis.
Let P (x, y, z) be any point in this surface. The x and y
coordinates of P on the surface are the same as those of M
on the directing curve, and therefore the coordinates of P
satisfy the equation x^ — Ax
-)- 4 2^2 _ 0. (Since z does
not appear in this equation, it
may have any value.) This
equation therefore is the
equation of the surface of the
cylinder.
In general, the equation af a
cylindrical surface whose axis
-X is parallel to one of the axes is
the same as the equation of the
generating curve in the plane of
the other two axes.
102. Spherical surface. — The equation of a sphere whose
center is at C {h, k, I) and whose radius is r is
(x - hy +iy- fe)'» +{z- i)" = r8. (45)
This equation results immediately from Art. 95.
EXERCISES
1. Describe the following surfaces: (a) a;^ — z" = 4. (b) x!' + ifi =
4a;. (c) y = cosx. {d)4:^ + ^ + 8y = 0.
2. Find center and radius of each of the following spheres:
(a) 3?-2x + ^-6y+^ + 2z-5 = 0.
(6) 4x2 + 42/i' + 4z2-4a;H-122/-20z= 1.
3. Find the equation of the sphere whose center is on the z-axis,
whose radius is 7, and which passes through the point (2, —3, 4).
4. The axis of a cylinder is parallel to the a>-axis and its directing
curve is a circle in the j/a-plane with radius 5, with center on the «-axis
aiid tangent to the y-asda. Find the equation of the cylinder.
A SURFACE OF REVOLUTION
209
6. Find the equation of the sphere whose diameter is the line joining
(2, 4, -3) and (2, -2, 1).
6. Find the equation of the sphere whose center is at ( — 1, 3, —5)
and which passes through the point ( — 3, 6, 1).
7. Find the equation of a sphere through the four points (0, 0, 0),
(-3, 0, 3), (0, 3, 11), and (0, -8, 0).
8. Find the equation of the sphere of radius 7 whose center is in
the 2/2-plane and which passes through the points (—2, —3, 3) and
(-3,6, -1).
9. Find the equation of the sphere with center at the origin and
which is tangent to the sphere x' — 12 x + i/^ + 4y + z^ — 6z + 24: = 0.
10. Find the equation of the sphere concentric with x^ — 2x -{■ y^ +
Qy + tfi — 8z + 1=0 and passing through the point (5, 1, —3).
11. Prove that if a point moves so that the sum of the squares of its
distances from (0, 0, 1) and ( — 1, 1, 0) is 7, its locus is a sphere. Find
its center and radius.
12. Find the equation of the locus of a point which moves so that its
distance from (—3, —6, 3) is twice its distance from the origin. Prove
that the locus is a sphere and find its center and' radius.
13. Find the equation of the locus of a point which moves so that its
distance from the s-axis is equal to its distance from the point (1, 0, 2).
Describe and construct the surface.
103. A surface of revolution is formed by revolving a
plane curve about an axis in its plane. If the equation of a
curve in one of
the coordinate
planes is given,
and if the axis
of revolution is
one of the coor-
dinate axes, the
equation of the
surface is readily-
found.
Thus, find the
equation of the surface formed by revolving the ellipse
ix^ + 9y^ = 36 about OX.
~X.
210
SOLID ANALYTIC GEOMETRY
Let the ellipse ABA'B' be revolved about OX. In order
to avoid confusion of the coordinates of any point on the
surface with the coordinates of the points on the generating
curve, let P {x', y', z') represent any point on the surface.
Then x' = ON, y' = NM, z' = MP. Pass a plane through
P perpendicular to OX. This section PKRS is evidently a
circle. In the triangle NMP,
NM" + MP" = nP or y'^ + s'^ = NP" = Wt = y\
It is now required to express y^ in terms of the coordinates
of P. From the equation of the generating eUipse,
2/2 = I (9 - x2) = 1(9 - a;'").
Hence y"" + s'^ = | (9 - x'")
or, dropping primes and simphfying,
y'
+
i+9
1.
Again, find the equation of
the conical surface formed by
revolving the line z = 2x about
OZ.
Let P {x'j y', z') be any point
on the surface of the cone.
Then x' = ON, y' = NM, z' =
MP- Pass planes DEFG and
■X MNRP through P perpendicu-
lar to OZ and OX respectively.
Then CR = ON = x', RP =
NM = ^'.
But CR' + RP' = CP% or x"" ^ y"" = cf =C^ = x\
From the equation of the generating line, x = i z = ^ z';
hence ^,2 + y'^ = i z'^
or, dropping primes and simplifying,
4a;=' + 42/2 = z\
EQUATIONS OP A CURVE 211
EXERCISES
1. Find the equations of the surfaces of revolution generated by re-
volving
(a) y = X about the a;-axis.
(b) 4 a;2 -(. 2^ = 16 about the y-aToa.
(c) X? = iz about the z-axis.
{d) a;2 _ 22 = 4 about the z-axis.
(e) y — X = 1 about the 2/-axis.
2. Find the equation of the surface generated by revolving the ellipse
3? /a? + 2/2/62 = 1
(o) about its major axis. This surface is called a prolate spheroid.
(6) about its minor axis. ' This surface is called an ohlate spheroid.
3. Find the equation of the surface generated by revolving about the
K-axis, the line z = 4 in the xz-plane.
4. Show that the surface obtained by revolving the parabola ^ = 4 a;
about the i-axis is the same as that obtained by revolving the parabola
g2 = 4 a; about the x-axis.
5. Find the equation of the surface generated by revolving the circle
s' + 2/' = o2 around the j/-axis. What curve in the xz-plane would
generate the same surface when revolved about the x-axis?
6. A circle in the xz-plane of radius 4, with center on the z-axis at a
distance 7 from the origin, is revolved about the z-axis. Find the
equation of the surface generated.
104. Equations of a curve. — Two surfaces intersect in
a curve. The equations of two surfaces when considered
simultaneously define the curve of intersection, since the
coordinates of any point on this curve of intersection satisfy
both equations.
If the equations of the two surfaces are combined so as to
obtain a third equation, this equation represents another
surface through the curve of intersection, and this together
with either of the given equations defines the curve of inter-
section.,- It is thus seen that the same curve may be repre-
sented in an infinite number of ways. In particular, if one
of the variables is eliminated between the equations of the
surfaces which define a curve, the resulting equation, con-
212 SOLID ANALYTIC GEOMETEY
taining but two variables, is a cyKnder with elements parallel
to one of the axes, and passing through the given Curve. This
is called a projecting cylinder.
106. The locus of an equation. — It is evident from
Art. 100 that every equation in one variable represents a plane
or series of planes parallel to one of the coordinate planes.
Also from Art. 101, it is evident that every equation in two
variables represents a cylindrical surface, the equaiion of whose
directing curve in one of the coordinate planes is the same as the
given equation and whose elements are perpendicular to the plane
of this curve.
In determining the loci of most other equations, a discus-
sion somewhat similar to that used in plane analytic geometry
is helpful. The principal points in such a discussion are:
1st. Symmetry.
2nd. Intercepts on the coordinate axes.
3rd. Intersections on the coordinate planes.
Ath. Intersections on planes parallel to the coordinate
planes.
