f A COURSE OF PURE GEOMETRY CAMBRIDGE UNIVERSITY PRESS C. F. CLAY, Manager LONDON Fetter Lane, E.C. 4 EDINBURGH 100 Princes Street NEW YORK: G. P. PUTNAM'S SONS BOMBAY, CALCUTTA, MADRAS: MACMILLAN AND CO., Ltd. TORONTO : J. M. DENT AND SONS, Ltd. TOKYO: THE MARUZEN-KABUSHIKI-KAISHA All rights reserved A COURSE OF PURE GEOMETRY CONTAINING A COMPLETE GEOMETRICAL TREATMENT OF THE PROPERTIES OF THE CONIC SECTIONS BY E. H. ASKWITH, D.D. Rector of Dickleburgh, Norfolk formerly Chaplain of Trinity College, Cambridge Cambridge at the University Press 1917 h n X. First Edition 1903 Reprinted 1911 Netc Edition 1917 PREFACE npHIS work is a revision and enlargement of ray Course of Pure Geometry published in 1903. It differs from the former edition in that it does not assume any previous know- ledge of the Conic Sections, which are here treated ah initio, on the basis of the definition of them as the curves of projection of a circle. This is not the starting point of the subject generally in works on Geometrical Conic Sections. The curves are usually defined by means of their focus and directrix property, and their other properties are evolved therefrom. Here the focus and directrix propert}^ is established as one belonging to the projections of a circle and it is freely used, but the fact that the conies are derived by projection from a circle and therefore possess all its projective properties is kept constantl}^ in the mind of the student. Many of the properties of the Conic Sections which can only be established with great labour from their focus and directrix property are proved quite simply when the curves are derived directly from the circle. Nor is the method employed here any more difficult than the prevalent one, though it is true that certain ground has to be covered first. But this is not very extensive and I have indicated (p. xii) the few articles which a Student should master before he proceeds to Chapter ix. Without a certain know- ledge of cross ratios, harmonic section, involution and the /> A /\ C\ *\ *% " elementary principles of conical projection no one can follow the argument here adopted. But these things are <piite easy, and the advantage gained by the student who from the be- ginning sees the Conic Sections whole, as he does when they are presented to his mind as the projections of a circle, more than compensates for any delay there may be through the short study of the necessary preliminaries. I hope that, thanks to the efficiency of the Readers of the Cambridge University Press, there are not many misprints to be found in this book. But if any are found I shall be grateful if I may be informed of the)u at the address given below. E. H. A. DiCKLEBURGH ReCTORY, SCOLE, Norfolk. September 1917. CONTENTS CHAPTER I SOME PROPERTIES OF THE TRIANGLE PAGE Definitions 1 Orthocentre 2 Nine-points circle ....... 3 Pedal line 6 Medians 8 CHAPTER II SOME PROPERTIES OF CIRCLES Poles and polars . . . . . .13 Conjugate points and lines ........ 15 Radical axis 17 Coaxal circles 20 Coninion tangents of two circles ....... 22 CHAPTER III THE USE OF SIGNS. CONCURRENCE AND COLLINEARITY Signs of lines, areas, angles ........ 28 Menelaus' and Ceva's theorems 31 Isogonal conjugates 36 Symmedians ...... ... 37 CHAPTER IV PROJECTION Ceneral principles .......... 41 Projection of angles into angles of given magnitude ... 42 One range of three points projective with any other .. . .45 Vlll CONTENTS CHAPTER V CEOSS-EATIOS PAGE Definition 47 Twenty-four cross-ratios reducible to six ..... 48 Projective property of cross-ratios 50 Equi-cross ranges and pencils mutually projective .... 54 CHAPTER VI PEESPECTIVE Definition ............ 58 Ranges and pencils in perspective '.60 nomographic ranges and pencils 62 Triangles in perspective 64 CHAPTER VII HARMONIC SECTION Definition and properties of harmonic ranges .... 71 Harmonic property of pole and polar of circle .... 74 Harmonic property of quadrilateral and quadrangle ... 75 CHAPTER VIII INVOLUTION Definition and criterion of involution range 80 Involution projective 83 Involution' properties of the circk' . . . . . . .84 Orthogonal involution 85 Pair of orthogonal X'ays in every involution pencil .... 86 CHAPTER IX THE CONIC SECTIONS Definitions . 90 Focus and directrix property ........ 91 Projective projjerties 92 Circle projected into another circle 93 Focus and directrix as pole and polar 94 CONTENTS IX PAGE Parallel chords 95 Focus and directrix property established 96 (1) Parabola 97 (2) Ellipse 101 (3) Hyperbola 102 Diameters and ordinates 106 CHAPTER X PEOPEBTIES COMMON TO ALL CONICS Intersection of chord and tangent with directrix .... Curves having focus and directrix property are the projections of a circle Pair of tangents. The Normal Latus rectum Carnot's theorem Newton's theorem Some applications Circle of curvature Conic through foiu- points of a quadrangl 108 110 111 113 114 116 117 US 120 121 CHAPTER XI THE PARABOLA Elementary properties 126 Tangent and normal . . . . . . . 127 Pair of tangents 130 Parabola escribed to a triangle 132 Diameters 134 Circle of curvature 138 Sum of focal distance; Tangent and normal Pair of tangents Director circle . Conjugate diameters Auxiliary circle . Equi conjugate diameters Circle of curvature CHAPTER XII THE ELLIPSE constant 144 145 150 150 151 153 156 158 X CONTENTS CHAPTER XIII THE HYPERBOLA I'ACiE Form of curve 163 Difference of focal distances constant 164 Tangent and normal 164 On the length of the conjugate axis . . . . . .166 Pair of tangents . . 168 Director circle 169 The conjugate hyperbola 170 Asymptotic properties 171 Conjugate diameters 174 Circle of curvature . . . . . . . . . .187 CHAPTER XIV THE EECTANGULAR HYPEEBOLA Conjugate diameters 192 Perpendicular diameters . , . 193 Rectangular hyperbola circumscribing a triangle .... 194 Chord and tangent properties 196 CHAPTER XV OllTHOGONAL PEOJECTION Principles 201 Fundamental propositions 202 The ellipse aa orthogonal projection of a circle .... 205 CHAPTER XVI CROSS-EATIO PEOPEETIES OF CONICS P {ABCD) constant 210 Pascal's theorem 214 Brianchon's theorem 215 Locus of centres of conies through four points .... 215 Involution range on a conic .216 CHAPTER XVII EECIPEOCATION Principles 220 Involution properties of quadrangle and quadrilateral . . . 224 Desargues' theorem and its reciprocal 227 CONTENTS XI • PAGE Reciprocation applied to conies ....... 228 Special case where the base conic is a circle 231 Coaxal circles reciprocated into confocal conies .... 234 A pair of self-conjugate triangles ....... 236 Reciprocal triangles .......... 237 CfTAPTER XVIII CmCULAE POINTS. FOCI OF CONICS Definition of circular points ........ 242 Analytical point of view ......... 243 Properties of conies obtained by using circular points . . . 244 The four foci of a conic ........ 246 Two triangles circumscribing a conic ...... 248 Generalising by projection ....... 249 CHAPTER XIX INVERSION Inversion of line and circle 256 Inversion of sphere ....... . . 258 Inversion of inverse points into inver.se points .... 259 Feuerbach's theorem .......... 262 CHAPTER XX SIMILARITY OF FIGURES Horaothetic figures .......... 267 Figiu-es directly similar but not homothetic 270 Circle of similitude for two circles 271 Figures inversely similar ......... 272 Miscellaneous E.xamples 276 Index .......' 285 J The student who may be using this work as a firsi?' t' . xl book on Geometrical Conic Sections will be able to j^ro^^^r" to Chapter IX after reading the following paragraphs of tW'i'Tst eight chapters : 13 to 16a, 29 to 35, 40 to 45, 48 to 53, 58, 67, 68, 69 ^- "'^ 77 to 87. CHAPTER I SOME PROPERTIES OF THE TRIANGLE 1 Definition of terms. iy ^1/(66% unless otherwise stated, will be meant straight 4. , The lines joining the vertices of a triangle to the middle ' the opposite sides are called its medians. (c) 'y the circumcircle of a triangle is meant the circle •^ps<5ing through its vertices. xhe centre of this circle will be called the circiimcentre of the triangle. The reader already knows that the circumcentre is the point of intersection of the perpendiculars to the sides of the tiiangle drawn through their middle points. (a) The incircle of a triangle is the circle touching the sides of th? triangle and lying within the triangle. The centre of this circle is the incentre of the triangle. The incentre is the point of intersection of the lines bisecting the angles of the triangle. (e) An ecv.xle of a triangle is a circle touching one side of *"' the other fewo sides produced. There are three ..tie of an ecircle is called an ecentre. An ecentre is the point of intersection of the bisector of (•i.e of tb^ angles and of the bisectors of the other two external angles. A. G. 1 1^ Sd.VK, PPOI'ERTIES OF THE TRIANGLE (/) Two triangles which are such that the sides and angles of the one are equal respectively to the sides and angles of the other will be called congruent. If ABC be congruent with A'B'C, we shall express the fact by the notation : A ABO = A A'B'C. 2. Proposition. 'The peiyendiculars from the vertices of a triangle on to the opposite sides meet in a point (called tJie orthocentre) ; and the distance of each vertex from the ortho- centre is trvice the perpendicular distance of the circumcentre from the side opposite to that vertex. Through the vertices of the triangle ^Z?6' draw lines parallel to the opposite sides. The triangle A'B'C thus formed will be similar to the triangle ABC, and of double its linear dimensions. Moreover, A, B, G being the middle points of the sides of A'B'C, the perpendiculars from these points to the sides on which they lie will meet in the circumcentre of A'B'C. But these perpendiculars are also the perpendiculars from A, B, C to the opposite sides of the triangle ABC Hence the first part of our proposition is proved. SOME PROPERTIES OF THE TRIANGLE 3 Now let P be the orthocentre and the circumcentre of ABC. Draw OD perpendicular to EC. Then since P is also the circumcentre of the triangle A'B'C. PA and OD are corresponding lines in the two similar triangles A'B'C, ABC. Hence ^P is twice OD. 3. Definition. It will be convenient to speak of the perpendiculars from the vertices on to the opposite sides of a triangle as the peiyendiculars of the triangle; and of the perpendiculars from the circumcentre on to the sides as the perpendiculars from the circumcentre. 4. Prop. The circle through the middle points of the sides of a triangle passes also through the feet of the perpendiculars of the triangle and through the middle points of the three lines joining the ortliocentre to the vertices of the triangle. Let D, E, F be the middle points of the sides of the triangle ABC, L, M, N the feet of its perpendiculars, the circum- centre, P the orthocentre. Join FD, DE, FL, LE. Then since E is the circumcentre of ALC, /.ELA= ZEAL. 1—2 SOME PROPERTIES OF THE TRIANGLE And for a like reason ZFLA= ZFAL. .-. zFLE= zFAE = Z FDE since AFDE is a parallelogram. .•. L is on the circumcircle of DEF. Similarly M and N lie on this circle. Further the centre of this circle lies on each of the three lines bisecting DL, EM, FN at right angles. Therefore the centre of the circle is at U the middle point, of OP. Now join DU and produce it to meet AP in X. The two triangles OUD, PUX are easily seen to be con- gr uent, so that UD = UX and XP = OD. Hence X lies on the circle through D, E, F, L, M, N. And since XP ^ (jD = \AP, X is the middle point of AP. Similarly the circle goes through Y and Z, the middle points of BP and CP. Thus our proposition is proved. 5. The circle thus defined is known as the nine-points circle of the triangle. Its radius is half that of the circumcircle, as is obvious from the fact that the nine-points circle is the circum- circle of DEF, which is similar to ABC and of half its linear SOME PROPERTIES OF THE TRIANGLE 5 diuiensions. Or the same may be seen from our figure wherein DX = OA, for ODXA is a parallelogram. It will be proved in the chapter on Inversion that the nine- points circle touches the incircle and the three ecircles of the triangle. 6. Prop. If the perpendicular AL of a tiiangle ABC he produced to meet the circumcircle in H, then PL = LH, P being the orthocentre. Join BH. Then Z HBL = z HAC in the same segment = Z LBP since each is the complement of /LACB. Thus the triangles PBL, HBL have their angles at B equal, also their right angles at L equal, and the side BL common. \-.PL = LH. 7. Prop. The feet of the perpendiculars fi^om any point Q on the circumcircle of a triangle ABC on to the sides of the triangle are collinear. Let R, S, T be the feet of the perpendiculars as in the figure. Join QA, QB. H SOME PROPERTIES OF THE TRIANGLE Q'fAS is a cyclic quadi'ilateral since T and S are right angles. .-. ^AT>S = zAQS = complement of Z QAS = complement of zQBG (since QAiJ, QBC aie supplementary) = z BQR = z BTR (since QBRT is cyclic). .'. RTS is a straight line. This line RTS is called the pedal line of the point Q. It is known also as the Simson line. The converse of this proposition also holds good, viz. : If the feet of the perpendiculars from a point Q on to the sides of a triangle are collinear, Q lies on the circumcircle of the triangle. For since QBRT and QTAS are cyclic, z BQR = z BTR = zATS = z AQS. .-. z QBR = z QAS, so that QBC A is cyclic. SOME PROPERTIES OF THE TRIANGLE 7 8. Prop. The pedal line of Q bisects the line joining Q to P, the orthocentve of the triangle. Join QP cutting the pedal line of Q in K. Let the perpendicular AL meet the circunicircle in H. Join QH cutting the pedal line in M and BC in N. Join PN and QB. Tlien since QBRT is cyclic, zQRT^zQBT = Z QUA in same segment = Z HQR since QR is parallel to J H. .'. QM = iMR. .-. J/ i> the middle point of QN. But Z PiVZ .= z ZiV^ since A 7\VZ = A HNL = z RNM = z i¥iei\^. .-. PA^ is parallel to /?7'. .-. QK'.KP = QM.MN. .-. QK = KP. 8 SOME PROPERTIES OF THE TRIANOLK 9. Prop. The three luedians of a triangle meet in a point, and this point is a point of trisection of each niedian, and also of the line joining the circumcentre and the orthocentre P. Let the median AB o{ the triangle ABC cut OP in G. Then from the similarity of the triangles GAP, GDO, we deduce, since AP = 20D, that AG = 2GB and PG = 2G0. Thus the median AD cuts OP in G which is a point of trisection of both lines. Similarly the other medians cut OP in the same point G, which will be a point of trisection of them also. This point G is called the median jmint of the triangle. The reader is probably already familiar with this point as the centroid of the triangle. 10. Prop. If AD he a median of the triangle ABC, then AB' + AC^^= 2 AD' + 2BD\ Draw AL perpendicular to BC. Then AG-' = A B' + BC - 2BC . BL and AD' = AB' + BD'-2BD.BL. These equalities include the cases where both the angles B and C are acute, and where one of them, B, is obtuse, provided SOME PROPERTIES OF THE* TRIANGLE 9 that BG and BL be considered to have the same or opposite signs according as they are in the same or opposite directions. Multiply the second equation by 2 and subtract from the first, then AC' - 2 AD' = BC - AB' - -IJWi . • . AB'+ AC' = 2 A /)-■ + BC - 2 BD' = 2AD'' + 2BD-, since BC = 2BD. 11. The proposition proved in the last article is only a special case of the following general one : //' D he a point in the side BC of a triawjle ABC such that lU) = ^ BC then n (n - 1) AB"- ■vAC- = n. AD' + (l - ^) BC. For proceeding as before, if we now multiply the second of the equations by n and subtract from the first we get A C -n.AD'={\-n)AB' + BC - n . BD\ .