Symmetry. — A locus is symmetrical with respect to
(a) one of the coordinate planes, if the variable corre-
sponding to the axis perpendicular to that plane
can be changed in sign without changing the
equation.
(b) one of the coordinate axes, if the variables correspond-
ing to the other two axes can be changed in sign
without changing the equation.
(c) the origin, if all three variables can be changed in
sign without changing the equation.
The proof is similar to that in Art. 13.
Intercepts on the coordinate axes. — These are found
by setting two of the variables equal to zero and solving for
the third.
THE LOCUS OF AN EQUATION
213
Intersections of a surface with the coordinate planes. —
These intersections are found by treating the eiquations of the
coordinate planes x = 0, y = 0, and 2 = simultaneously
with the equation of the given surface. These curves are
called the traces.
Intersections of a surface by planes parallel to the
coordinate planes. — Represent these planes by x = k,
y = ki, z = ICi. These taken simultaneously with the equa-
tion of the given surface determine the curves of intersection.
By giving k, ki, ki different values, the general form and
limits of the surface are determined.
Frequently it is sufficient to discuss the set of planes parallel
to but one coordinate plane.
□.LUSTRATIVE EXAMPLE
Discuss and construct the locus oi x' — y' — ^ = i.
1st. This surface is evidently symmetrical with respect to the coordi-
nate planes, the coordinate axes and the origin.
2nd . Intercepts
on a>axis are ±2.
There is no inter-
cept on y-ajds or
s-axis.
3rd. Let x = 0,
then s/i! + g2 = -4.
Therefore the sur-
face does not inter-
sect the 2/g-plane.
Let y=0, then x' —
s^ = 4. Therefore,
the trace is an hy-
perbola. QARQ' A' B' 0- Y
in the a;2-plane. Let z = 0, then x' — y'^ = 4.
h3T)erbola BACB'A'C in the xy-plsme.
ith. To find the intersection of the plane x = k with the surface,
substitute a; = fc in the equation of the surface. The result is 'f + ^ =
]fi _4j which is a cylinder whose trace in the 2/2-plane is the circle
Therefore, the trace is an
214
SOLID ANALYTIC GEOMETRY
J/' + z* = A;^ — 4 and whose elements are parallel to the axis of x. The
curve of intersection of the surface by the plane is the same as the curve
of interse ction of the cylinder by the plane, that is, a circle of radius
Vk^ — 4. If — 2 < /k < 2, the radius is imaginary and there is no in-
tersection and if fc > 2 or < — 2, the intersection is a circle whose radius
increases without limi t as k increases without limit.
Without considering planes parallel to the other coordinate planes,
the surface can be sketched as above.
106. Quadric surfaces. — The general equation of second
degree in three variables is Ax^ + By^ + Cz^ + Dxy + Eyz
+ Fxz -\- Gx + Hy + Kz + L = 0. The surface repre-
sented by this equation is called a quadric surface. Some
special forms of this equation which are of frequent occurrence
will be discussed here.
107.
Theempsoidf, + ^ + ^=l.
,,. X' , y" . X' z^
ellipses -, + p=l,^ + -.
1st. This surface is
symmetrical with respect
to the origin, the coor-
dinate planes,- and the
coordinate axes.
2nd. The intercepts on
the axes are x = ±0,
y = ±6, z = ±c.
Srd. The traces in the
coordinate planes are the
l,andg + 5=l.
^th. The intersection with the plane a; = fc is the ellipse
or
X = k,
• = 1, X = k.
^ {a? - ¥) -„ (o2
k^)
THE HYPERBOLOID OF ONE SHEET
215
It is seen that as k increases numerically, the semi-axes
- Va^ — fc'' and - Va^ — k^ decrease from b and c respectively
when fc = to when k = ±o. If fc > a or < —a the
ellipse is imaginary. The ellipsoid then Ues between the
planes x = ±a.
A similar discussion shows that the sections made by planes
parallel to the other coordinate planes are also ellipses, and
that the figure lies between the planes z = ±c and j/ = ±6.
108. The hyperboloid of one sheet
1st. The surface is symmetrical with respect to the origin,
the coordinate planes,
and the coordinate axes.
2nd. The intercepts
on the axes are x= ±a,
y = zhb. There is no
intercept on the z-axis.
3rd. The traces in the
coordinate planes are
the ellipse i + ^ = 1,
the hyperbola -5 5
u C
= 1 and the hyperbola j^ — ^ = 1.
4:th. The intersection with the plane 2 = fc is the ellipse
+
r
= 1, X = k.
^, {(? + ¥) ^(c^ + fc^)
It is seen that as fc increases numerically from to x , the
216
SOLID ANALYTIC GEOMETRY
semi-axes of the ellipse increase without limit, and therefore
the smf ace extends indefinitely in the direction of the z-axis.
The intersections in planes parallel to the other coordinate
planes are hyperbolas.
109. The hyperboloid of two sheets
1st. The smface is symmetrical with respect to the origin,
the coordinate planes, and the coordinate axes.
2nd. The intercepts on the x-axis are ±o. There are no
intercepts on the other axes.
3rd. The traces in the coordinate planes are the hyperbolas
-t; — rj = 1 and -x ; = 1
a'=
c"
There is no trace in the
a' ¥
2/z-plane.
ith. The intersection with the plane x = kis the ellipse
p r^ ~ 1) X = k.
If —a<k<a, the ellipse is imaginary. If k increases
numerically from a to oo , the semi-axes increase indefinitely.
THE ELLIPTIC PARABOLOID
217
The intersections in planes parallel to the other coordinate
planes are hyperbolas.
110. The elliptic paraboloid — + j^ = cz.
1st. The surface is symmetrical with respect to the yz-plsme
and to the a;0-plane, but not to the xy-Tpla.ne. It is sym-
metrical with respect to the z-axis only. It is not sym-
metrical with respect
to the origin.
2nd. The surface
intersects the three
axes at the origin
only.
3rd. The traces in
the coordinate planes
are the point ellipse
T
ifi
-„ + |j = 0, and the
parabolas — ^ = cz and
P = '^'-
4th. The intersection with the plane z = k (where k has
the same sign as c) is the ellipse
y
a%c ' h%c
= 1, z = k.
As h increases from to oo , the semi-axes increase from
to 00 and therefore the surface extends indefinitely in the
direction of the z-axis, lying entirely above the a;2/-plane when
c is positive and entirely below if c is negative.
The intersections in the planes parallel to the other coordi-
nate planes x = ky and j/ = A^ are parabolas whose vertices
recede from the xiz-plane as fci and kz increase in numerical
value.
218
SOLID ANALYTIC GEOMETRY
111. The hyperbolic paraboloid
y -
ca.
or b"
The following discussion considers c positive.
1st. The surface is symmetrical with respect to the yz-
plane and to the xz-plane, but not to the xy-pla,ne. It is
symmetrical with respect to the 2-axis only. It is not sym-
metrical with re-
spect to the ori-
gin.
2nd. The sur-
face intersects
the three axes at
the origin only.
3rd. The traces
in the coordi-
nate planes are
the intersecting
straight lines x^/a^ — y^jh"^ = and the two parabolas x^lo?
= cz and j/^/6^ = — cz.
Mh. The intersection in the plane z = k'\s the hyperbola
J^ t_^->
d'ch ¥ck '
z = k.
If k is positive, the hyperbola in the plane z = k has its
principal axis parallel to the a;-axis and its vertices on the
x^
parabola -5 = cz. These vertices recede indefinitely as k in-
creases from to + 00 . If fc is negative, the hyperbola in the
plane z = k has its principal axis parallel to the y-ajda and its
vertices on the parabola t^ = —cz.
In a similar manner, the intersection of the surface by the
plane x = fci is the parabola ^ = —cz + — ,x = ki, which has
THE CONE
219
its vertex on the parabola -^ = cz. The intersection of
the surface by the plane y = ki is the parabola -^ = c +
Tj , 2/ = fe, which has its vertex on the parabola -yA— ~ c^.
The surface extends indefinitely along all axes.
112. The cone 4 + 6-^ = 0.
a^ Ir (?
Isi. The surface is symmetrical with respect to the origin,
the coordinate planes, and the co-
ordinate axes.
2nd. The surface intersects the
three axes at the origin only.
3rd. The trace in the x2/-plane is
the point elhpse -^ + rj = 0, and
in the a;z-plane the intersecting
straight lines -j 2=0? ^nd iii the
yz-plane the intersecting straight
lines \-
= 0.
4:th. The intersection with the plane 2 = A; is the eUipse
^27,2 ' h21„2 -^J * "'•
As fc increases nimierically, the semi-axes of the ellipse in-
crease and the surface extends indefinitely in the direction of
the z-axis.
The intersections in the planes x — hx and y = "ht are
hyperbolas with vertices on the straight lines -5 — ;5 = and
t* 1/
-5=0, respectively.
220 SOLID ANALYTIC GEOMETRY
EXERCISES
1. Discuss and construct the surfaces represented by the following
equations:
(o) 4a;2 + 92/24.2:2 = 36.
(6) 4x2-9dj|^z'i = 36.
(c) 4^2 ^ allM:j:.,
(d) ^ — a;2 — / i 4.
(e) 4 x2 - 22 = 4 2/.
(/) 93/2 -4x2 -z2 = 36.
(ff) 4 s2 - 9 2/2 - 4 z2 = 0.
W 2/2/62 - zVo2 - z2/c2 = 1.
2. Find the equation of the locus of a point which moves so that the
sum of the squares of its distances from the x- and z-axes equals 4.
Discuss and construct the locus.
3. A point moves so that the sum of its distances from two fixed
points is constant. Prove that the locus is an ellipsoid.
Hint. — Take the straight line through the two points as x-axis and
the point halfway between as origia.
4. A point moves so that the differaice of its distances from two fixed
points is constant. Prove that the locus is an hyperboloid.
6. Find the equation of the locus of a point which moves so that its
distance from the x^-plane increased by 1 is equal to 1/V2 times its
distance from the point (0, 0, —4).
6. Prove that the sections of the paraboloid x'^/a? + ^/62 = cz by
planes parallel to the 2/>pIane are equal parabolas; also those parallel
to the iz-plane.
^ «i2 02
7. Discuss and construct the locus of — +^ := 1. Show that
Or Or c^
sections parallel to the xy-TphxiB are circles. What curve revolved about
the z-axis would generate this surface?
8. A point moves so that the sum of the squares of its distances from
two perpendicular lines is constant. Prove that the locus is an ellipsoid.
113. The normal form of the equation of a plane. — A
plane is determined in position if the length of a perpen-
dicular from the origin upon the plane and the direction
angles of this perpendicular are known. This perpendicu-
lar from the origin to the plane is called the normal to the
plane.
NORMAL FORM OP THE EQUATION OF A PLANE 221
The normal form of the equation of a plane is
X cos a + y cos p + » cos 7 = p, (46)
where p is the perpendicular distance frorn, theoTjigin to the plane,
and a, /3, and 7 the direction angles of tha^gf ^^ fjdicular.
Proof. — Let ABC be any plane and^i*tfcffl^perp_endicular
from the origin upon it be the Une OQ whi'(3i makes the angles
a, ;3, and 7 with
the axes of x, y,
and z, respec-
tively. The di-
rection OQ from
the origin to the
plane is always
taken as posi-
tive, also a, j8,
and 7 are con-
sidered positive
angles.
Let P (x, y, z)
represent any
point in the plane and draw its coordinates ON = x,
NM = y, and MP = z.
Project ONMP and OP on OQ. By theorem II, Art. 97,
projection ON + projection NM + projection MP = pro-
jection OP. By theorem I, Art. 97, this becomes
X cos a-\- y cos /3 + z cos 7 = p.
This equation is seen to be of first degree.
114. The general equation of first degree '
Ax + By + Cs+D =
represents a plane.
Proof. — Consider the equations
Ax + By + Cz + D = (1)
and a; cos a -1- 2/ cos jS + 2: cos 7 — p = 0. (2)
222 SOLID ANALYTIC GEOMETRY
Equation (2) represents a plane. Equation (1) also rep-
resents a plane if it differs from equation (2) only by a con-
stant multiplier as K. If, then,
KA = cos a, KB = cos/3, KC = cos 7, and KD = —p,
it is desired to show that K can be determined.
By Art. 98, cos'' a + cos^ /3 + cos^ 7=1 and, therefore,
1
K^A^ + Km^ + K^C^ = 1 or X =
itVA^ -H £2 -h C2
This shows that equation (2) represents a plane in which
A C
cos a = , , cos 7 =
±VA2 + B2 + C2' ± V42 + S2 _|_ (72
cos/3 = — 7=======, V =
± VA^ + 52+^' ± VA" + £2 _(- c='
Since p is always positive, the sign of the radical will be
opposite to that of the constant term. ■
115. Plane determined by three conditions. — Of the
four coefficients in the equation
Ax + By + Cz + D = (1)
only three are independent, and therefore three conditions
are sufficient to determine three of them in terms of a
fourth. After substituting these values, the equation can
be divided by the fourth coefficient.
116. The -equation of a plane in terms of its intercepts.
— The equation of a plane in terms of a, b, and c, the intercepts
on the axes, is
^ + 1 + 1 = 1. (47)
a b c
Let the intercepts of a plane on the axes of x; y, and z be
a, b, and c, respectively. The three points (a, 0, 0), (0, b, 0),
THE DISTANCE FROM A PLANE TO A POINT 223
and (0, 0, c) on the plane are therefore known and the method
suggested in Art. 115 appUes.
Substituting the coordinates of these three points in equa-
tion (1) of that article,
Aa + D = 0, Bb + Di=0, Cc + D = 0.
Whence
a' b' c
Substituting in (1),
a c
V
Dividing by D and transposing,
a c
117. The angle between two planes. — The cosine of
the angle between two planes, whose equations are of the form
Ax + By + Cz + D = and'AiX + B^y + Ciz + Di =
is given by the equation
cos = , — , ■ • (48)
The angle between two planes is evidently the same as the
angle between their normals. Substituting in the formula
of Art. 99, the values of cos a, cos /3, and cos y found in Art.
114, the above formula results immediately.