-. (71 -\)AB' + AC'=n. AD' + BC - n (^ BC = n.AD' + (l-^\BC 12. Prop. The distances of the points of contact of the incircle of a triangle ABC ivith the sides from the vertices A, B,C are s — a, s — b,s — c respectively; and the distances of the points of contact of the ecircle opposite to A are s, s — c. 10 SOME PROPERTIES OF THE TRIANGLE .V -b respective/ 1/ ; a, b, c being the lengths of the sides apposite to A , B, G and s half the sum of them. ■ Let the points of contact of the incircle be L, M, N. Then since AM=AN', CL = CM and BL = BN, .•. AM+ BC = h&,U the sura of the sides = s, .-. AM = s -a. Similarly BL^BN^s-b, and GL = CM=s-c. Next let L', M', N' be the points of contact of the ecircle opposite to A. Then • AN' = AB-\-BN' = AB + BL' and AM' = AG + GM' = AG + GL. .-. since AM' =^AN', 2AN' = AB + AG+BG=2s. .'. AN' = s, and BL' = BN' = s-c, and GL' = GM' =s-b. Cor. BL' = GL, and thus LIJ and BG have the same middle point. SOME PROPERTIES OF THE TRIANGLE EXERCISES 1. Defining the pedal triangle as that formed by joining the feet of the perpendiculars of a triangle, shew that the pedal triangle has for its incentre the orthocentre of the original triangle, and that its angles are the supplements of twice the angles of the triangle. 2. A straight line PQ is drawn i)arallel to AB to meet the circumcircle of the triangle ABC in the points P and Q, shew that the pedal lines of P and Q intersect on the perpendicular from C on AB. 3. Shew that tlie pedal lines of three points on the circumcircle of a triangle form a triangle similar to that formed by the three points. 4. The j^edal lines of the extremities of a chord of the circum- circle of a triangle intersect at a constant angle. Find the locus of the middle point of the chord. 5. Given the circumcircle of a triangle and two of its vertices, prove that the loci of its orthocentre, centroid and nine-points centre are circles. 6. The locus of a point which is such that the sum of tlie squares of its distances from two given points is constant is a sphere. 7. A\ B\ C are three points on the sides BC, CM, AB of a triangle ABC. Prove that the circumcentres of the triangles ABC, BC'A', CA'B' are the angular points of a triangle which is similar to ABC. 8. A circle is described concentric with the circumcircle oi the triangle ^^C, and it intercepts chords A^A.,, B^Bo, C^Co on BC, ('A, AB respectively; from Ai perpendiculars A^b^, A^c^ are drawn to CA, AB respectively, and from A.^, i^,, B.,, C,, Co similar perpen- diculars are drawn. Shew that the circumcentres of the si.\ triangles, of which Ab^Ci is a typical one, lie on a circle concentric with the nine-points circle, and of radius one-half that of the original circle. 9. A plane quadrilateral is divided into four triangles b\- its internal diagonals ; shew that the quadrilaterals having for angular points (i) the orthocentres and (ii) the circumcentres of the foui- 12 SOME I'H(>pp:hties of the triangle triangles are similar parallelograms; and if their areas be Aj and Ao, and A be that of the quadrilateral, then 2A + A^ = 4Ao. 10. Prove that the line joining the vertex of a triangle to that point of the inscribed circle which is farthest from the base passes through the point of contact of the escribed circle with the base. 11. Given in magnitude and position the lines joining the vei'tex of a triangle to the points in which the inscribed circle and the circle escribed to the base touch the base, construct the triangle. 12. Prove that when four points A, B, C, D lie on a circle, the orthocentres of the triangles BCD, CDA, BAB, ABC lie on an equal circle. 13. Prove that the pedal lines of the extremities of a diameter of the cii'cumcircle of a triangle intersect at right angles on the nine-points circle. 14. ABC is a triangle, its circumcentre ; OB perpendicular to BC meets the circumcircle in K. Prove that the line through B perpendicular to AK will bisect KP, P being the orthocentre. 15. Having given the circumcircle and one angular point of a triangle and also the lengths of the lines joining this point to the orthocentre and centre of gravity, construct the triangle. 16. If AB be divided at in such a manner that I. AO = m. OB, and if P be any point, prove / . AP-' + m . BP- = {l + m) OP- + I . AO' + m . BO'. If a, b, c be the lengths of the sides of a triangle ABC, find the locus of a point P such that a . PA^ + b . PB'^ + c . PC" is constant. 13 CHAPTER II SOME PROPERTIEH OF CIRCLES 13. Definition. When two points P and F' lie on the saine radius of a circle whose centre is and are on the same side of and their distances from are such that <JP . OP' = square of the radius, they are called inverse points with respect to the circle. The reader can already prove fur himself that if a pair of tangents be drawn from an external point P to a circle, centre 0, the chord joining the points of contact of these tangents is at right angles to OP, and cuts OP in a point which is the inverse of P. 14. The following proposition will give the definition of the polar of a point with respect to a circle : Prop. The locus of the points of intersection of pairs of tangents drawn at the extremities of chords of a circle, which pass through a fixed point, is a straight line, called the polar of that point, and the point is called the pole of the line. Let A be a fixed point in the plane of a circle, centre 0. Draw any chord QR of the circle to pass through A. Let the tangents at Q and R meet in P. Draw PL perpendicular to OA. Let OP cut QR at right angles in M. 14 SOME PROPERTIES OF CIRCLES Then PMLA is cyclic. .-. OL.OA = OM .OP = sqnAYO of radius. .•. /y is a fixed })()int, viz. the inverse of A. Thus the locus of F is a straight line perpendicular to OA, and cutting it in the inverse point of A. 15. It is clear from the above that the polar of an external point coincides with the chord of contact of the tangents from that point. And if we introduce the notion of imaginary lines, with Avhich Analytical Geometry has furnished us,, we may say that the polar of a point coincides with the chord of contact of tangents real or imaginarj^ from that point. We may remark here that the polar of a point on the circle is the tangent at that point. Some writers define the polar of a point as the chord of contact of the tangents drawn from that point ; others again define it by means of its harmonic property, which will be given in a later chapter.' It is unfortunate that this difference of treatment prevails. The present writer is of opinion that the method he has here adopted is the best. 16. Prop. //■ the polar of A goes through B, then the polar of B goes through A. Let BL be the polar of A cutting 0x4 at right angles in L. Draw AM at right angles to OB. SOME PROPERTIES OF CIRCLES Then OM.OB=OL.OA = sq. of radiiu .-. AM \s the polar of B, that is. A lies on the polar o{ B. 15 Two points such that the polar of each goes throuoji thi' other are called conjugate points. The reader will see for himself that inverse points with respect to a circle are a special case of conjugate points. We leave it as an exercise for the student to prove that if /, VI be two lines such that the pole of / lies on ni, then the pole of m will lie on /. Two such lines are called conjuc/ate lines. From the above property for conjugate points we see that the polars of a number of coUinear points all pass through a common point, viz. the pole of the line on which they lie. For if A, B, C, D, &c., be points on a line p whose pole is P ; since the polar of P goes through A, B, C, kc, .'. the polars of A, B, C, kc, go through P. We observe that the intersection of the polars of two points is the pole of the line joining them. IG SOME PROPERTIES OF CIRCLES 16a. Prop. //" OP and OQ be a pair of conjugate lines of a circle whic/t meet the polar of in P and Q, then the triangle OPQ is such that each vertex is the pole of the opposite side, and the centre of tJie circle is the orthocentre of the triamgle. For the pole of OQ must lie on the polar of 0, and it also lies on OP, since OP and OQ are conjugate lines. Thus OQ is the polar of P. Similarly OP is the polar of Q. Q Also the lines joining C the centre to 0, P, Q are perpen- dicular respectively to the polars of those points, and therefore C is the orthocentre of the triangle. 17. Prop. If P and Q he any two poiiits in the plane of a circle whose centre is 0, then OP : OQ=perp. from P on polar of Q : perp. from Q on polar of P. Let P' and Q' be the inverse points of P and Q, through which the polars of P and Q pass. Let the perpendiculars on the polars be PM and QN ; draw PT and QR perp. to OQ and OP respectively. Then we have OP . OP' = OQ . OQ', since each is the square of the radius, and OR . OP = 0T.OQ since PRQT is cyclic, OQ' _OP_OZ- OQ' -or ^ PM ■'■OP'~OQ~OR~OP' - OR ~ QN ' SOME PROPERTIES OF CIRCLES Thus the proposition is proved. 17 p N/>^ / ^ / /t \ 1 ^ y \ This is known as Salmons theorem. 18. Prop. The locus of points from which the tangents to two given coplanar circles are equal is-a line (called the radical axis of the circles) perpendicular to the line of centres. Let PK, PF be equal tangents to two circles, centres A and B. Draw PL perp to AB. Join PA, PB, AK and BF. Then PK' = AP' - AK' = PL' + AD - AK\ and PF' = I'B' - BF' = PfJ + LB' - BF', ..AL'-AK' = LB'-BF', .-. AL' - LB' = AK' - BF', .■ . {AL - LB) {AL + LB) = AK'- BF'. Thus if be the middle point of AB, we have 20iy . AB = difterence of sqq. of the radii, .-. L is a fixed point, and the locus of P is a line perp. to AB. 18 SOME PROPERTIES OF CIRCLES Since points on the common chord produced of two inter- secting circles are such that tangents from them to the two circles are equal, we see that the radical axis of two intersecting circles goes through their common points And introducing the notion of imaginary points, we may say that the radical axis of two circles goes through their common points, real or imaginary. 19. The difference of the squares of the tangents to two coplanar circles, from any point P in their plane, varies as the perpendicular from P on their radical axis. Let PQ and PR be the tangents from P to the circles, centres A and B. Let PN be perp. to radical axis NL, and PM to AB; let be the middle point of AB. Join PA, PB. Then PQ'- PR' = PA' -AQ'- (PB' - BR') = PA' - PB' - AQ' + BR' = AM'-MB'-AQ' + BR' ^203I.AB-20L.AB (see § 18) = 2AB.LA[=2AB.NP. SOME PROPERTIES OF CIRCLES This proves the proposition. 19 We may mention here that some writers use the term " power of a point " with respect to a circle to mean the square of the tangent from the point to the circle. 20. Prop. The radical axes of three coplanar circles taken in pairs meet in a point. Let the radical axis of the circles A and B meet that of the circles A and C in P. 2—2 20 SOME PROPERTIES OF CIRCLES Then the tangent from P to circle G = tangent from P to circle A = tangent from P to circle B. .• . P is on the radical axis of B and C. 21. Coaxal circles. A system of coplanar circles such that the radical axis for any pair of them is the same is called coaxal. Clearly such circles will all have their centres along the same straight line. Let the common radical axis of a system of coaxal circles cut their line of centres in A. Then the tangents from A to all the circles will be equal. Let L, L be two points on the line of centres on opposite sides of A, such that AL, AL' are equal in length to the tangents from A to the circles ; L and L' are called the limiting points of the system. , They are such that the distance of any point P on the radical axis from either of them is equal to the length of the tangent from P to the system of circles. SOME PROPERTIES OF CIRCLES 21 For if C be the centre of one of the circles which is of radius r, PD = PA' + AD = PA^' +AG''-r''- = PC - r^ = square of tangent, from P to circle C The two points L and L' may be regarded as the centres of circles of infinitely small radius, which belong to the coaxal system. They are sometimes called the i^omt circles of the system. The student will have no difficulty in satisfying himself that of the two limiting points one is within and the other without each circle of the system. It must be observed that the limiting points are real only in the case where the system of coaxal circles do not intersect in real points. For if the circles intersect, A will lie within them all and thus the tangents from A will be imaginary. Let it be noticed that if two circles of a coaxal system inter- sect in points P and Q, then all the circles of the system pass through P and Q. 22. Prop. The liniitiny points of a system of coaxal circles are inverse points ivith respect to every circle of the si/stein. Let C be the centre of one of the circles of the system. Let L and L' be the limiting points of which L' is without the circle C. 22 SOME PROPERTIES OF CIRCLES Draw tangent L'T to circle C; this will be bisected by the radical axis in P. Draw TN perpendicular to line of centres. Then L'A:AN=L'F:PT, .-. L'A = AN, .'. N coincides with L. Thus the chord of contact of tangents from L' cuts the line of centres at right angles in L. Therefore L and L' are inverse points. 23. The student will find it quite easy to establish the two following propositions : Eve7y circle passing through the limiting points cuts all the circles of the system orthogonally. A common tangent to two circles of a coaxal system subtends a right angle at either limiting jmint. 24. Common tangents to two circles. In general four common tangents can be drawn to two coplanar circles. Of these two will cut the line joining their centres ex- ternally; these are called direct common tangents. And two will cut the line joining the centres internally ; these are called transverse common tangents. P We shall now prove that the common tangents of two circles cut the line joining their centres in ttuo points which divide that line internally and externally in the ratio of the radii. Let a direct common tangent PQ cut the line joining the centres A and B in 0. Join AF, BQ. SOME PROPERTIES OF CIRCLES 28 Then since P and Q are right angles, the triangles APO, BQO are similar, .-.AO-.BO^AP.BQ. Similarly, if P'Q' be a transverse common tangent cutting AB in 0', we can prove AO' : O'B = ratio of the radii. We have thus a simple construction for drawing the common tangents, viz. to divide AB internally and externally at 0' and in the ratio of the radii, and then from and 0' to draw a tangent to either circle ; this will be also a tangent to the other circle. If the circles intersect in real points, the tangents from 0' will be imaginary. If one circle lie wholly within the other, the tangents from both and 0' will be imaginary. 25. Through the point 0, as defined at the end of the last paragraph, let a line be drawn cutting the circles in RS and R'S' as in the figure. Consider the triangles OAR, OBR'. We have OA:OB = AR: BR', also the angle at is common to b(jth, and each of the remaining angles at R and R' is less than a right angle. Thus the triangles are similar, and OR:OR'=AR:BR', the ratio of the radii. In like manner, by considering the triangles OAS, OBS', in which each of the angles S and *S" is greater than a right angle, we can prove that OS: 0/Sf' = ratio of radii. 24 SOME PROPERTIES OF CIRCLES We thus see that the circle B could be constructed from the circle A by means of the point by taking the radii vectores from of all the points on the circle A and dividing these in the ratio of the radii. On account of this property is called a centre of similitude of the two circles, and the point R' is said to correspond to the point B. The student can prove for himself in like manner that 0' is a centre of similitude. 26. In order to prove that the locus of a point obeying some given law is a circle, it is often convenient to make use of the ideas of the last paragraph. If we can prove that our point P is such as to divide the line joining a fixed point to a varying point Q, which describes a circle, in a given ratio, then we know that the locus of P must be a circle, which with the circle on which Q lies has for a centre of similitude. For example, suppose we have given the circumcircle of a triangle and two of its vertices, and we require the locus of the nine-points centre. It is quite easy to prove that the locus of the orthocentre is a circle, and from this it follows that the locus of the nine-points centre is a circle, since, if be the circumcentre (which is given) and P the orthocentre (which describes a circle) and U the nine-points centre, U lies on OP and OU = \0P ; therefore the locus of U is a circle, having its centre in the line joining to the centre of the circle on which P lies. 27. Prop. The locus of a point which moves in a plane so that its distances from two fixed points in that plane are in a constant ratio is a circle. Let A and B be the two given points. Divide AB internally and externally at C and D in the given ratio, so that C and D are two points on the locus. Let P be any other point on the locus. SOME PROPERTIES OF CIRCLES 25 Then since AP:P£ = AC:GB=AD:BD, .•. PC and PD are the internal and external bisectors of the /-APB. .'. CPD is a right angle. .•. the locus of P is a circle on CB as diameter. Cor. 1. If the point P be not confined to a plane, its locus is the sphere on CD as diameter. Cor. 2. If the line AB be divided internally and externally at C and D in the same ratio, and P be any point at which CD subtends a right angle, then PC and PD are the internal and external bisectors of Z APB. 28. If on the line 00' joining the two centres of similitude of circles, centres A and B, as defined in § 25, a circle be described, it follows from § 27 that if C be any point on this circle, CA : CB = radius of A circle : radius of B circle. The circle on 00' as diameter is called the circle of sindli- tude. Its use will be explained in the last chapter, when we treat of the similarity of figures. 26 SOME PROPERTIES OF CIRCLES EXERCISES 1. If P be any point on a given circle A, the square of the tangent from P to another given circle B varies as the perpendicular distance of P from the radical axis of A and B. 2. If A, B, be three coaxal circles, the tangents drawn from any point of C to A and B are in a constant ratio. 3. If tangents drawn from a point P to two given circles A and B are in a given ratio, the locus of P is a circle coaxal with A and B. 4. If ^, jB, C &c. be a system of coaxal circles and X be any other circle, then the radical axes oi A, X ; B, X ; C, X &c. meet in a point. 5. The square of the line joining one of the limiting points of a coaxal system of circles to a point P on any one of the circles varies as the distance of P from the radical axis. 6. If two circles cut two others orthogonally, the radical axis of either pair is the line joining the centres of the other pair, and passes through their limiting points. 7. If from any point on the circle of similitude (§ 28) of two given circles, pairs of tangents be drawn to both circles, the angle between one pair is equal to the angle between the other pair. 8. The three cii'cles of similitude of three given circles taken in pairs are coaxal. 9. Find a pair of points on a given circle concyclic with each of two given pairs of points. 10. If any line cut two given circles in P, Q and F', Q' respectively, prove that the four points in which the tangents at F and Q cut the tangents at F' and Q' lie on a circle coaxal with the given circles. 11. A line FQ is drawn touching at P a circle of a coaxal system of which the limiting points are K, K', and Q is a point on the line on the opposite side of the radical axis to P. Shew that if T, T' be the lengths of the tangents drawn from F to the two con- centric circles of which the common centre is Q^ and whose radii are respectively QK, QK\ then T:T' =PK.FK'. SOME PROPERTIES OF CIRCLES 27 12. is a fixed point on the circumference of a circle C, P any other point on C ; the inverse point ^ of f is taken with respect to a fixed circle whose centre is at 0, prove that the locus of Q is a straight line. 13. Three circles Cj, Co, C^ are such that the chord of inter- section of Ca and C'g passes through the centre of C^ , and the chord of intersection of C\ and Cj through the centre of Co ; shew that the chord of intersection of Cj and C.j passes through the centre of CV 14. Three circles A, B, C are touched externally by a circle whose centre is P and internally by a circle whose centre is Q. Shew that PQ passes through the point of concurrence of the radical axes oi A, B, C taken in pairs. 15. AB is a. diameter of a circle S, any point on AB or AB produced, C a circle whose centre is at 0. A' and B' are the inverse points of A and B with respect to C. Prove that the pole with respect to C of the polar with respect to S of the point is the middle point of A'B'. 16. A system of spheres touch a plane at the same point 0, prove that any plane, not through 0, will cut them in a system of coaxal circles. 17. A point and its polar with respect to a variable circle being given, prove that the polar of any other point A passes through a fixed point B. 18. .4 is a given point in the plane of a system of coaxal circles ; prove that the polars of A with respect to the circles of the system all pass through a fixed point. 28 CHAPTER III THE USE OF SIGNS. CONCURRENCE AND COLLINEARITY 29. The reader is already familiar with the convention of signs adopted in Trigonometry and Analytical Geometry in the measurement of straight lines. According to this convention lengths measured along a line from a point are counted positive or negative according as they proceed in the one or the other direction. With this convention we see that, if A, B, C he three points in a line, then, in whatever order the points occur in the line, AB+BC = AG. AC B If C lie between A and B, BC is of opposite sign to AB, and in this case AB + BC does not give the actual distance travelled in passing from A to B, and then from B to G, but gives the final distance reached from A. From the above equation we get BC = AC-AB. This is an important identity. By means of it we can reduce all our lengths to depend on lengths measured from a fixed point in the line. This process it will be convenient to speak of as inserting an origin. Thus, if we insert the origin 0, AB^OB-OA. THE USE OF SIGNS. CONCURRENCE AND COLLINEARITY 29 30. Prop. If M he the middle point of the line AB, and be any other point in the line, then 20M=0A + 0B. O A M B For since AM=MB, by inserting the origin we have OM-OA = OB-OM, .-. 20M=0A+0B. 31. A number of collinear points are said to form a range. Prop. If A, B, G, D he a range of four points; then AB.CD + BC.AD+CA.BD = 0. A BCD For, inserting the origin A, we see that the above == AB{AD - AC) + {AG - AB) AD - AC {AD - AB), and this is zero. This is an important identity, which we shall use later on. 32. If A, B, C be a range of points, and any point out- side their line, we know that the area of the triangle OAB is to the area of the triangle OBC in the ratio of the lengths of the bases AB, BG. O Now if we are taking account of the signs of our lengths AB, BG and the ratio AB : BG occurs, we cannot substitute for this ratio A OAB : A 0^6* unless we have some convention respecting 30 THE USE OF SIGNS the signs of our areas, whereby the proper sign of AB : BC will be retained when the ratio of the areas is substituted for it. The obvious convention is that the area of a triangle PQR shall be accounted positive or negative according as the triangle is to the one or the other side as the contour PQR is described. Thus if the triangle is to our left hand as we describe the contour PQR, we shall consider A PQR to be a positive mag- nitude, while APRQ will be a negative magnitude, for in describing the contour PRQ the area is on our right hand. With this convention we see that in whatever order the points A, B, C occur in the line on which they lie, AB:BC=AOAB:AOBG, or = AAOB-.ABOG. It is further clear that with our convention we may say AOAB+ AOBC= AOAC, and AOAB- AOAC= AOCB, remembering always that A, B, G are coUinear. 33. Again, we know that the magnitude of the area of a triangle OAB is ^OA .OB sin AOB, and it is sometimes con- venient to make use of this value. But if we are comparing the areas OAB, OBC by means of a ratio we cannot substitute ^OA .OB sin AOB and ^ OB. OC sin BOG I BPA for them unless we have a further convention of signs whereby the sign and not merely the magnitude of our ratio will be retained. The obvious convention here again will be to consider angles positive if described in one sense and negative in the opposite sense; this being effective for our purpose, since sin(— ^r) = — sin x. CONCURRENCE AND COLLINEARITY 31 In this case Z APB = - Z BPA. The angle APB is to be regarded as obtained by the revolution of PB round P from the position PA, and the angle BPA as the revolution of PA round P from the position PB; these are in opposite senses' and so of opposite signs. With this convention as to the signs of our angles we may argue from the figures of § 32, AB _ AAOB _ I OA.OBsinZAOB BC ~ "KbOG ~ ^OB.OCsm^ BOG (the lines (JA, OB, OC being all regarded as positive) _0A sin z^ OS ~" 00 • sin Z BOO ' AB In this way the sign of the ratio j^p is retained in the process of transformation, since sin ZAOB and sin Z 50C are of the same or opposite sign according as AB and BC arr of the same or opposite sign. The student will see that our convention would have been useless had the area depended directly on the cosine of the angle instead of on the sine, since cos {— A) = + cos(.I). 34. Test for coUinearity of three points on the sides of a triangle. The following proposition, known as Menelaus' theorem, is of great importance. The necessary and sufficient condition that the points D, E, F on the sides of a triangle ABC opposite to the vertices A, B, C respectively should be collinear is AF.BD.CE = AE.CD.BF, regard being had to the signs of these lines. All these lines are along the sides of the triangle. We shall consider any one of them to be positive or negative according as the triangle is to our left or right respectively as we travel along it. 32 THE USE OF SIGNS We will first prove that the above condition is necessary, if D, E, F are collinear. Let p, q, r be the perpendiculars from A, B, C on to the line DEF, and let these be accounted positive or negative according as they are on the one or the other side of the line DEF. With this convention we have ^J^ ^ BD q CE r BF q' CD r' AE p' Hence AF.BD.CE BF.CD.AE lat is, AF.BD.GE = AE.GD . BF. Next let D, E, F be three points on the sides such that AF.BD.CE = AE.CD.BF, then shall D, E, F be collinear. Let the line DE cut AB in F', .-. AF' .BD.GE=AE. CD . BF', Ar_AF •'■ BF'' BF ' CONCURRENCE AND COLLINEARITY .■.(AF+ FF') BF= AF (BF + FF'), .■.FF'(BF-AF) = 0, .■.FF' = 0, .'. F coincides with F'. Thus our proposition is completely proved. 33 35. Test for concurrency of lines through the vertices of a triangle. The following proposition, known as Ceva's theorem, is fundamental. The necessary and suffi,cient condition t/uit the lines AD, BE, 34 TEIE USE OF SIGNS CF drawn through the vertices of a triangle ABC to meet the opposite sides in D, E, F should be concurrent is AF.BD.CE = -AE.CD. BF, the same convention of signs being adojyted as in the last proposition. First let the lines AD, BE, OF meet in P. Then, regard being had to the signs of the areas, AF_ is AFC _ AAFP _ AAFC- AAFP _ AAPC BF~ ABFC ~ A BFP ~ ABFC - ABFP ~ ABPG ' BD ^ ABDA _ ABDP ^ ABDA- ABDP _ ABPA GD~ A CD A ~ A CDP " ACDA -A GDP ~ XCPA ' CF _ AGEB _ A CE P _ ACEB - A CEP _ AGPB AE ' AAEB ~ AAEP ~ AAEB-AAEP ~ AAPB ' AF.BD.CE AAPC ABPA AGPB '■ AE. CD. BF~ AGFA' AAPB ABPG = (-l)(-l)(- 1)^-1. Next let D, E, F be points on the sides of a triangle ABC such that AF.BD.CE = -AE. CD . BF, then will AD, BE, CF be concurrent. B D C Let AD, BE meet in Q, and let GQ meet AB in F'. CONCURRENCE AND COLLINEARITY 35 .. AF' .BD .CE=-AE.GD.BF'. Ar_AF •'• BF'~ BF' .•.(AF + FF')BF = {BF + FF')AF. ..FF'(BF-AF) = 0. .■.FF' = 0. .'. F' and F coincide. Hence our proposition is completely proved. 36. Prop. // D, E, F he three points on the sides of a triangle ABC opposite to A, B, C respectively. AF.BD.GE s'mA CF sin BA L) sin CBE A E . CD . BF sin A BE sin CA D sin BCF " For BD ABAD _^AB.AD s'm BAD _AB sm BAD CD ~ 'KOAD " fAC\~ADliin^UAlJ ~~ AG 'sin GAD' with our convention as to sign, and AB, AC being counted positive. Similarly and AF BF GE AE AG ^xnACF BlJ'smBZ'F BG sin CBE AB sin ABE' AF.BD.G E _ sin AG F sin BA D sin CBE ' ' AE.GD .BF~ sin ABE sin GAD sni BGF '.]() THE USE OF SIGNS Cor. The necessary and sufficient condition that AD, BE, OF should be concuiTent is s in ^Ci^ sin BAD sin CBE _ _ sin A BE sin GA D sin BCF ~ A If be the point of conciirrence this relation can be written in the form sin ^^0 s in ^(70 sin CM sin Z(X)sin GBO sin BA6~~ ' this being easy to remember. 37. Isogonal conjugates. Two lines AD, AD' through the vertex A of a triangle which are such that zBAD = zD'AC (not z GAD') are called isogo7ial conjugates. Prop. //■ AD, BE, GF he three concun-ent lines tli rough the vertices of a triangle ABG, their isogonal conjugates AD', BE', GF' will also he concurrent. For Similarly and sin BAD _ sin D' AG _ sin GAD' sin GAD ~ sin DAB ~ siu^BAD' ' sin G BE _ si n ABE ' sin ABE ~ sin GBE' sin AGF _ s in BGF' sin BGF ~ sin AGF' ' CONCURRENCE AND COLLINEARITY 37 sin CAD' sin ABE' sin BCF ' sin BAD' sin CBE' sin ACF' sin 5^i) sin (75£; sin ACF sin G'^Z> sin ^i?A' sin BGF •. .4/)', i?A"', CF' are concurrent. = -1, 38. The isogonal conjugates of the medians of a triangle are called its symmedians. Since the medians are concurrent, the symmedians are concurrent also. The point where the symmedians intersect is called the symmedian point of the triangle. The student will see that the concurrence of the medians and perpendiculars of a triangle follows at once by the tests of this chapter (§§ 35 and 36). It was thought better to prove them by independent methods in the first chapter in order to bring out other properties of the orthocentre and the median point. 