ABC
If two planes are parallel, -r = s" = 7=^ > ^^^ */ ^^^V "'"^
Ai Bi Ci
perpendicular, AAi + BBi + CCi = 0. The proof is left
to the student.
118. The distance from a plane to a point. — By finding
the equation of a plane parallel to the given plane and pass-
ing through the given point, and computing the difference of
224 SOLID ANALYTIC GEOMETRY
the distances from the origin to the planes, it is found that
the distance from the plane Ax + By + Cz + D = to the
point Pi {xi, j/i, Zi) is
VAo + 3" + ^
in which the sign of the radical is opposite to the sign of D.
EXERCISES
1. Reduce the following equations to intercept and normal forms:
(0) 7x-2y-2z + U =0.
(6) 2a; + 62/-3z-42 = 0.
2. Find the equations of the planes which satisfy the following con-
ditions:
(a) passing through the points (1, 1, 0), ( — 2, 1, 2), and (4, 0, 1).
(b) parallel to the plane 7x + 2y + Wz + 25 =0, and passing
through (3, 1, -2).
(c) perpendicular to the plane 3x + 2y — z + 11 =0 and pass-
ing through the points (1, 0, 1) and ( — 1, 1, 1).
(d) a>intercept = 5, 2/-intercept = 3, and z-intercept = —7.
(e) distance from origin to plane = 5, cos a = f , and cos P = —i.
(/) passing through the point (1, 5, 6) and perpendicular to each
of the planes ix — 5y + 2z = 5 and x — y + e = 3.
(g) passing through the points (1, —2, 3) and (5, 0, 3) and at a
distance of 3 from the origin.
(h) at a distance of 2 from the origin, the normal making equal
angles with the axes.
(i) perpendicular to the line joining the points (4, 3, 1) and (1, 3, 5)
at its middle point.
(j) containing the z-axis and the point (xi, j/i, zi).
(fc) passing through the line of intersection of the planes ix + y +
2 z = 3 and 2x + y + z = 1, and perpendicular to the
plane 3a; + 4^ — 2z = 7.
Hint. — The equation of the plane through the line of intersection of
the two given planes is4a;-)-j/-|-2z — 3 + \(2a;-l- y + z — 1) =0.
(1) perpendicular to the Une through the points (4, 3, 1) and
(2, 4, —1), and five units from the origin.
THE EQUATIONS OF A STRAIGHT LINE 225
3. Prove that the planes .x + y + z — 1=0, 2x + y — g=0,
X + 6y + 4z + 1 =0, and 5x + y — i = meet in a point.
4. Prove that the four points (8, 15, 4), (2, 1, 0), (0, 3, 2), and (2, 3, 1)
lie in a plane.
6. • Find the distance from the origin to the plane through (0, —3, 2),
(2, 1, 2), and (5, 3, 0). In which octant does the foot of the normal lie?
6. Find the angles between the following planes:
(a) ix — 7y + iz = 5 and 3 a; + 4 2/ = 17.
(6) Zx-2y + 6z = 7 and 4:X-Sy + 12z = 0.
7. Find the equations of the planes bisecting the angles between the
planes ix — 7y + iz + l5=0 and 2x — y — 2z — 5 =0.
8. So determine K that the plane 3 a + Ky + 12 « = 26 shall be
(a) two units from the origin.
(6) perpendicular to the plane x + Qy — z = 5.
119. The general equations of a straight line. — Two
planes intersect in a straight line. It has been shown that
the locus of an equation of first degree is a plane and that the
curve of intersection of two surfaces is defined by considering
their equations simultaneously, hence:
The locus of two equations oj first degree
Ax + By + Cs-\-D = 0, A^x + Biy + CiS + Di =
is a straight line.
120. The equations of a straight line through a given
point and in a given direction. — The equations of a straight
line parsing through the given point Pi {xi, yi, Zi) and having
direction angles a, ;S, and y are
^Lnj^^y-yi^^-^K (50)
cos a cos p cos y
Proof. — Let P {x, y, 2) be any other point on the line, then
by Art. 98,
X — Xi a y ~ yi Z — Zl
cosa= — T — , COS |8 = " , " , cos 7 = , .
226 SOLID ANALYTIC GEOMETRY
Solving each of these equations for d and equating,
X- xi ^ y — yi ^ z — zi
cos a cos jS cos 7
If instead of cos a, cos |8, and cos 7, numbers a, h, and c
proportional to them are given, it is readily seen that the
equation will take the form
x^zJ^ _ y-yi ^ g-gi. (51)
a o c
121. The equations of a straight line through two given
points. — The equations of a straight line through the two
points Pi (xi, yi, Zi) and Pi {xi, ys, Za) are
*-^^ = y-y^ = fjm. (52)
xg - xi ya- yi ss - «x
Proof. — Substituting cos a = 5LZ_E1 ^ cos ^ - ^' ~ ^'
cos 7 = — T — ^, in equation (50), and dividing by d, the
suit is
X — xi ^ y — yi ^ g — zi
% - a;i yi — yi Z2 — Zi
122. The projection form of the equations of a straight
line. — A plane through a Une perpendicular to one of the
coordinate planes is called a projecting plane of the line.
If between two equations in the form Ax + By + Cz + D
= and Aix + Biy + Ciz + Di = 0, one of the variables is
eliminated, an equation in two variables results. This from
Art. 101 is a cyhndrical surface with elements parallel to that
axis which corresponds to the variable eliminated, and with
its trace in the plane of the other two axes. The equation is
of first degree and the cyhndrical surface is therefore a plane.
This plane is the projecting plane of the line. Two such
DIRECTION ANGLES OF A LINE 227
planes will determine the line. By eliminating x, y, and z in
turn, the projecting planes of- the hne x-\-2y + Zz = Q,
2a; — 2/ — 3^ = 5 are found tobe52/ + 92 = 7, 5x — Ss
= 16, and Z x -\- y = 1\.
123. Direction angles of a line. — - If the equations of a
line are given in the form Ax -{- By -\- Cz -\- D = Q, Aix -\-
Bxy + Ciz + Di = 0, the direction cosines of the line may be
foimd by a process illustrated in the following example.
Find the direction cosines of the line x-\-Zy — 2z = 2,
3x — 2y — 4:z = 5. Having determined two of the pro-
jecting planes to be x — 8y = 1 and 11 j/ — 2 z = 1, the
values of y may be equated, giving
x-1 _y_ 22 + 1 ^j. x-1 _y _z + ^
8 1 11 8 1 V-
This is in the form of equation (51) and therefore the direc-
ti on cosines are p roport iona l to 8, 1, and V"- Dividing by
V'8'' + 1" •+ (-y-)" = h "^381, the direction cosines are found
to be ^^ 2_ 1]_
^ VSST' V38l' V381
EXERCISES
1. Find the equations of the hnes through the following pairs of
points: (o) (0, 0, 0) and (1, 2, 2). (6) (1, 4, 0) and (3, -2, 3). (c)
(1, 5, 1) and (-6, 1, 5).
2. Find the coordinates of the points in which each of the above Unes
cuts the coordinate planes.
3. Find the equations of the Unes determined by the following con-
ditions:
2
(a) passing through the point (3, 0, 1) and having cos a = — ^
3 V 5
J o 5
and cos B = ;=•
3V5
(5) passing through the point (5, —3, 1) and perpendicular to the
plane Sjp - 6y + 3z - 7 = 0.
228 SOLID ANALYTIC GEOMETRY
(c) passing through the pomt (3, 0, -1) and parallel to the line
X — 5 y + U ^ z_
3 4 12*
(d) passing through the origin and perpendicular to the lines
£ZLi = £±6 = ^ and X + 3 2/ - 2 - 3 = 0, 3 a; + 5 J, +
z - 1 = 0.
4. Prove that the following pairs of lines are perpendicular:
. , x-1 y g + 2 , x + \ _ y-3 _ z + 5
(b) 2x -y + z -2 =0,4tx + y - 4:z - 4: = 0&iidx + 7y —
z + 8=0, x + 3y + 3z + i = 0.
6. Find the angle between the following pairs of lines:
. . X y + 1 2 + 2 ^, X _y _ z-7
(b)x-4:y-3z-4 = 0, 2x-2y + 3z + l=0a,nd4,x +
4:y + 3z-4:=0, ix + y-Qz-1 =0.
6. Prove that the lines 5x + 8y-z + 20=0, 5x-8y + 3z-
32 = and 4a;+2/ + z-2=0, 4a; + 22/ + 3z-l =0 meet in a
point and find the angle between them.
7. Prove that the points (1, 0, —3), (4, 1, — 1), and (7, 2, 1) lie on a
Une.
8. Prove that the line 2a; + 6j/ + 8-2 = 0, 2a;-32/-2a-2
= is parallel to the plane 2x — 3y — 2z = l.
Hint. — Prove that the hne is perpendicular to the normal to the
plane.
9. Prove that the three planes 4a; + j/ — z + 3=0, 12 a; — j/ — a
— 5 = 0, and ix — 3y-{-z — 11 =0 meet in a common Une. Find
its equation and direction cosines.
10. Find K such that the lines x-3y + 3=0,x + y-z~l =0
and(7 + X)a;-72/ + 7z-28-X=0, 6a; + 73-6 = 0are per-
pendicular.
11. Find£suchthat (4, 15, K), (1, K, 2), and(-2, -1, -3) are col-
' linear.
12. Prove that the line 2a; + 62/ + z-2 = 0, 2a;-32/-2z-2
= is perpendicular to the plane 3x — 2y + 6z = l.
13. Find the projecting planes of the Une 3x + 2y + z = l, x —
4 y - 2 z = 3.
DIRECTION ANGLES OF A LINE 229
14. Find the equations of the planes satisfying the following con-
ditions:
/g ^ 1 7/ I 1 g
(a) determined by the parallel lines — = — = = - and
x + 1 _ y-3 ^ z + 1
-2 4 -4 ■
(6) determined by the intersecting lines — ^ — ■ = T^ = y and
a: + 2 ^ y + 4 _ s + l
3 2 1 ■
(c) containing the points (1, — 1, 2) and the line
a- 1 ^ y+2 ^ s-7
2 4 2 '
ANSWERS
Art. 10. Pages 21 and 22.
2. y + V3x=3 + 2V3.
4. 4,y — 2x = 5.
6. X — y = 7.
6. x + 3y =U.
7. x+y =7.
9. 2x ~y = 5.
11. (b) x^ -Qx+y-Ay = 12.
12. a;2 - 4a; + ^2 - 6j/ + 4 = 0.
13. 3?-2x+y^ -4:y =20.
16. a? -22; +2/! -62/ = 15.
17. a?-8a;+3/'+8j/+22 =0.
18. a? - 4s + 2/2 - 82/ + 10 = 0.
Art. 13. Page 40.
3. (a> 1? = 6y -9.
(c) 2/^ + 8t/+4a; + 20 =0.
(d) 2/2 -21/ -4x4-13 =0.
(e) 3a;2 + 32/2 - 18x - Uy + 10 = 0.
(/) 9x2 + 252/2 =225.
(g) 162/2-9x2 = 144.
Alt. 14. Page 41.
1. (6, 8) and (-J^,-^).
2. (1, ^ 2) .
5. (4, 3), (-4, -3), (3, 4), (-3, -4).
7. (2, 3) and (-y, i^).
8. (2, 2) and (-1,^).
10. (0,0), (^-2-. ^j and ^-^, — ^j-
11. (a, 2 o) and (9 a, -6 o).
17. ¥.
231
232 ANSWERS
19. h -1^00. _
„. 5 V85 5Vl3
Art. 19. Pages 49 and 60.
1. (fi)x + Zy = 2. {J)Zx + y = \2. {.g)x + y = 6.
8.x-y==8, x + 2y = l7, and 5x + y + li = 0.
9. X + y = 5 and y — x = 1.
10. x-y = 6, 5x-2y = 31, and x + 2y = 7.
11. y — X = 2 and x + y = 6.
12. (a) 2y = x.
(6) 21 + 2/ = 0.
(c) 2/ (3 - Vs) + a; (s Vs + l) = 0.
Art. 21. Pages 61 and 62.
1. (a) 3x -y = 7.
6,3x + 7y = 2, x + y + 2=Q, and y = - 1.
6. 2x + y + 5 =0.
l.'x — 6y = Q and 3x — 5y = 1.
9. y = 1, 12x + y = 21, and 3 s + 2 j/ = 7.
11. 2x-32/ = 12, 3a; -2/ = 4, and x + 22/ + l=0.
12. a; = 4.
Art. 24. Page 66.
3. y + xVz = 10.
A. (o) I + 2/ = 5 V2. (6) X + 2^ + 5 V2 = 0.
6. (6) 2/+fV^=3.
7. x + yV3=3.
Art. 26. Page 68.
2. X - 2 2/ = 0, p = 0.
3. x+y + 9 = 0, p = - f V2.
Art. 26. Page 61.
1. -¥• 2. 2. 3. 4, ^, -^.
Art. 27. Page 63.
1. 2x = 3and2/ = 0. 2. 7x +2/ = 6 and 72/ - x = 6.
ANSWERS 233
Miscellaneous Exercises. Pages 63 and 64.
3. (a) 4 a; - 3 2/ = 0.
(6) 5x + 12y + 12 = 0.
4. (a) 5y-12x = 52.
(6) 12x-5y = 26.
6. y -2x = 5 Vs. ,
7. 14.
8. -^, 3 V2, and 5.
V26
9. (4,2).
10. {a) x + y = 0, 7x-y = 24, and y = -3.
(b) 17x-17y = i, 7x + 17y = 0, and 6a; = 1.
Ic) x + y = 0, 3 X - 9 2/ + 16 = 0, and 2 a; - 2/ + 4 = 0.
13. 11 a; - eSy = 456.
14. (a) 3V13. (6) 3V5.
Art. 28. Page 67.
6. Zx+y=—5, 3y — x==5, and x + 2y = 5.
6. 90°, tan-' i, and tan-' 3.
8. 2^ - 3 = ^"^ \! (a; - 2) and 2/ - 3 = ^~ (a; - 2).
1-^3 V3 + 1
10. 2, - 1 = ^^^ (a; - 2), 2/ - 1 = ?^1^ (^ - 2),
Art. 29. Page 70.