39. We will conclude this chapter by introducing the student to certain lines in the plane of a triangle which are called by some writers antiparallel to the sides. Let ABC be a triangle, D and F points in the sides 38 THE USE OF SIGNS AB and AC such that zADE = zBCA and therefore also Z AED = Z CBA. The line BE is said to be antiparallel to BC. It will be seen at once that DBGE is cyclic, and that all lines antiparallel to BG are parallel to one another. It may be left as an exercise to the student to prove that the sym median line through A of the triangle ABC bisects all lines antiparallel to BG. EXERCISES 1 . The lines joining the vertices of a triangle to its circumcentre are isogonal conjugates with the perpendiculars of the triangle. 2. The lines joining the vertices of a triangle to the points of contact with the opposite sides of the incircle and ecircles are respectively concurrent. 3. ABC is a triangle; AD, BE, GF the perpendiculars on the opposite sides. If AG, BH and CK be drawn perpendicular to EF, FD, DE respectively, then AG, BH and 6' /i will be concurrent. 4. The midpoints of the sides BG and GA of the triangle ABC are D and E : the trisecting points nearest B of the sides BC and BA respectively are U and K. CK intersects AD in L, and BL inter- sects AH in M, and CM intersects BE in N. Prove that iV is a trisecting point of BE. CONCURRENCE AND COLLINEARITY 39 0. If perpendiculars are drawn from the orthocentre of a triangle ABC on the bisectors of the angle A, shew that their feet are collinear with the middle point of BC. 6. The points of contact of the ecircles with the sides BC, CA, AB of a triangle are respectively denoted by the letters D, E, F with suffixes 1, 2, 3 according as they belong to the ecircle opposite A, B, or C. BE.,, CF^ intersect at P ; BE^, CF^ at Q : E^F^ and BC Sit X; F^D, and CA at Y ; D,E, and AB at Z. Prove that the groups of points A, P, D^, Q : and X, Y, Z are respectively collinear. 7. Parallel tangents to a circle at A and ^are cut in the points C and D respectively by a tangent to the circle at E. Prove that AD, BC and the line joining the middle points oi AE and BE are concurrent. 8. From the angular points of any triangle ABC lines AD, BE, CF are drawn cutting the opposite sides in D, E, F, and making equal angles with the opposite sides measured round the triangle in the same direction. The lines AD, BE, CF iorm a triangle A'B'C. Prove that A' B . B'C . CA ^ A'C^ B'A . CB ^ BC.CA.AB AE . BF . CD AF . BD.CE AD. BE. CF ' 9. Through the symmedian point of a triangle lines are drawn antiparallel to each of the sides, putting the other two sides. Prove that the six points so obtained are equidistant from the symmedian point. [The circle through these six points has been called the cosine circle, from the property, which the student can verify, that the intercepts it makes on the sides are pi'oportional to the cosines of the opposite angles.] 10. Through the symmedian point of a triangle lines are drawn parallel to each of the sides, cutting the other sides. Prove that the six points so obtaiued are equidistant from the middle point of the line joining the symmedian point to the circumcentre. [The circle through these six points is called the Lemoine circle. See Lachlan's Modern Pure Geometry, § 131.] 1 1. AD, BE, CF are three concurrent lines through the vertices of a triangle ABC, meeting the opposite sides in D, E, F. The circle circumscribing DEF intersects the sides of ABC again in D', E', F\ Prove that AD', BE', CF' are concurrent. 40 THE USE OF SIGNS. CONCURRENCE AND COLLIN EARITY 12. Prove that the tangents to the circumcircle at the vertices of a triangle meet the opposite sides in three points which are colHnear. 13. If AD, BE, CF through the vertices of a triangle ABC meeting the opposite sides in D, E, F are concurrent, and points D', E', F' be taken in the sides opposite to A, B, C so that DD' and BC, EE' and CA, FF' and AB have respectively the same middle point, then AD' , BE', CF' are concurrent. 14. If from the symmedian point aS' of a triangle ABC, per- pendiculars SD, SE, SF be drawn to tlie sides of the triangle, then S will be the median point of the triangle DEF. 15. Prove that the triangles formed by joining the symmedian point to the vertices of a triangle are in the duplicate ratio of the sides of the triangle. 16. The sides BC, CA, AB of a triangle ABC are divided internally by points A', B', C so that BA' :A'C=CB' : B'A = AC' : C'B. Also B'C produced cuts BC externally in A". Prove that BA" : CA" = CA" : A'B\ 41 CHAPTER IV PROJECTKDX 40. If V be any point in space, and A any other point, then if VA, produced if necessary, meet a given plane tt in ^', A' is called the projection of A <»n the plane tt by means of the vertex V. It is clear at once that the projection of a straight line on a plane tt is a straight line, namely the intersection of the plane TT with the plane containing V and the line. If the plane through V and a certain lim- be parallel to the TT plane, then that line will be projected to infinity o;i the tt plane. The line thus obtained on the tt plane is called the line at infinity in that plane. 41. Suppose now we are projecting points in a plane p by means of a vertex V on to another plane tt. Let a plane through F parallel to the plane vr cut the plane p in the line AB. This line AB will project to infinity on the plane tt, and for this reason AB is called the vanishing line on the plane p. The vanishing line is clearly parallel to the line of inter- section of the planes p and tt, which is called tlie axis of projection. 42. Now let EDF be an angle in the plane p and let its lines DE and BF cut the vanishing line AB in E and F, then the angle EDF will project on to the tt plane into an angle of magnitude EVF. 42 PROJECTION For let the plane VDE intersect the plane ir in the line de. Then since the plane VEF is parallel to the plane tt, the intersections of these planes with the plane VDE are parallel ; that is, de is parallel to VE. Similarly df is parallel to VF. Therefore Aedf^/.EVF. Hence we see that mu/ angle in the plane p projects on to the TT plane into an angle of magnitude equal to that subtended at V hy the portion of the vanishing line intercepted by the lines containing the angle. 43. Prop. By a proper choice of the vertex V of projec- tion, any given line on a plane p can be projected to infinity, while two given angles in the plane p are projected into angles of given magnitude on to a plane ir properly chosen. Let ^J5 be the given line. Through AB draw any plane p'. Let the plane ir be taken parallel to the plane p'. Let EDF, E'D'F' be the angles in the plane p which are to be projected into angles of magnitude a. and /3 respectively. Let E, F, E', F' he on AB. PROJECTION 43 On EF, E'F' in the plane p describe segments of circles containing angles equal to a and /3 respectively. Let these segments intersect in V. Then if V be taken as the vertex of projection, AB will project to infinity, and EDF, E'D'F' into angles of magnitude a and yS respectively (§42). Cor. 1. Any triangle can he projected into an equilateral triangle. For if we project two of its angles into angles of 60'^ the third angle will project into 60° also, since the sum of the three angles of the triangle in projection is equal to tw^o right angles. Cor. 2. A quadrilateral can be projected into a square. Let A BCD be the quadrilateral. Let EF be its third diagonal, that is the line joining the intersections of opposite pairs of sides. Let AC and BD intersect in Q. Now if we project EF to infinity and at the same time 44 PROJECTION project Z s BA D and BQA into right angles, the quadrilateral will be projected into a square. For the projection of EF to infinity, secures that the pro- jection shall be a parallelogram ; the projection of Z BAD into a right angle makes this parallek)gram rectangular ; and the projection of Z AQB into a right angle makes the rectangle a square. 44. It may happen that one of the lines DE, D'E' in the preceding paragraph is parallel to the line AB which is to be projected to infinity. Suppose that DE is parallel to AB. In this case we must draw a line FV in the plane p' so that the angle EFV is the supplement of a. The vertex of projection V will be the intersection' of the line FV with the segment of the circle on E'F'. If D'E' is also parallel to AB, then the vertex F will be the intersection of the line i^Fjust now obtained and another line i^'F so drawn that the angle E'F'V is the supplement of yS. 45. Again the segments of circles described on EF, E'F' in the proposition of § 43 may not intersect in any real point. In this case F is an imaginary point, that is to say it is a point algebraically significant, but not capable of being presented to the eye in the figure. The notion of imaginary points and lines which we take over from Analytical Geometry into our present subject will be of considerable use. PROJECTION 45 46. Prop. A range of three points is projective with any other range of three points in space. AV— /B Let A, B, C be three collinear points, and A', B', C" three others not necessarily in the same plane with the first three. Join A A'. Take any point V in A A'. Join VB, VC and let them meet a line A'DE diawn ihroucrh o A' in the plane VAC in D and E. Join DB', EC. These are in one plane, viz. the plane con- taining the lines A'C and A'E. Let DB', EC meet in V. Join V'A'. Then by means of the vertex V, A, B, C can be projected into A', D, E ; and these by means of the vertex V can be projected into A', B' , C. Thus our proposition is proved. 47. The student must understand that when we speak of one range being projective with another, we do not mean necessarily that the one can be projected into the other by a single projection, but that we can pass from one range to the other by successive projections. 46 PROJECTION A range o^ four points is not in general projective with any other range of four points in space. We shall in the next chapter set forth the condition that must be satisfied to render the one projective with the other. EXERCISES 1. Prove that a system of parallel lines in a plane jy will project on to another plane into a system of lines through the same point. 2. Two angles such that the lines containing them meet the vanishing line in the same points are projected into angles which are equal to one another. 3. Shew that in general three angles can be projected into angles of the same magnitude a. 4. Shew that a triangle can be so projected that any line in its plane is projected to infinity while three given concurrent lines through its vertices become the perpendiculars of the triangle in the projection. 5. Explain, illustrating by a figure, how it is that a point E lying on a line PQ, and outside the portion PQ of it, can be pro- jected into a point r lying between jj and q, which are the projections of P and Q. 6. Any three points ^i, B^, C\ are taken respectively in the sides BG, CA, AB of the triangle ABC ; B^C^ and BC intersect in F; Ci^i and CA in G ; and A,B, and AB in H. Also FH and BB^ intersect in A/, and FG and CCj in N. Prove that MG, NH and BC are concurrent. 7. Prove that a triangle can be so projected that three given concurrent lines through its vertices become the medians of the triangle in the projection. 8. If A Ay, BBy, CC-^ be three concurrent lines drawn through the vertices of a triangle ABC to meet the opposite sides in A-^B^C^ ; and if B-^C-y meet BC in A^, C\Ay meet CA in B.^, and A^B^ meet AB m Co,; then A^, B^, C\ will be collinear. [Project the concurrent lines into medians.] 9. If a triangle be projected from one plane on to another the three points of intersection of corresponding sides are collinear. 47 CHAPTER V CROSS-RATIOS 48. Definition. 1( A, B, C, D he a range of points, the ratio . ^^„ is called a cross-ratio of the four points, and is conveniently represented by (A BCD), in which the order of the letters is the same as their order in the numerator of the cross-ratio. Some writers call cross-ratios ' anharmonic ratios.' This is hoAvever not a fortunate term to use, and it will be best to avoid it. For the term ' anharmonic ' means not harmonic, so that an anharmonic ratio should be one that is not harmonic, whereas a cross-ratio may be harmonic, that is to say may be the cross- ratio of what is called a harmonic range. The student will better appreciate this point when he comes to Chapter VII. 49. The essentials of a cross-ratio of a range of four points are : (1) that each letter occurs once in both numerator and denominator ; (2) that the elements of the denominator are obtained by associating the first and last letters of the numerator together, and the third and second, and in this particular order. AB.CD . • , , ■ . -T-p. — TTp IS not a cross-ratio but the negative oi one, tor ^^ — jYn, though not appearing to be a cross-ratio as it ,. . BA. CD , stands, becomes one on rearrangement, tor it = jyj- — y^^ , that BD . CA is (BAGD). 48 CROSS-RATIOS Since there are twenty-four permutations of four letters taken all together, we see that there are twenty-four cross-ratios which can be formed with a range of four points. 50. Prop. The twenty -four cross-ratios of a range of four points are equivalent to six, all of tvhich can he expressed in terms of any one of them. Let {ABGD) = \. First we observe that if the letters of a cross-ratio be inter- changed in pairs simultaneously, the cross-ratio is unchanged. = (ABGD), = (ABCD), = (ABCD). Hence we get (A BGD) = {BADG) = (GDAB) = {DGBA ) = X...(1). Secondly we observe that a cross-ratio is inverted if we interchange either the first and third letters, or the second and fourth. .-. {ADGB)==iBGDA)^{GBAD) = {DABG) = l ...(2). A, These we have obtained from (1) by interchange of second and fourth letters ; the same result is obtained by interchanging the first and third. For {BADG) = BA.DG ^ BG.DA^ AB. AD. GD GB (GDAB) -- GD . AB CB.AD AB. AD. GD .GB (DGBA)- DG.BA DA.BC AB. AD CD .GB Thirdly, since by § 31 AB.GD-\-BG. AD + CA. BD = 0, AB.CD ■AD. GB L 1 ^^^ .BD = 0. ^ ' AD .GB ^ ^ GA.BD ^ ^- AD.CB AC. BD ■■{ACBD). AD. BG . CROSS-EATIOS 49 Thus the interchange of the second and third letters changes A, into 1 -\. We may remark that the same result is obtained by interchanging the first and fourth. Thus from (1) (ACBD) = (BDAC) = {(JADB) = (DBCA) = 1 - \ ... (3), and from this again by interchange of second and fourth letters, {ADBC) = {BCAD) = {CBDA) = (BACB) = ^ ^ . . . (4). In these we interchange the second and third letters, and get (A BBC) = (BACD) - (GDBA ) = (IJCAB) 1 ^ And now interchanging the second and fourth we get (ACDB) = (IWCA) = (CABD) = (DBAC) = ^-" ^ ... (6). A, We have thus expressed all the cross-ratios in terms of X. And we see that if one cross-ratio of four coUinear points be equal to one cross-ratio of four other collinear points, then each of the cross- ratios of the first range is equal to the corre- sponding cross-ratio of the second. Two such ranges may be called equi-cross. 51. Prop. If A, B, C be three separate collinear points, and D, E oilier points in their line such that (ABCD) = {ABCE), then D must coincide luith E. ^ . AB.CD AB.GE Forsmce AD7CB = AE~CB' .-. AE.CD = AD.CE. .