3. (o) 2/ = 2a; + 5 Vs.
(c) 2/= 2a: + 12.
(d) 2/ = 2x-3V5.
4. (a) 2/ -3a; = 10.
(6) 2a; -62/ = 2.
(c) y — 2x = 6 and a; — 4y = 4.
(d) a; +2/ + 6=0.
(e) a; + 22/ + 8 = 0.
6. (a) x + 2y = iV5.
(6) 4a; + 32/ = 20.
(c) 42/ - 3a; = 20 and 42/ + 3a; = 20.
(d) a; +2/ = 4^2.
234 ANSWERS
6. x + y = i.
7. J/ + 2a; = 6 and y + 8s = 12.
8. 2a; + 2/ + 10 = 0.
9. a; + 2/ = ± 2 V2.
Miscellaneous Examples on Chapter HI. Pages 70, 71 and 72.
1. 3x + 22/ = leandy = 2x + l.
2. x-5y = 9 and y + 2 = 0.
i. The equations of two sides of one triangle are y — 2 = = (a; — 3)
V3
and 2/ + 4 = — ;= (x — 3), and of the other, y — 2 = —= (x — 3) and
v3 V3
8. a; - 32/ + 10 = and 3a; +2, = 0; (A, -A) and (fj, fj).
9. (6) 5a: + 25^ = 143.
(c) 103 a; + 44 2/ = 91.
(d) 3x+2/ = 7.
11. 42/ -7a; + 16 = and 292/ + 28a; + 26 = 0.
16. (a) C = -3^.
(d) O = 25 42 + 25 M
(/) 3A + 5fi + C = 0.
18. 3 a; - 2 2/ = 5.
Art. 39. Page 93.
4. (c) V2a;2 - a; = 0.
(d) -5a;2 + 202/'+8V5a;-16V52/ = 25.
6. 2/ V2 = 1.
Art. 41. Page 97.
2. (o) 2a;2 = 9.
(c) 4x2 + 2/' =4.
8. (a) a;2 + 32/* +4x V2 -6^ -\/2 = 2.
(b) x^ -2/2 -2a; - 22/ = 4.
4. (1, -1), 2x - 2/ = 0, and X + 22/ = 0.
Art. 42. Page 99.
4. (d) xy = 2.
(e) (x2 + 2/2)2 = o2 (x2 - j/2 ).
(?) x2 + 2,2 = a (Vx2 + 2/2 _ a;),
(i) X (x2 + 2/2) = 2 02/2.
6. (o) p2 = 40.
(6) p2cos2e = 16.
ANSWERS 235
Art. 43. Page 101.
2i (a) a? - ix + y" -2y = 20.
(c) a;2 + 2a; +2/2 - 6^ + 2 = 0.
(d) x' -10x+y^-12y + 25= 0.
%fc Art. 46. Pages 106 to 109.
3. (a) x' + 2x + t/'-Sy + l=0.
(6) x^ + y<' -,6y = 16.
id) x^ + 2x + y^ -6y = 90. .
(f) 3a;2 -44a; + 32/2 + 1122/ + 41 =0.
(ff) a2 - 6 a; + ^2 _ 8 2/ + 9 = 0.
i. X — y + i = 0. *
5. 2a; -3 2/ = 5; -f.
6. 2 a; + 3 y = 13.
8. (a) a;'' -12 a; + 2/" -8 2/ = 48 and a;' + 16a; + 2/' + 20^ + 64 = 0.
(c) a;' + 6a; + 2/24.62/ + 9 =0 and a;« + 2a; +2/2 - 22/ + 1 = 0.
(d) ^2 + 4a; + 2/2-22/ = 20.
(e) 5a;2 + 52/^ + 20a; - 8O2/ + 308 = and 5a;2 - 20a; + 5y^ = 52.
(/) a;2 + 2/2 + 4(2/-a;) V2 + 8 = 0.
10. x2 - 6a; + ^2 - 82/ = 0.
11. a;2-2a; + 2/2-102/ + l = and a;' - 34a; + 2/2 - 1702/ + 289 = 0.
12. x2 + 2/2 + 82/ = 9.
Art. 48. Page 116.
2. (o) y^ = -9a;.
(6) 2/2 + 42/ + 4a; + 12 =0.
(c) a;2 = 12^ + 36.
(d) a;2-2x-12 2/ + 13 = 0.
(e) (2/ -1)2 = 4 a;.
(/) a;2 + 92/2 + 6a;2/ -56a; + 522/ + 14 =0.
3. 2/ = -2 a;.
6. a;2 + 2/2 - 5 2/ = 0.
6. (a) a;2-4a; = 42/ + 16.
(&) a;2 -4a; = iy -12.
7. 72/ = 24a; -36.
8. 42/ + 3a; + 3 = 0;(-|, -A).
Art. 60. Page 119.
1. (-2, -2); (-2,, -i); 6; 2/ = -J; a; + 2 = 0.
2. (3, 3); (1, 3); 8; x = 5; y = 3.
3. (-1, 2); (-1, W, i;l2g = 29; a; = -1.
236 ANSWERS
4. (-1,1); (-1,1); 2; 82/ =11; a; =-1.
5. (if); (1, i); 1; 2a; = i; 22, = 5.
6. (-i, 1); (-1^, 1); f ; 12 X + 13 = 0; 2/ = 1.
Art. 64. Pages 126 and 127.
1. (o) (±5, 0); (±4, 0); 4x = ±25; ^s^
(b) (±4, 0); (±2, 0); x = ±8; 6.
(c) (0, ±5); (0, =t3); 3 2/ =±25; ^.
2.
(o)
36 + 32 ^- ^^' 36 + 27 ^-
«ffl+l-'-
(d)
!+¥-• <')S+^--
</) l+l-l-
(?)
i+?-'- (« S+S-
. «f+f=..
3.
(a)
'■t"'+*r''-i. m
(x + 4)» (2/ + 2)»
25 ' 16
'•' f + ^ =
1.
4.
(o)
fe±a!+H^=i. <»
{x + iy (n + iy
9 ' . 25 ~ ^•
(c)
^tf^+'s^-- »
{x - 2)2 (2/ - 1)2 .
16 ' 12
7. x2 + 2/2 = 25; (4, ±3) and (-4, ±3).
8. 152/ -8a; = 24 and 15;/ + 8a; + 24 = 0, ±ff
9. 25a;2H-9 2/2 = 225.
Art. 66. Pages 130 and 131.
1. (a) (-2, 1); (_-2, 3) and (-2, -1); (-2, 5) and (-2, -3); 4
and 2 VS; 6; 2/ = 9 and 2/ = -7.
(6) (2, -1); (2, 2) and (2, -4); (2, 3) and (2, -5); 4, and \/7
I; 3 2/ = 13 and 3 2/ = -19.
(c) (1, 4); (1 ± 2_V2, 4); (5, 4) and (-3, 4); 4 and 2 Vl; 4
X = 1 ±4 V'2.
(d) (-3,1); (-3 ±5^3,1); (7, 1) and (-13, 1); 10 and 5; 6
3a; = -9 ±20 Vs.