-. {AD + DE) CD = AD {CD -f- DE). .-. DE{AD-CD) = 0. .-. DE.AG = 0. .-. DE = for AC^O, that is, D and E coincide. 50 CROSS-RATIOS 52. Prop. A range of four points is eqni-cross with its projection on any plane. Let the range A BCD be projected by means of the vertex V'mto A' B'G'D'. rhen AB AD .CD .CB AAVB AA\^D' ACVD AGVB' regard b( iing had to the signs of the areas, \VA. VB sin AVB ^VC . VD sin CVD \VA . VD sin AVD'^VG. VB sin GVB' regard being had to the signs of the angles, sin AVB sin CVD sin ^FD sin GVB' Similarly A'B ' . CD ' A'D' . CB' Fig. 1. sm J.' Fi^' sin CVD' sinA'VD'smCVB'- Now in all the cases that arise s in^^F^^sinO^Fi)^ _ sin ^ VB sin G VJ) QinA'VD' sin CVB' ~ sin A VD sin GVB ' CROSS-RATIOS 51 This is obvious in fig. 1. In fig. 2 sin A'VB' = sin B'VA, these angles being supplementary. = - sin ^ VB, and sin A' VD' = sin D'VA = Fig. 2. Further Fig. 3. sin C"FZ)' = sin CVD, and sin C'VB' = sin CVB. In fig. 3 sin ^'F^' = sin. I Fi^, sin CVD' = sin CVD, sin A'VD' = sin Z>'F^ = - sin .4 VD, sin C'Fi?' = sin ' BVC =- sin CVB. Thus in each case {A'B'C'D') = {ABCD). 4—2 52 CROSS-RATIOS 53. A number of lines in a plane which meet in a point V are said to form a pencil, and each constituent line of the pencil is called a ray. V is called the vertex of the pencil. Any straight line in the plane cutting the rays of the pencil is called a transversal of the pencil. From the last article we see that if VP^, VP.„ VP,, VP^ form a pencil and any transversal cut the rays of the pencil in A, B, C, D, then (ABCB) is constant for that particular pencil; that is to say it is independent of the particular transversal. It will be convenient to express this constant cross-ratio by the notation V {P,P^P,P,). .0 Pi" .-""Pi ,^''^ P?^''' .,.^v' We easily see that a cross-ratio of the projection of a pencil on to another plane is equal to the cross-ratio of the original pencil. For let V(P^, P., Pj, P,) be the pencil, the vertex of projection. CROSS-RATIOS 53 Let the line of intersection of the p and tt planes cut the rays of the pencil in A, B, C, D, and let V be the projection of V, V'P;, of VP^, and so on. Then ABCD is a transversal also of v'{p;,p:,p:,p:). .-. V(P,P,P,P,) = (ABCD)= V'{P/P.:P,'P:). 54. We are now in a position to set forth the condition that two ranges of four points should be mutually projective. Prop. // ABCD be a range, and A'B'G'D' another range such that {A'B'C'I)') = (ABCD), then the tivo ranges are projective. V B/ Join ^^4' and take any point V upon it. Join VB, VC, VD and let these lines meet a line through A' in the plane VAD in P, Q, R respectively. Join PB', QC and let these meet in V. Join V'A', and V'R, the latter cutting A'D' in X. 54 CROSS-RATIOS Then (ABCD) = (A'PQR) = (A'B'C'X). But (A BCD) = {A'B'C'D') by hypothesis. .■.{A'B'G'X)={A'B'G'D'). .'. X coincides with D' (§ 51). Thus, by means of the vertex V, ABCD can be projected into A'PQR, and these again by the vertex V into A'B'C'D'. Thus our proposition is proved. 55. Def. Two ranges ABCDE... and A' B'C'D'E' ... are said to be homographic when a cross-ratio of any four points of the one is equal to the corresponding cross-ratio of the four corresponding points of the other. This is conveniently expressed by the notation (ABCDE...) = {A' B'C'D'E'...). The student will have no difficulty in proving by means of I 54 that two homographic ranges are mutually projective. Two pencils V(F,Q,R,S,T...) and V (P' , Q', R', S', T' ...) are said to be homographic when a cross-ratio of the pencil formed by any four lines of rays of the one is equal to the corresponding cross-ratio of the pencil formed by the four corre- sponding lines or rays of the other. 56. Prop. Two homographic pencils are mutually pro- jective. For let PQRS..., P'Q'R'S'... be any two transversals of the two pencils, V and V the vertices of the pencils. Let PQ"R"S"... be the common range into which these can be projected by vertices and 0'. Then by means of a vertex K on OV the pencil ViP,Q,R,S...) can be projected into (P, Q", R", S" ...); and this last pencil can, by a vertex L on 00', be projected into 0' (P, Q", R", S"...), that is, 0' (P', Q', R', &'...)■ and this CROSS-RATIOS 55 again by means of a vertex M on O'V can be projected into V'{P',Q',R\S'...). 57. We will conclude this chapter with a construction for drawing through a given point in the plane of two given parallel lines a line parallel to them, the construction heinrj effected bi/ means of the ruler only. Let A(o, AiOi' be the two given lines, &> and to' being the point at infinity upon them, at which they meet. Let P be the given point in the plane of these lines. Draw any line AC to cut the given lines in A and C, and take any point B upon it. Join PA cutting Aico' in A^. Join PB cutting A^w' in Bi and Aco in B.,. Join PC. Let A^A and BoC meet in Q. 56 CROSS-RATIOS Let QB meet GP in 0. Let A^O and AC meet in D. PD shall be the line required. For {A^B.Coi') = B {A,B,Cco') = (A^BUoo) = ^i (Ao^BUo)) = {BB,PB,) = C(BB,PB,) = {AQPA,) = (AQPA,) = {ABGD). .-. P (A,B,Cco') = P (ABCD). .'. PD and Pco' are in the same line, that is, PD is parallel to the given lines. EXERCISES 1. If {ABCD) = -^ and B be the point of trisection of AD towards A, then C is the other point of trisection of AD. 2, Given a range of three points A, B, C, find a fourth point D on their line such that (ABCD) shall have a given value. CROSS-RATIOS , 57 3. If the transversal ABC be parallel to OD, one of the rays of a pencil {A, B, C, D), then 0(ABCD) = ^. 4. If {ABCD) = {ABG'D'), then {ABCC') = {ABDD'). 5 If A, B, C, D he a. range of four separate points and {A£GD) = {AI)CB), then each of these ratios = — 1. 6. Of the cross-ratios of the range formed by the circumcentre, median point, nine-points centre and orthocentre of a triangle, eight are equal to — 1, eight to 2, and eight to -J-. 7. Any plane will cut four given planes all of which meet in a common line in four lines which are concurrent, and the cross-ratio of the pencil formed by these lines is constant. 8. Taking a, b, c, d to be the distances from to the points A, B, C, D all in a line with 0, and X=(a-d){h-c), iJ.= (b- d) (c - a), v = (c - d) (a - b), shew that the six possible cross-ratios of the i-angcs that can be made up of the points A, B, C, D are /u, I' V A A /A J' /x A I' /i. A 58 CHAPTER VI PERSPECTIVE 58. Def. A figure consisting of an assemblage of points P, Q, R, S, &c. is said to be in perspective with another figure consisting of an assemblage of points P', Q', R', S', &c., if the lines joining corresponding points, viz. PP', QQ', RR', &c. are concurrent in a point 0. The point is called the centre of perspective. It is clear from this definition that a figure when projected on to a plane or surface is in perspective with its projection, the vertex of projection being the centre of perspective. It seems perhaps at first sight that in introducing the notion of perspective we have arrived at nothing further than what we already had in projection. So it may be well to compare the two things, with a view to making this point clear. Let it then be noticed that two figures luhich are in the same plane may be in perspective, whereas we should not in this case speak of one figure as the projection of the other. In projection we have a figure on one plane or surfice and project it by means of a vertex of projection on to another plane or surface, whereas in perspective the thought of the planes or surfaces on which the two figures lie is absent, and all that is necessary is that the lines joining corresponding points should be concurrent. So then while two figures each of which is the projection of the other are in perspective, it is not necessarily the case that of two figures in perspective each is the projection of the other. 59. It is clear from our definition of perspective that if two ranges of points be in perspective, then the two lines of the ranges must be coplanar. PERSPECTIVE 59 For if A, B, C, &c. are in perspective with A', B', C , &c., and be the centre of perspective, A'B' and AB are in the same plane, viz. the plane containing the lines OA, OB. It is also clear that ranges in perspective are homographic. But it is not necessarily the case that two homographic ranges in the same plane are in perspective. The following proposition will shew under what condition this is the case. 60. Prop. // tivo homographic ranges in the same plane he such that the point of intersection of their lines is a point corresponding to itself in the two ranges, then the ranges are in perspective. For let {ABODE. ..) = {AB'C'B'E'. . .). Let BB', CC meet in 0. 60 PERSPECTIVE Join OD to cut AB' in D". Then {AB'G'D') = {ABCB) = {AB'C'D"). .•. D' and B" coincide. (§ 51.) Thus the line joining any two corresponding points in the two homographic ranges passes through ; therefore they are in perspective. 61. Two pencils V(A, B, Q, D ...) smd V {A', B', C',D'...) will according to our definition be in perspective when V and V are in perspective, points in VA in perspective with points in V'A' , points in VB in perspective with points in V'B' and We can at once prove the following proposition : If tiuo pencils in different planes he in pej'spective they have a common transversal and are homographic. Let the pencils be V{A,B, C,D...) and V (A', B',0',D'...). Let the point of intersection of VA and V'A', which are coplanar (§ 59), be P ; let that of VB, V'B' be Q ; and so on. PERSPECTIVE 61 The points F, Q, R, S, &c. each lie in both of the planes of the pencils, that is, they lie in the line of intersection of these planes. Thus the points are coUinear, and since V(ABCD...) = (PQRS..:)^ V'iA'B'C'D'...), the two pencils are homographic. The line PQRS... containing the points of intersection of corresponding rays is called the (uis of perspective. 62. According to the definition of perspective given at the beginning of this chapter, two pencils in the same plane are always in perspective, with any point on the line joining their vertices as centre. Let the points of intersection of corresponding rays be, as in the last paragraph, P, Q, R, S, &c. We cannot now prove P, Q, R, S... to be collinear, for indeed they are not so necessarily. But if the points P, Q, Szc. are collinear, then we say that the pencils are coaxal. If the pencils are coaxal they are at once seen to be honio- graphic. 63. It is usual with writers on this subject to define two pencils as in perspective if their corresponding rays intersect in collinear points. The objection to this method is that you have a different definition of perspective for different purposes. We shall find it conducive to clearness to keep rigidly to the definition we have already given, and we shall speak of two pencils as coaxally in perspective if the intersections of their corresponding rays are collinear. As we have seen, two non-coplanar pencils in perspective are always coaxal ; but not so two coplanar pencils. Writers, when they speak of two pencils as in perspective, mean what we here call ' coaxally in perspective.' 62 PERSPECTIVE 64. Prop. If two homographic pencils in the same plane have a corresponding ray the same in both, they are coaxally in perspective. Let the pencils be V (A, B, G, D, &c.) and V (A, B', C, D', &c.) with the common ray V'VA. Let VB and V'B' intersect in 0, FC and V'C in y, VD and VD' in S, and so on. B' B 0' Let 7^ meet V'VA in a, and let it cut the rays VD and VD' in §1 and h^ respectively. Then since the pencils are homographic, V(ABGD)= V'(AB'C'D'). Therefore Sj and S.. coincide with S. Thus the intersection of the corresponding rays VD and VD' lies on the line /Sj. Similarly the intersection of any two other corresponding rays lies on this same line. Therefore the pencils are coaxally in perspective. PERSPECTIVE 63 65. Prop. If ABC..., A'B'C'..., he two coplanar homo- graphic ranges not having a common corresponding point, tlien if tiuo pairs of corresponding points he cross-joined {e.g. AB' and A'B) all the points of intersection .so obtained are collinear. Let the lines of the ranges intersect in P. Now according to our hypothesis P is not a corresponding point in the two ranges. It will be convenient to denote P by two different letters, X and Y', according as we consider it to belong to the ABC... or to the A'B'C ... range. Let X' be the point of the A'B'G'... range corresponding to A" in the other, and let Y be the point of the ABC... range corresponding to ]"' in the other. Then {ABCX Y...) = [A'B'C X'Y'. . . ). .-.A' {ABCXY. ..) = A (A'B'CX'Y'...). These two })encils have a common ray, viz. Axi', therefore by the last proposition the intersections of their corresponding rays are collinear, viz. A'B, AB'; A'C, AC; A'X, AX'; A'Y,AY'; and so on. From this it will be seen that the locus of the intersections of the cross-joins of A and A' with B and B', C and C and so on is the line X'Y. 64 PERSPECTIVE Similarly the cross-joins of any two pairs of corresponding points will lie on A'T. This line X'Y is called the liomogntpJiic axis of the two ranges. This propositi(jn is also true if the two ranges have a common corresponding point. The proof of this may be left to the student. 66. The student may obtain practice in the methods of this chapter by proving that if V {A, B, C.) and V {A', B' , G'...) be two homographic coplanar pencils not having a common corresponding ray, then if we take the intersections of VF and Vq, and of VQ and V F' {VF, V'F'; and VQ, V'Q' being any two pairs of corresponding lines) and join these, all the lines thus obtained are concurrent. It will be seen when we come to Reciprocation that this proposition follows at once from that of § Qb. TRIANGLES IN PERSPECTIVE 67. Prop . // the vertices of two triangles are in perspective, the intersections of their corresponding sides are collinear, and conversely. (1) Let the triangles be in different planes. Let be the centre of perspective of the triangles ABC, A'B'C. Since BC, B'C are in a plane, viz. the plane containing OB and OC, they will meet. Let A' be their point of intersection. Similarly GA and G'A' will meet (in Y) and AB and A'B' (in Z). Now A'', V, Z are in the planes of both the triangles ABG, A'B'G'. Therefore they lie on the line of intersection of these planes. Thus the first part of our proposition is proved. PERSPECTIVE 65 Next let the triangles ABC, A'B'C be such that the inter- sections of corresponding sides {X, Y, Z) are collinear. Since BC and EC meet they are coplanar, and similarly for the other pairs of sides. Thus we have three planes BCOB', CAA'C, ABB' A', of which AA', BB', CO' are the lines of intersection. But three planes meet in a point. Therefore AA', BB', CC are concurrent, that is, the triangles are in perspective. (2) Let the triangles be in the same plane. First let them be in perspective, centre 0. Let A", Y, Z be the intersections of the corresponding sides as before. Project the figure so that A'^Fis projected to infinity. A. G. 6 QG PERSPECTIVE Denote the projections of the different points by corres- ponding small letters. We have now oh : oh' = oc : oc since be is parallel to h'c' = oa : oa' since ca is parallel to c'a'. .". ah is parallel to n'b'. .'. z is at infinity also, that is, X, y, z are collinear. .*. A', F, Z are collinear. Next let A^, F, Z be collinear; we will prove that the triangles are in perspective. Let A A' and BB' meet in 0. Join OC and let it meet A'C in G". Then ABC and A'B'C" are in perspective. .-. the intersection of BG and B'G" lies on the line YZ. But BG and B'G' meet the line YZ in X by hypothesis. .•. B'G" and i?'C" are in the same line, i.e. G" coincides with C". Thus ABG and A' B'G' are in perspective. 68. Prop. The necessary and sufficient condition that the coplanar triangles ABG, A' B'G' should he in perspective is AB, . AB., .GA,. GA^ . BC, . BG., = AG,.AG,.BA,. BA, . GB, . GB,, PERSPECTIVE 67 A^, A2 being the points in which A'B' and A'C meet the non-corresponding side BC, B„ Bo being the points in ivhich B'C and B'A' meet the non- con-esponding side CA, Ci, Co being the points in which CA' and C'B' meet the non- corresponding side AB. First let the triangles be in perspective; let XYZ be the axis of perspective. 5—2 = 1. 68 PERSPECTIVE Then since X, B^ , Co are collinear, A B, . CX . BO, •'• AC.BX.CB, Since Y, C\ , Ao are collinear, AY.CA,.BC\ * ■ AC,.BA.,.CY Since Z, Ai, Bo are collinear, AB^.CA,.BZ _ 'az.ba,.cb~ Taking the product of these we have AB, . AB., . GA, . GA,. BC, . BC,. AY.GX . BZ AC, . AG., . BA, . BAo^ . C% . CB., . AZ . BX . CY~ But X, Y, Z are collinear, AY.GX.BZ • • AZ.BX.CY~ .-. AB,.AB,.GA,.CA,.BC\.BC., =^AG,.AC,.BA,. BA, . CB, . GB,. Next we can sheAv that this condition is sufficient. For it renders necessary that AY.CX .BZ AZ.BX.GY' .". X,Y,Z are collinear and the triangles are in perspective. Cor. If the triangle ABC be in perspective with A'B'C', and the points A,, Ao, B,, Bo, Cj, Co be as defined in the above proposition, it is clear that the three following triangles must also be in perspective with ABC, viz. (1) the triangle formed by the lines A^Bo, B^C,, CoA^, (2) „ „ „ „ A,B„ B,Co_, G,A„ (3) „ „ „ „ A,B„ B,C\, C,A,. PERSPECTIVE 69 EXERCISES 1. ABC, A'B'C' are two ranges of three points in the same plane; i?C" and 5'C intersect in ^i, CA' and C A in B^, and AB' and A'B in Cj ; prove that A^, B^, Cj are collinear. 2. ABC and A'B'C are two coplanar triangles in perspective, centre 0, through any line is drawn not in the plane of the triangle ; ^S* and .S" are any two points on this line. Prove that the triangle ABC by means of tlie centre S, and the triangle A'B'C by means of the centre S', are in perspective with a common triangle. .3. Assuming tliat two non-coplanar triangles in perspective are coaxal, prove by means of Ex. 2 that two coplanar triangles in per- spective are coaxal also. 4. If ABC, A'B'C be two triangles in perspective, and if BC and B'C intersect in A^, CA' and CA in B^, AB' and A'B in Cj, then the triangle A^B^C^ will be in perspective with each of the given triangles, and the three triangles will have a common axis of perspective. 5. When three triangles are in perspective two by two and have the same axis of perspective, their three centres of perspective are collinear. 6. The points Q and Ji lie on the straight line AC, and the point Ton the straight \ine_AD ; VQ meets the straight line AB in Z, and VR meets AB in Y ; X is anotlier point on AB ; XQ meets AD in U, and XR meets AD in II', pro\e that YU, ZW, JC are concurrent. 7. The necessary and sutHcient condition that the coplanar triangles ABC, A! B'C should be in perspective is AV.Bc' . Ca = Ac' . Ba . Cb', where a, b', c denote the sides of the triangle A'B'C opposite to A' , B' , C respectively, and Ab' denotes the perpendicular from A on to b'. [Let B'C and BC meet in X ; CA' and CA in Y ; A'B' and AB in Z. The condition given ensures that X, Y, Z are collinear.] 70 PERSPECTIVE , 8. Prove that the necessary and sufficient condition that the coplanar triangles ABC, A'B'C should be in perspective is sin ABC sin ABA' sin BCA' sin BC B' sin CAB' sin CAC _ sin A CB' sin A CA' sin CBA' sin CBC ' sin BA C ' sin BAB' ~ ' [This is proved in Lachlan's Modern Pure Geometry. The student has enough resources at his command to establish the test for himself. Let him turn to § 36 Cor., and take in turn ■aX A', B', C and at the centre of perspective. The result is easily obtained. Nor is it difficult to remember if the student grasps the principle, by which all these formulae relating to jjoints on the sides of a triangle are best kept in mind — the principle, that is, of travelling round the triangle in the two opposite directions, (1) AB, BC, CA, (2) AC, GB, BA.'\ 9. Two triangles in plane perspective can be projected into equilateral triangles. 10. ABC is a triangle, /j, /o, L^ its ecentres opposite to A, B, C respectively. IJ^ meets BC in A^, /.j/j meets CA in B^ and /j/, meets AB in Ci, prove that A^, B^, C'j are collinear. 11. \i AD, BE, CF and AD', BE', CF' be two sets of con- current lines drawn through the vertices of a triangle AP>C and meeting the opposite sides in D, E, F and D' , E', F', and if EF and E'F' intersect in X, FD and F'D' in 7, and DE and U E' in Z, then the triangle XYZ is in perspective with each of the triangles ABC, DEF, D'E'F'. [Project the triangle so that AD, BE, CF become the per- pendiculars in the projection and AD' , BE', CF' the medians, and then use Ex. 7.1 71 CHAPTER VII HARMONIC SECTION 69. Def. Four collinear points A, B, C, D are said to form harmonic range if {ABCB)=-l. We have in this case AB.CD = -1. AJJ . CB AB AB-AC^ AB-A G '' AD~ AD-AC AG- AD' thus AC is a harmonic mean between AB and AD. Now reverting to the table of the twenty-four cross-ratios of a range of four points (§ 50), we see that if (A BCD) = — 1, then all the follow ing cross-ratios = — 1 : ' "^^ \aBCD\ (BADC), (GDAB), (DCBA), (ADGB), (BCDA), (CBAD), (DABC). Hence not only is AC a harmonic mean between AB and AD, but also BD is a harmonic mean between BA and BC, DB „ „ „ DC and DA, CA „ „ „ CB and CD. We shall then speak of A and C as harmonic conjugates to B and D, and express the fact symbolically thus : (AC,BD) = -1. 72 HARMONIC SECTION By this we mean that all the eight cross-ratios given above, and in which, it will be observed, A and C are alternate members, and B and D alternate, are equal to — 1. When (AC, BD) = - 1 we sometimes speak of D as the fourth harmonic of J., i? and C ; or again we say that AG is divided harmonically at B and D, and that BD is so divided at A and C. Or again we may say that G is harmonically conjugate with A with respect to B and D. A pencil P {A,B, G, D) of four rays is called harmonic when the points of intersection of its rays with a transversal form a harmonic range. The student can easily prove for himself that the internal and external bisectors of any angle form with the lines containing it a harmonic pencil. 70. Prop. // {AG, BD) = -1, and be the middle jmnt of AG, then OB.OB=OG' = OA\ B For since {ABGD) = - 1, .-. AB.GD = -AD.GB. Insert the origin 0. .-. (OB- OA)(OJJ- OG) = -(OD - OA)(OB -OG). But 0A=- OG. .•; (OB+OC)(OD-OG)=-(OD+OG)iOB-OG). .-. OB.OD + OG.OD-OB.OG- OG"- = - OD . OB + OG . OD - OB . OG + 0G\ .-. 20B.On = 20G'. .-. 0B.0D = 0G"- = A0'' = 0A\ Similarly if 0' be the middle point of BD, 0'G.O'A = 0'B'=0'D\ Cor. 1. The converse of the above proposition is true, viz. that if ABGD be a range and the middle point of ^Cand OG' = OB . OD, then (AG, BD) = - 1. This follows by working the algebra backwards. HARMONIC SECTION 73 Cor. 2. Given three points A, B, C in a line, to find a point D in the line such that (AB, CD) = - 1 we describe a circle on AB as diameter, then D is the inverse point of C. 71. Prop. // {A C, BD) = — 1, the circle on A C as dia meter will cut orthogonally evenj ciixle through B and D. Let be the middle point of AC and therefore the centre of the circle on A C. Let this circle cut any circle through B and D in P ; then OB.OD=OC'=OP\ Therefore OP is a tangent to the circle BrD\ thus the circles cut orthogonal 1}-. Similarly, of course, the circle on BD will cut orthogonally every circle through A and C. Cor. 1. If ABCD he a range, and if the circle on AC as diameter cut orthogonally some one circle passing through B and D,then (AC,BD) = -1. For using the same figure as before, we have OB.OD = OP'=:OC\ .-. {AC,BD) = -1. Cor. 2. If fioo ciixles cut orthogonally, any diameter of one is divided harmo)iically by the other. 74 HARMONIC SECTION 72. Prop. If P(AB, CD) = -1 and APB be a right angle, then PA and PB are the bisectors of the angles betiueen PC and PP. Let any transversal cut the rays PA, PB, PC, PD of the harmonic pencil in A, B, C, D. Tht • • AD.BC .-. AC:AD=CB:BD. .•. as P lies on the circle on AB as diameter we have by §27 PC:PD = CB:BD=AG:AR .". PA and PB are the bisectors of the angle GPD. 73. Prop. If on a chord PQ of a circle two conjugate points A, A' with respect to the circle be taken, then {PQ,AA') = ~l. Draw the diameter CD through A to cut the polar o^ A, on which A' lies, in L. Let be the centre. Then by the property of the polar, OL.OA=^ 0C-\ .-. {CD,LA) = -1. Therefore the circle on CD as diameter (i.e. the given circle) will cut orthogonally every circle through A and L (§ 71). But the circle on A A' as diameter passes through A and L. Therefore the given circle cuts orthogonally the circle on A A' as diameter. HARMONIC SECTION 75 But the given circle passes through P and Q. .'. (PQ,AA') = -1. This harmonic property of the circle is of great importance and usefulness. It may be otherwise stated thus : Chords of a ciixle through a point A are harmonically divided at A and at the point of intersection of the chord with the polar of A. 74. Prop. Each of the three diagonals of a plane quadn- lateral is divided harmonically by the other two. Let AB, BC, CD, DA be the four lines of the quadrilateral ; A, B, C, D, E, F its six vertices, that is, the intersections of its lines taken in pairs. Then AC, BD, EF ave its diagonals. Let PQR be the ti-iangle formed by its diagonals. Project EF to infinity. Denote the points in the projection by corresponding small letters. bp . dq _ bp bq . dp dp Then bci 1, q being at x = -1. 76 HARMONIC SECTION Similarly {AFCR) = -l. Also (FQER ) = n {FQER) = (.4 FOR) = -1. Thus we have proved {AC,FR) = -l, (BD,FQ) = -1, {EF,QR) = -l. Cor. The circumcircle of FQR will cut orthogonally the three circles described on the three diagonals as diameters. Note. It has been incidentally shewn in the above proof that if M be the middle point of AB, a the point at infinity on the line, (/i^, il/a)) = -l. 75. The harmonic property of the quadrilateral, proved in the last article, is of very great importance. It is important too that at this stage of the subject the student should learn "to take the 'descriptive' view of the quadrilateral; for in 'descriptive geometry,' the quadrilateral is not thought of as a closed figure containing an area ; but as an assemblage of four lines in a plane, which meet in pairs in six points called the vertices ; and the three lines joining such of the vertices as are not already joined by the lines of the quadrilateral are called diagonals. By opposite vertices we mean two that are not joined by a line of the quadrilateral. 76. A (|uadrilateral is to be distinguished from a quadrangli A quadrangle is to be thought of as an assemblage of fou points in a plane which can be joined in pairs by six straight I HARMONIC SECTION 77 lines, called its sides or lines ; two of these sides which do not meet in a point of the quadrangle are called opposite sides. And the intersection of two opposite sides is called a diagonal point. This name is not altogether a good one, but it is suggested by the analogy of the quadrilateral. Let us illustrate the leading features of a quadrangle by the accompanying figure. ABCD is the quadrangle. Its sides are AB, BC, CD, DA, AC and BD. AB and CD, AG and BD, AD and BC are pairs of opposite sides and the points P, Q, R where these intersect are the diagonal points. The triangle PQR may be called the diagonal triangle. The harmonic property of the quadrangle is that the two sides of the diagonal triangle at each diagonal point are harmonic conjugates IV ith respect to the two sides of the quadrangle meeting in that point. ■ The student will have n(^ difficulty in .seeing that this can be deduced from the harmonic property of the quadrilateral proved in § 74, On account of the harmonic property, the diagonal triangle associated with a quadrangle has been called the harmonic triangle. 78 HARMONIC SECTION EXERCISES 1 . If M and JS^ be points in two coplanar lines AB, CJ), shew that it is possible to project so that M and N' project into the middle points of the projections of AB and CJJ. 2. AA^, BB^, CC-i are concurrent lines through the vertices of a triangle meeting the opposite sides in yli, B^, 6\. B^C^ meets BG in A.,; C, .Ij meets CA in Z>._, ; A^B^ meets AB in Cgj prove that {BC, A,A.^ = - 1, {CA, B,B,) = -l, {AB, C,C.^^-l. 3. Prove that the circles described on the lines A^A.,, B^B.^, C\ Co (as defined in Ex. 2) as diameters are coaxal. [Take P a point of intersection of circles on A-^Ac^, B^B,,, and shew that CjCo subtends a right angle at F. Use Ex. 2 and § 27.] 4. The collinear points A, D, C are given: CB is any other fixed line through C, E is a fixed point, and B is any moving point on CE. The lines A E and BD intersect in Q, the lines CQ and DE in R, and the lines BR and AC in P. Prove that P is a fixed point as B moves along G E. 5. From any point M in the side BC of a triangle ABC lines MB' and MC are drawn parallel to AC and AB respectively, and meeting AB and AC in B' and C". The lines BC and GB' intersect in P, and AP intersects B'C in M'. Prove that M'B' : M'C'^MB : MC. 6. Pairs of harmonic conjugates {DD'), {EE'), {EF') are respectively taken on the sides BC, CA, AB of a triangle ABC with respect to the pairs of points {BC), {CA), {AB). Prove that the corresponding sides of the triangles DEE and D'E'E' intersect on the sides of the triangle ABC, namely EE a.nd E'F' on BC, and so on. 7. The -lines VA', VB', VC bisect the internal angles formed by the lines joining any point V to the angular points of the triangle ABC ; and A' lies on BC, B' on CA, G' on AB. Also A", B", C" are harmonic conjugates of A', B', 6" with respect to B and G, C and A, A and B. Prove that A", B", C" are collinear. 8. AA-^^, BB^, CC^ are the perpendiculars of a triangle ABC ; A^Bj^ meets AB in Cg; X is the middle point of line joining A to the orthocentre ; C'jX and BB^ meet in 7\ Prove that C.2T is perpendicular to BC. HARMONIC SECTION 79 9. ^ is a fixed point without a given circle and P a variable point on the circumference. The line ^i^ at right angles to AP meets in i^ the tangent at P. If the rectangle FAPQ be completed the locus of ^ is a straight line. 10. A line is drawn cutting two non-intersecting circles; find a construction determining two points on this line such that each is the point of intersection of the polars of the other point with respect to the two circles. 11. A^, B^, C\ are points on the sides of a triangle ABC opposite to A, B, C. A^, B.,, 0-2. are points on the sides such that Jj, A.^ are harmonic conjugates with B and C ; B^, Bo with C and A ; C\, C^ with A and B. If A.,, B.^, C, are collinear, then must AA^, BB^, CCi be concurrent. 12. AA^, BBy, CCi 3-re concurrent lines through the vertices of a triangle ABC B^C^ meets BC in A.