(e) (-1, -3); (2, -3) and (-4, -3); (5, -3) and (-7, -3)
6 and 3 VS; 9; x = 11 and x= -13.
if) (-3, 1); (-3, 3) and (-3, -1); (-3, J) and (-3, -5); 6
and 4 V2; ^/; 2/ = 19 and 2/ = -17.
2. 16x2 + 25 2/! = 400.
ANSWERS 237
Art. 69. Pages 136 and 137.
1. (a) 6 and 10; ^; (0, ±3); (O, iVsi); y = ±4—
** V34
(6) 8 and 4 VS; 6; (±4, 0); (±2 V7, 0); a: = ±
8
vr
„ . - i« 32/2 3.2 162/2 ,
2- (a) 4--4- = l; ('')9--8r = l'
^''^ 16 20 ^' ^''^ 36 28 ~ ^'
Ce) ^_f! = i. fn ^-2^'-l•
(.)f-f = l. —
4. (a) ^^'-^^' = l;(-l,3±V4i);(-l,7)and(-l-l);\^.
(6) ^^1^'--^^^' = 1; (-2±2Vl3, -3); (2,-3) and
(-6, -3); 18.
(c) ^^4^ - ^^^ = 1; (-1, 11) and (-1, -5); 24.
6. (a) — g 2^ 1, (6) -^g jg— = 1;
(2/ - ly (x + 2)' _ (j/ + 3)' (a + ly .
/o^ (^ - 7)' fa + D' n. (f. (y - 2)2 (x - 2)2
(e) —4 12~ - ^' (■^ —9 9 1-
*• 16 20 ^■
. /)2 cos2 e p2 sin2 9 _
02 62
Art. 61. Page 140.
1. (o) (1, 2); 2 and 3; (1 ± vl3, 2); (3, 2) and (-1, 2); 9.
(6) (3, -2); 2 and 4; (3, - 2 ±2 Vs); (3, -4) and (3, 0); 16.
(c) (3, -1); 1 and 3; (3, -1 ± Vio); (3, 0) and (3, -2); 18.
(d) (1, -1); Sand V3;(l±2 V3, -1); (4, -l)and(-2, -1);2.
5238 ANSWERS
(e) (i -i); Vr and V2; (1, -J) and (-f, -|); (J ± V7, -|);
4
V7-
0) (0, 1); I and 4; (0, 1 ± i V29); (0, V) and (0, -|); 20.
2. xy + 5x-3y = 7.
3. X? -y^ = 100.
25 39
(a; - 1)!! 13 (y -!)' _.
12 192
Art. 64. Pages 143 and 144.
1. ?/=±3x; g-| = l. 2.iy=^3x;^.
3. 9y'- 4a;2 = 36; 4 and 6; 9; (o, ±Vl3); y = ±-^-
6. 4x2-12 2/2 = 75. 6. i x" - y' = S&.
7. 43/2H-82/-x2 + 2x - 1 =0. 16. 4x^-9^2 = 64.
Alt. 66. Pages 149 and 160.
2. (a) 3 X - 2/ = 2; 3 2/ + s = 4.
(6) 2/ + 4x = 13; X-42/ + 18 =0.
(c) 3x -2/ = 2; x + 3v = 14.
(d) 2/-2x = 3; x + 22/ = 1.
(e) 4x + 52/ = 25; 25x -20y = 64.
(/) 2/ = 4; X = 2.
(ff) 15x - 2/ = 36; X + 152/ = 138.
(h) X-32/+4 = 0; 3x + ^ = 28.
Art. 67. Pages 164 and 166.
1. (o) 4x-y = 2; 4y+x = 9; i; 8.
(6) 3x+42/ = 25; 4:x-3y = 0; —V; -3.
(c) X - 2/ = 4; X + 2/ = 4; 0; 0.
id) 4x + 2/ = 8; x-4y + 15 = 0; -1; -16.
(e) 3x + y = 7; x-32/ = 9; f; 6.
(/) 4x + 32/ = 16; 162/-12x = 27; -|; -4.
(ff) y - X = 10; 2/ + X = 4; 7; 7.
2. 2/ = a;; 2/ = -x.
3. 3x + 42/ = 7; 4x-32/ = l.
ANSWERS 239
4- (a) 2/ = 3; a; = 2.
(6) x + y + l=:0; x~y=Z.
(c) 15a; + 4l2/= -9; 123a; -45^ + 455=0.
(d) a; - 3 2/ = 4; 3 a: + j/ = 12.
7. ;^. 8. tan-i(-^).
10. a; + 2/ = ±V34; x-y = ±V34.
14. (a) (1, -2). (&) (1, 2).
Art. 68. Pages 166 and 167.
1. (a) 4a; + 3y = ±25; (±4, ±3).
(b) y = x + 2; (0,2).
(c) 4 a: +2/ = ±8; (±1, ±4).
(d) y = 2x-3;y = 2x- 9; (0, -3); (4, -1).
3. Sx + Wy = ±25. •
4. 2x -y = 5.
6. bx + ay = zkdbV2.
7. 3a; + 2^ = 0; 3a; + 22/ + 4 = 0; 2a; = 32/; 6a; - 9 2/ = 5.
Art. 70. Pages 162 and 163.
2. (a) a; + 12 2/ + 16 = 0. 4. (2,2). 6. (1,-6).
9. a; = ±o. 10. y + 5 = 0; ix + 3y = 25.
Art. 72. Pages 168 and 169.
3. 2/ = 4. 6. 2/ = 2;,2/ - a; = fc.
7.
(±2 V2, T2 V2)
8. X — y = 2. 9. 42/ + a; = 5.
Art. 76. Page 176.
1.
(a) 2/' -4a; + 4 =
= 0. (6) 5a;2 + V5a; = 2.
(c) a;= + 22/-^2 = 0.
2.
«f+¥-..
(6) J - 1' = 1. (c) a? - j/« = a
(d) 52/^! = 9.
(e) 2/" = 3 a;.
Art. 76. Page 176.
1. (o) a;* - 2/2 — a; + 2/ = 0.
(6) 3^ + xy + 2y'-3x-4:y + 2=0.
(c) 7a;''-7a;2/ + 32/i' + 21a;-82/ = 28.
2. xy + y'' — 3x — y = 6.
240 ANSWERS
3. x^ -2xy + y' = 1.
4. Sx^ + 4:xy + y = 0.
6. 4a;2 -82; + 2/2 _ 102/ + 4 = 0.
6. 2/2 - 62/ - a;2 + 8s + 9 = 0.
7. ix' — 4:xy +y^ — ix — y = 0.
Art. 96. Page 201.
1. (-2,1,4). 2. (1,6,-3).
3. 1 : 2. 6. 2 : 1; (3, -3, 0).
6. (-7, 6, -2). 7. (0, 6, 0).
Art. 98. Pages 204 and 20S.
3. ±|V3.
4. ±^, t4=. ±-P=-
Vl4 Vl4 Vl4
6. ±1, ±i =Fi; -8, -4, 1.
8. (4, 120°, 135°, 60°) or (4, 120°, 135°, 120°); (-2, -2^2,2) or
(-2, -2 V2, -2).
12. (10, -1, -2).
Art. 99. Page 206.
5
1. COS-'^f. 2. C0S-1±- ;=•
2V1O5
4 ±— L= ±— !=• =F ^ • 5 V-
V107' V107' Vl07
Art. 100. Page 207.