^, C,Ji meets CA in B.,, A^B^ meets AB in Cj. Prove that the circles on A^A^, B^B.^, C^C^ as diameters all cut the circumcircle of ABC orthogonally, and have their centres in the same straight line. [Compare Ex. 3.] 13. If .1 and B be conjugate points of a circle and J/ the middle point of AB, the tangents from M to the circle are of length MA. 14. If a system of circles have the same pair of points con- jugate for each circle of the system, then the radical axes of the circles, taken in pairs, are concurrent. 15. If a system of circles have a common pair of inverse points the system must be a coaxal one. 16. and 0' are the limiting points of a s3'stem of coaxal circles, and A is any point in their plane ; shew that the chord of contact of tangents drawn from A to any one of the circles will pass through the other extremity of the diameter through A of the circle AGO'. 80 CHAPTER Yin INVOLUTION 77. Definition. If be a point on a line on which lie pairs of points A, A-^ ; B, Bi ; C, C'l ; &c. such that OA.OA, = OB.OB, = OC.OC,= = k, the pairs of points are said to be in Involution. Two associated points, such as A and A^, are called conjugates; and sometimes each of two conjugates is called the 'mate' of the other. The point is called the Centre of the involution. If k, the constant of the involution, be positive, then two conjugate points lie on the same side of 0, and there will be two real points K, K' on the line on opposite sides of such that each is its own mate in the involution ; that is OK^ = OK'- = k. These points K and K' are called the double points of the involution. It is important to observe that K is not the mate of K' ; that is why we write K' and not iLj. It is clear that {AA^, KK') = — 1, and so for all the pairs of points. If ^ be negative, two conjugate points will lie on opposite sides of 0, and the double points are now imaginary. If circles be described on AA-^, BB^, CCi, &c. as diameters they will form a coaxal system, whose axis cuts the line on which the points lie in 0. INVOLUTION , 81 K and K' are the limiting points of this coaxal system. Note also that for every pair of points, each point is inverse to the other with respect to the circle on KK' as diameter. It is clear that an involution is completely determined when two pairs of points are known, or, what is equivalent, one pair of points and one double point, or the two double points. We must now proceed to establish the criterion that three pairs of points on the same line may belong to the same involution. 78. Prop. The necessary and siifficient condition that a pair of points C, C'l should belong to the involution determined by A,\-i,; B, B, is {ABCA,)={A,B,C\A). First we will shew that this cijnditinn is necessary. Suppose C and C\ do belong to the involution. Let be its centre and k its constant. . •. OA . OA, = OB . OH, = OC . 0C\ = k. / k k \ / k k \ [ob,~oaJ \oa~ocJ [oA OaJ \0n\ ~ 00 J _ ( OB,- OA,) {OA -0C\) ^ A,B,.C,A ^ {OA- OA,) {OB, - OC,) 'A,A . C\B, y^^^^^'^^^)- Thus the condition is necessary. [A more purely geometrical proof of this theorem will be given in the next paragraph.] Next the above condition is sufficient. For let {ABC A,) = (A,B,C,A) and let C be the mate of C in the involution determined by A,A,i B,B,. A. G. 6 «2 INVOLUTION .■.{A,B,C\A) = {A,B,G'A). .*. Cj and 6" coincide. Hence the proposition is established. Cor. 1. If A, A,; B,B,; C,C,; D, D, belong to the same involution {ABGI)) = {AACM. Cor. 2. If K, K' be the double points of the involution {AA.KK') = {A.AKK') and (ABKA,) = {A,B,KA). 79. We may prove the first part of the above theorem as follows. If the three pairs of points belong to the same involution, the circles on AA^^, BB^, CCi as diameters will be coaxal (§ 77). Let P be a point of intersection of these circles. Then the angles AFA,, BPB,, CPC\ are right angles and therefore P(ABCA,) = F(A,B,C\A). .-. (ABCA,) = (A,B,C,A). The circles may not cut in real points. But the proposition still holds on the principle of continuity adopted from Analysis. 80. The proposition we have just proved is of the very greatest importance. INVOLUTIOX 83 The criterion that thr^ee pairs of points belong to the same involution is that a cross-ratio formed with three of the points, one taken from each pair, and the mate of any one of the three should be equal to the corresponding cross-ratio formed by the mates of these four points. It does not of course matter in what order we wi'ite the letters provided that they correspond in the cross-ratios. We could have had {AA,CB) = {A,AC,B,) or {AA,C\B) = {A,ACBy). All that is essential is that of the four letters used in the cross-ratio, three should form one letter of each pair. 81. Prop. A range of points in involution projects info a range in involution. For let A, A^; B, B^; C, C\ be an involution and let the projections be denoted by corresponding small letters. Then {ABC A ,) = {A, B, C, A ). But {ABC A,) = {abca,) and (zi,B,C^A) = {aAciCi). .'. {abca^) = {a^biCia). .•. a, tti : 6, 61 ; r, d form an involution. Note. The centre of an involution does not project into the centre of the involution obtained by projection; but the double points do project into double points. 82. Involution Pencil. We now see that if we have a pencil consisting of pairs of VP, VF; VQ, VQ'; VU, VR' &c. such that any transversal cuts these in pairs of points A, A, ; B,B,\ C, C, &c. forming an involution, then every transversal will cut the pencil so. Such a pencil will be called a Pencil in Involution or simply an Involution Pencil. 6—2 84 INVOLUTION The double lines of the involution pencil are the lines through V on which the double points of the involutions formed by different transversals lie. Note that the double lines are harmonic conjugates with any pair of conjugate rays. From this tact it results that if VD and VU he the double lines of an involution to ivhich VA, VA^ belong, then VD and VD' are a pair of conjugate lines for the involution wJtose double lines are VA, VA^. 83. We shall postpone until a later chapter, when we come to deal with Reciprocation, the involution properties of the quadrangle and quadrilateral, and pass now to those of the circle which are of great importance. We shall make considerable use of them when we come to treat of the Conic Sections. 84. Involution properties of the circle. Prop. Pairs of points conjugate for a circle which lie along a line form a range in involution of ivhich the double points are the points of intersection of the line with the circle. Let P and Q be a pair of conjugate points on the line I. Let be the pole of /. Thus OPQ is a self-conjugate triangle, and its orthocentre is at G, the centre of the circle (§ 16 a). Let CK be the perpendicular from C on I. Then PK.KQ= KO.KC. .-. KP.KQ=OK.KC. INVOLUTION 85 Thus P and Q belong to an involution whose centre is K, and whose constant is OK . KG. Its double points are thus real or imaginary according as PQ does or does not cut the circle. If PQ cut the circle in A and B, then OK . KG = KA- = KB\ thus we see that A and B are the double points of the involu- tion. It is obvious too that A and B must be the double points since each is its own conjugate. The following proposition, which is the reciprocal of the foregoing, is easily deduced from it. Prop. Pairs of conjugate lines for a circle, luhich pass through a point form an involution pencil of which the double lines are the tangents from the point. For pairs of conjugate lines through a point will meet the polar of in pairs of conjugate points, which form an involution range, the double points of which are the points in which the polar of cuts the circle. Hence the pairs of conjugate lines through form an involution pencil, the double lines of which are the lines joining to the points in which its polar cuts the circle, that is the tangents from 0. If be within the circle the double lines are not real. 85. Orthogonal pencil in Involution. A special case of an involution puneil is that in which each of the pairs of lines contains a right angle. That such a pencil is in involution is clear from the second theorem of § 84, for pairs of lines at right angles at a point are conjugate diameters for any circle having its centre at that point. But we can also see that pairs of orthogonal lines VP, VP^; ^^Q, V^Qi &c. are in involution, by taking any transversal t to cut these in A, A^; B, B^ &c. and drawing the perpendicular VO on to t ; then 0A.0A, = -0V'=0B.0B,. 86 INVOLUTION Thus the pairs of points belong to an invohition with imaginary double points. Hence pairs of orthogonal lines at a point form a pencil in involution with imaginary double lines. Such an involution is called an orthogonal involution. Note that this property may give us a test whether three pairs of lines through a point form an involution. If they can be projected so that the angles contained by each pair become right angles, they must be in involution. 86. Prop. In every involution pencil there is one pair of rays mutually at right angles, nor can there he more than one such pair unless the involution pencil he an orthogonal one. Let P be the vertex of the pencil in involution. Take any transversal I, and let be the centre of the involution range which the pencil makes on it, and let k be the constant of this involution. Join OP, and take a point P' in OP such that OP . OP' = k. Thus P and P' will be on the same or opposite sides of according as /..• is positive or negative. Bisect PP' in M and draw J/C at right angles to PP' to meet the line I in C. Describe a circle with centre C and radius OP or CP' to cut I in A and A^. The points ^1 and A^ are mates in the involution on I for OA .OA,= OP.OP' = k. INVOLUTION 87 Also the angle APA^ being in a semicircle is a right angle. Hence the involution pencil has the pair of rays PA and PA^ mutually at right angles. In the special case where is the middle point of PP', C and coincide. In the case where PO is perpendicular to /, the line through 2T, the middle point of PP', perpendicular to PP' is parallel to L, and the point C is at infinity. In this case it is PO and the line through P parallel to I which are the pair of orthogonal rays, for and the point at infinity along I are mates of the involution range on /. Thus every involution pencil has one pair of orthogonal rays. If the pencil have more than one pair of rays at right angles, then all the pairs must be at right angles, since two pairs of rays completely determine an involution pencil. 87. Prop. An invohdion pencil projects into a pencil in involution, imd ani/ involution can be projected into an orthogonal involution. For let the pairs of rays of an involution pencil at in the p plane meet the line of intersection of the p and ir planes in A., A^; B, Bi] C, C^ &c. and let 0' be the projection of 0. Then .4, A^ ; B, B^; C, Cj &c. is an involution range. .-. 0' (A, A^; B, B^; C, C\ &c.) is an involution. Again, as we can project two angles in the p plane into right angles, we may choose two angles between two pairs of 88 INVOLUTION rays of an involution to be so projected. Then the pencil in the projection must be an orthogonal one. It may be remarked too that at tlie same time that we project the involution pencil into an orthogonal one we can project any line to infinity (§ 43). Note. xA.s an orthogonal involution has no real double lines, it is clear that if an involution pencil is to be projected into an orthogonal one, then the pencil thus projected should not have real double lines, if the projection is to be a real one. An involution pencil with real double lines can only be projected into an orthogonal one by means of an imaginary vertex of projection. The reader will understand by comparing what is here stated with § 43 that the two circles determining V in that article do not intersect- in real points, if the double points of the involution range which the involution pencil in the p plane intercepts on the vanishing line be real, as they are if the involution pencil have real double lines. EXERCISES 1. Any transversal is cut by a system of coaxal circles in pairs of points which are in involution, and tlie double points of the involution are concyclic with the limiting points of the system of circles. 2. If A", K' be the double points of an involution to which A, A^; B, By belong ; then A, B^; A^, B ; K, A" are in involution. 3. If the double lines of a pencil in involution be at right angles, they must be the bisectors of the angles between each pair of conjugate rays. 4. The corresponding sides BC, B' C ifcc. of two triangles ABC, A.' B'C in plane perspective intersect in P, Q, R respectively; and AA', BB', CC respectively intersect the line PQR in P\ Q', K'. Prove that the range {PP' , QQ\ BR') forms an involution. EXERCISES 89 5. The centre of the circumcircle of the triangle formed by the three diagonals of a quadrilateral lies on the radical axis of the system of circles on the three diagonals. 6. Shew that if each of two pairs of opposite vertices of a quadrilateral is conjugate with regard to a circle, the third pair is also ; and that the circle is one of a coaxal system of which the line of collinearity of the middle points of the diagonals is the radical axis. 7. The two pairs of tangents drawn from a point to two circles, and the two lines joining the point to their centres of similitude, form an involution. 8. Prove that there are two points in the plane of a given triangle such that the distances of each from the vertices of the triangle are in a given ratio. Prove also that the line joining these points passes through the circumcentre of the triangle. 90 CHAPTER IX THE CONIC SECTIONS 88. Definitions. The Conic Sections (or Conies, as they are frequently called) are the curves of conical, or vertical, pro- jection of a circle on to a plane other than its own. They are then the plane sections of a cone having a circular base. It is not necessary that the cone should be a right circular one, that is, that its vertex should lie on the line through the centre of the circular base and at right angles to it. So long as the cone has a circular base (and consequently too all its sections parallel to the base are circles), the sections of it are called conic sections. 89. The conic sections are classified according to the relation of the vanishing line to the projected circle. If the vanishing line touches the circle, the curve of projection is called a parabola; if the vanishing line does not meet the circle, the curve is called an ellipse; and if the vanishing line cuts the circle the curve of projection is called a hyperbola. In other words a parabola is the section of a cone, having a circular base, by a plane parallel to a generating line of the cone. By a ' generating line ' is meant a line joining the vertex of the cone to a point on the circumference of the circle which forms its base. An ellipse is a section of the cone by a plane such that the plane parallel to it through the vertex cuts the plane of the base in a line external to it. A hyperbola, is a section of the cone by a plane such that the parallel plane through the vertex cuts the base of the cone. THE CONIC SECTIONS 91 The curves are illustrated by the following figure and it should be observed that the hyperbola consists of two branches, and that to obtain both these branches the cone must be pro- longed on both sides of its vertex. 