2. x-3y + 2z = 4. 5. 2/^ - 22/ - 6a; + 10 = 0.
Art. 102. Pages 208 and 209.
2. (o) (1, 3, -1); 4. (6) (J, -if); 3.
3. a;2 + 2/* +Z2 + 42 = 45 and x" + 2/' + z2 - 203 + 51 = 0.
4. 2/'' + z' - 10 z = 0.
5. x'-ix+y' -2y + t!' + 2g = 7.
6. a;'' + 2a;+2/2 -62/4-z2 + 10z = 14.
7. x2 - 8 X + 2/2 + 82/ + z= - 14z = 0.
8. x2 + 2/2 + z2 + 6z = 40 and x^ + (^ - ^^y + (s - W)' = 49-
9. x2 + 2/2 + z2 = 4.
10. x2 - 2 X + 2/2 + 6 2/ + z2 - 8 z = 55.
12. x2-2x+2/2-42/ + z2 + 2z = 18; (1,2, -1); 2 V6.
13. x2-2s-4z + 5 =0.
ANSWERS 241
Art. 103. Page 211.
1. (O) y2 + Z2 = X2. (6) 4x^+y2 + igi = 16.
(c) x2 + 2/2 = 4z. (d) x' +y' -^ = 4,
(e) a;2+32_^2 + 22/ = l.
_ , , a;' j/2 z^ 3:2 ,,2 ,2
^ («)^2 + ^+P = l. (&)|+| + |i=l-
3. 2/2 + z2 = 16. 6. a;2 + 2/2 + z2 - I4z + 33 = 0.
Art. 112. Page 220.
2. a;2 + 21/2 + z2 = 4. 6. a:' + 2/2 - z2 + 4z + 14 = 0.
Art. 118. Pages 224 and 226.
2. (o) 2a; + 9y + 3z = 11.
(6) 7x + 2y + 10z = 3.
(c) x + 2y + 7z = 8.
(d) 21 a! + 35 2/ - 15 z = 105.
(e) 2a; -y ±2z = 15.
(/) 3a; + 22/-z = 7.
(ff) z = 3 and 3x-Qy + 2z=21.
{h) x-\-y + z = ±2 V3.
(i) 8z - 6a; = 9.
(j) xyi — xiy = 0.
{k) 2x-y + z = Z.
6. I.
6. (a) cos-i H-
(6) cos-^ If.
7. x4-22/ — 5z = 15 and 5x — 5y— z = 0.
8. (a) X = ±4.
(6) K = l.
Alt. 123. Pages 227, 228 and 229.
^- (") 1 - 2 = 2 (^) -2- = ^:6~ = 3'
W 7 - 4 - _4 ■
2. (6) (1,4,0), a, 0,2), (0,7, -f).
3. (a) -2- - 5 - -^T' ^"^ ~1~ - "^2" ~ ~~T~'
M ^~^ - ^ - g + 1 Ml ? _ M - -i_
W 3 4 12 ^'^^2 6 -l'
242 ANSWERS
6. (a) 45°. (6) cos-i— ^E=-
10. 2.
11. 7.
13. 7a; = 5, 14^ + 72 + 8 = 0.
li. (a) 2x + y = l. (b) x + y-5z + l=0.
(c) 11 X — 5 2/ — z = 14.
INDEX
The numbers refer to the pages
Abscissa, 2.
Angle, between two lines, 7,
205.
between two planes, 223.
Area of a triangle, 15.
Asymptote, 33.
of the hyperbola, 140.
Bisectors of angles, 61.
Circle, equations of, 100.
Cissoid, 195.
Cone, 219.
Conic sections, 110.
rectangular equation of. 111.
Conic through five points, 175.
Confocal conies, 169.
Conjugate diameters, 167.
Conjugate hjrperbolas, 142.
Coordinate planes, 197.
Coordinates, Cartesian, 2.
rectangular, 3.
polar, 73.
rectangular in space, 197.
polar in space, 203.
Cycloid, 190.
Cylindrical surface, 207.
Diameter, definition, 163.
of parabola, 165.
of eUipse, 163.
conjugate, 167.
properties of, 166.
Directed lines, 1.
64, Direction angles of a line, 227.
Direction cosines, 203.
Directrix of conic, 110.
Discussion of equation,
rectangular coordinates, 27.
polar coordinates, 77.
in space, 212.
Distance, between two points, 5,
198.
from a line to a point, 58.
from a plane to a point, 223.
Eccentricity of a conic, 110.
Ellipse, equations of, 120-125.
construction of, 128.
EUipsoid, 214.
Elliptic paraboloid, 217.
Epicycloid, 191.
Equation of a locus,
rectangular coordinates, 18.
polar coordinates, 76.
in space, 206.
Exponential curve, 177.
Extent, 30, 78.
Focal chord, 113.
Focus of conic, 110.
Function and variable, 24.
General equation of second de-
gree, 171.
243
244
INDEX
Harmonic division, 158.
Hyperbola, equations of, 132-
135.
construction of, 138.
equilateral, 143.
Hjrperbolic paraboloid, 218.
Hyperboloid, of one sheet, 215.
of two sheets, 216.
Hypocycloid, 192.
of four cusps, 193.
Inclination of a Une, 7.
Intercepts, 27, 77, 212.
Intersection of loci,
• rectangular coordinates, 40.
polar coordinates, 87.
in space, 211.
Latus rectum, of parabola, 113.
of ellipse, 124.
of hyperbola, 135.
Locus of an equation,
in rectangular coordinates, 22.
in polar coordinates, 77.
in space, 212.
Locus by factoring, 42.
Locus through intersections of
two loci, 44.
Logarithmic curve, 179.
Multiple representation in polar
coordinates, 83.
Normal equation of a line, 54.
reduction to normal form, 56.
Normal to a curve, definition,
146.
equation of, 149.
length of, 152.
Ordinate, 2.
Parabola, equations of, 112-114.
construction of, 117.
Parallel lines, condition for, 8.
Parametric equations, 186, 190.
Periodic functions, 183.
Perpendicular lines, condition for,
9.
Plane, equations of, 220-223.
Plotting, in rectangular coordi-
nates, 23.
in polar coordinates, 77.
straight Unes, 53.
Point of division, 11, 200.
Polar coordinates, 73.
Pole and polar, definition, 159.
equation, 159-161.
important properties, 161.
Projectiles, 193.
Projections, 201.
Quadric surfaces, 214.
Radical axis, 103.
Rotation of axes, 92.
Sine curve, 181, 183..
Slope of a hne, 8.
Spherical surface, 208.
Spirals, 85.
Straight line, equations of, 47-51,
54.
in space, 225-227.
systems of, 67.
Subnormal length, 152.
Subtangent length, 152.
Surfaces of revolution, 209.
Symmetry, 28, 77, 212.
Tangent curve, 184.
Tangent to a curve, definition,
146.
INDEX 245
Tangent to a curve, equation in Transcendental equations, 177.
terms of point of contact. Transformation of coordinates, 90,
145-154. 97.
equation in terms of slope. Translation of axes, 91.
155.
length of, 152. Witch of Agnesi, 194.
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