90. Focus and directrix property. Every conic section, or projection of a circle, possesses, as we shall presently shew, this property, namely that it is the locus of a point in a plane such that its distance from a fixed point in the plane bears to its distance from a fixed line, also in the plane, a constant ratio. The fixed point is called the focus of the conic, the fixed line is called the directrix, and the constant ratio the eccentricity. It Avill be proved later that the eccen- tricity is unity, less than unity, or greater than unity, according as the conic is a parabola, an ellipse, or a hyperbola. 92 THE CONIC SECTIONS 91. Text books on Geometrical Conic Sections usually take the focus and directrix property of the curves as the definition of them, and develop their properties therefrom, ignoring for the purpose of this development the fact that every conic section, even when defined by its focus and directrix property, is all the while the projection of some circle. This is to be regretted. For many of the properties which can only be evolved with great labour from the focus and directrix property are proved with great ease when the conies are regarded as the projections of a circle. We shall in the next chapter shew how easy it is to prove that plane curves having the focus and directrix property are the projections of a circle. 92. Projective properties. The conic sections, being the projections of a circle must possess all the projective properties of the circle. (1) They will be such that no straight line in their plane can meet them in more than two points, and from points which are the projections of such points in the plane of the circle as lie without the circle, two and only two tangents can be drawn, which will be the projection of the tangents to the circle. (2) The conic sections will clearly have the ' pole and polar property ' of the circle. That is, the locus of the intersections of tangents at the extremities of chords through a given point will be a line, the point and line being called in relation to one another pole and polar. The polar of a point from which tangents can be drawn to the curve will be the same as the line through the points of contact of the tangents. This line is often called ' the chord of contact,' but strictly speaking the chord is only that portion of the line intercepted by the curve. The polar is unlimited in length. (3) If the polar of a point A for a conic goes through B, then the polar of B must go through A. Two such points are called conjugate points. (4) Also if the pole of a line I lie on another line I', the pole of I' will lie on /, two such lines being called conjugate lines. THE CONIC SECTIONS 93 (5) The harmonic property of the pole and polar which obtains for the circle must hold also for the conic sections since cross-ratios are unaltered by projection. (6) As an involution range projects into a range also in involution, pairs of conjugate points for a conic which lie along a line will form an involution range whose double points will be the points (if any) in which the line cuts the curve. (7) Similarly pairs of conjugate lines through a point will form an involution pencil whose double lines are the tangents (if any) from the point. 93. Circle projected into another circle. The curve of projection of a circle is under certain con- ditions another circle. Prop. If in the curve of projection of a circle the pairs of conjugate lines through the point P luhicJi is the projection of the pole of the vanishing line form an orthogonal involution, then the curve is a circle having its centre at P. For since the polar of P is the line at infinity, the tangents at the extremities of any chord through P meet at infinity. But since the involution pencil formed by the pairs of conjugate lines through P is an orthogonal one, these tangents must meet on a line through P perpendicular to the chord. Hence the tangents at the extremities of the chord are at right angles to it. Thus the curve has the property that the tangent at every point of it is perpendicular to the radius joining the point to P. That is, the curve is a circle with P as centre. Cor. a circle can be projected into another circle luith any point within it projected into the centre. For we have only to project the polar of the point to infinity and the involution pencil formed by the conjugate lines through it into an orthogonal involution. Note. The point to be projected into the centre needs to be within the circle if the projection is to be a real one (see Note to § 87). 94 THE CONIC SECTIONS 94. Focus and directrix as pole and polar. Prop. //'" in the plane of the curve of projection of a circle there exist a point S such that the involution pencil formed hi/ the conjugate lines through IS is an orthogonal one, then 8 and its polar are focus and directrix for the curve. Let P and Q be any two points on the curve, and let the tangents at them meet in T. Join ST cutting PQ in R, and let PQ meet the polar of S in F. Join SF, and draAv PAI, QN perpendicular to the polar of S. Then ST is the polar of F, for the polar of F goes through S since that of S goes through F, and it also goes through T since the polar of T goes through F. .'. SF and ST are conjugate lines. But by hypothesis the conjugate lines at *S^ form an orthogonal involution. .-. TSF is a right angle. And by the harmonic property of the pole and polar {FR,PQ) = -l. .-. ST and SF are the bisectors of the angle PSQ,{^ 72). .-. SP:SQ = FP:FQ = PM : QN (by similar triangles). .-. SP : PM = SQ : QA\ THE CONIC SECTIONS 95 Thus the ratio of the distance of points on the curve from S to their distance from the polar of S is constant, that is S and its polar are focus and directrix for the curve. Note. If the polar of <S' should happen to be the line at infinity then the curve of projection is a circle (§ 93). We may here remark that the circle may be considered to have the focus and directrix property, the focus being at the centre, and the directrix the line at infinity. The eccentricity is the ratio of the radius of the circle to the infinite distance of the points on the circle from the line at infinity. 95. Parallel chords. Before we go on to establish the focus and directrix property of the curves of projection of a circle, which we shall do by shewing that for all of them there exists at least one point S, the pairs of conjugate lines through which form an orthogonal involution, we will establish a very important general propositi(jn about parallel chords. Prop. In cwy conic section (or curve of projection of a circle) the locus of the middle points of a system of parallel chords is a straight line, and the tangents at the points cohere this line meets the curve are parallel to the chords. Moreover the tangents at the extremities of each of the jicirallel chords ivill intersect on the line which is the locus of the middle points of the chords, and every line parallel to the chords and in the plane of the curve is conjugate with this line containing the middle jwints. Let QQ' be one <>f the chords of the system and ^1/ its middle point. The chords may be considered as concurrent in a point R at infinity which is the projection of a point r on the vanishing line ; and we have {QQ',iMR) = -l. •■• (qq',mr)^-l that is m is on the polar of r. Thus as the locus of the points m is a line, that of the points M is so too. 96 THE CONIC SECTIONS Let P be a point in which this loous meets the curve, and let P be the projection of p, then as the tangent at p goes through r that at P must go through R ; that is the tangent at P is parallel to QQ'. Further as the tangents at q and q' meet on the polar of 7\ those at Q and Q' will meet on the projection of the polar of r, that is on the line which is the locus of the middle points of the chords. Also every line through r will have its pole on the polar of r, and therefore every line through R in the plane of the conic will have its pole on the line PM, that is, every line parallel to the chords is conjugate with the locus of their middle points. In other words, the polar of every point on the line which is the locus of the middle points of a system of parallel chords is a line parallel to the chords. 96. Focus and directrix property established. We are now in a position to establish the focus and directrix property of the conic sections, defined as the projections of a circle. We shall take the parabola, ellipse and hyperbola separately, and in each case prove a preliminary proposition respecting their axes of symmetry. Prop. A parabola (or the projection of a circle touched by the vanishing line in its plane) has an axis of symmetry which meets the curve in tivo points one of luhich is at infinity. THE CONIC SECTIONS 97 Let the vanishing line touch the circle in co. In the plane through V, the vertex of projection, and the vanishing line draw Vr at right angles to Vco, meeting the vanishing line in r. Draw the other tangent ra to the circle. Now /■ is the pole of aw, and therefore if pp' be any chord which produced passes through 7- and which cuts aco in n then (pp', 7ir) = -l. Thus, using c<^rresponding capital letters in the projection, and remembering that rno} will project into a right angle since ru) subtends a right angle at V, we shall have a chord FF' at right angles to Ail and cutting it at N so that {FF\ NR) = -1. But R is at infinity. .-. FN=^^F\ Thus all the chords perpendicular to Ail are bisected by it, and the curve is therefore symmetrical about this line, which is called the axis of the parabola. The axis meets the curve in the point A, called the vertex, and in the point H which is at infinity. As raay projects into a right angle (for rw subtends a right angle at V) the tangent at A is at right angles to the axis. Finally the curve touches the line at infinity at O. 97. Prop. ^4 parabola (or projection of a circle touched by the vanisliing line) has the focus and directrix property, and the eccentricity of the curve is unity. A. G. 7 THE CONIC SECTIONS Let P be any point on the curve of projection. Let PNP' be the chord through P, perpendicular to the axis, and cutting it in iY. The tangents at P and P' will intersect in a point T on the line of the axis (§ 95). Then as T is the pole of I^P' .'. as n is at infinity, TA = AN. Now let the tangent at P meet that at A in ]' and draw YS at right angles to PF to meet the axis in S. The polar of S will be at right angles to the axis (§ 95). Let it be XM cutting the axis in X, the tangent at P in Z, and the line through P parallel to the axis in M. Join SP, SZ. Now as *S' is the pole of XM (xs, An) = -i. .-. XA= AS, and as TA = AN we have TS = XN = MP. But TY.YP=^TA.AN=^\, so that A TYS ^APYS, and TS=PS. Thus PS = PM, that is P is equidistant from >S^ and the polar of S. THE CONIC SECTIONS 99 Further, since ST is equal and parallel to PM and SP = PM, SPMT is a rhombus and PT bisects the angle SPM. Thus ASPZ = AMPZ, and Z ZSP = Z ZMP = a right Z . Now Z is the pole of SP, for the polar of Z must go through S (since that of S goes through Z) and through P since -Z'P is a tangent at P. Hence SZ and SP are conjugate lines for the curve, and they are at right angles. So also are ST and the line through >S' at right angles to it (§ '95). Therefore the involution pencil formed by the pairs of con- jugate lines through *S' is an orthogonal one. Thus *S' and its polar XM are focus and directrix for the curve (§ 94), and the eccentricity is SP : PM which is unity. 98. Prop. The projection of a circle not met by the vanishing line in its plane is either a circle or a closed curve having two axes of symmetry, mutually perpendicular, on which are intercepted by the curve chords of unequal length. ^\ ^'. / ^__ B R/ X^^^ M \ p r 1 ^ ^ -^ i \ sP' C N Let c be the pole of the vanishing line. Let pp be any chord through c of the circle. Then j^p' is divided harmonically at c and its intersection with the polar of c. 7—2 100 THE CONIC SECTIONS Using corresponding capital letters in the projection, we shall have that the chord FF' through C is divided harmonically at G and its intersection with the polar of C, which is the line at infinity. .-. FC =CF'. Thus every chord through C is bisected at C. For this reason the point C is called the centre of the curve, and the chords through it are called diameters. The tangents at the extremities of any diameter are parallel, for the tangents at the extremities of chords of the circle through c meet in the polar of c, which is the vanishing line. First suppose that the involution pencil formed by the pairs of conjugate lines through G is an orthogonal one ; then the curve is a circle (§ 93). Next suppose that the involution pencil is not an orthogonal one; then there must be one and only one pair of conjugate lines through G mutually at right angles (§ 86). Let the curve intercept on these lines chords A A' and BB'. Draw the chords FQ and FR perpendicular to A A' and BB' respectively cutting -them in N and Jli, and let O and H' be the points at infinity on the lines of A A' and BB'. Then as fl' is the pole oi AA', and FQ passes through O', {FQ,Nn') = -l. .-. FN=NQ. Similarly FM = MR. Thus the curve is symmetrical about each of the two lines AG A', BGB', which are called the axes. We shall now shew that AA' and BB' cannot be equal. Let the tangents at A and B meet in K, then GAKB is a rectangle, and GK bisects AB. But GK bisects the chord through G parallel to AB, for every chord through G is bisected at G. Hence GK an'd the line through G parallel to AB are con- jugate lines (§ 95). THE CONIC SECTIONS lOl But these lines would be at right angles if CA = CB. And thus if GA and GB were equal the involution pencil formed by the pairs of conjugate lines through G would be an orthogonal one ; which is contrary to hypothesis. Hence AA' and BB' cannot be equal. We shall suppose A A' to be the greater of the two. ^^' is called the major axis and BB' the minor axis. Then 99. Prop. An ellipse (or curve of projection, other than a circle, of a circle not met hij the vanishing line in its plane) has the focus and directrix property and the eccentricity of the curve is less than unity. Assuming that the projection is not a circle we have as shewn in § 98 two axes of symmetry AA', BB' of which AA' is the greater. With centre B and radius equal to GA describe a circle cutting the major axis in S and S'. The polars of S and S' are perpendicular to AA' (§ 95). Let these be XF and X' F' , cutting A A' in X and X' and the tangent at B in F and F'. Now since the polar of S goes through F, that of F goes through S ; but the polar of F goes through B, since FB is a tangent. .-. SB is the polar of F, and SF, SB are conjugate lines. We will shew that they are mutually at right angles. 102 THE CONIC SECTIONS ^ince S is the pole of A''^ (^^', /S'X) = -1. .-. CS.GX = CA"-. Now draw SK parallel to CB to meet BF in K. Then BK . KF = OS . SX = CS (CX - OS) = GA"- - CS-' =^SB'-GS'=SK' .'. FSB is a right angle. < B \(] \^ aI s 3 S' A' Thus Ave have two pairs of conjugate lines through >S' mutually at right angles, namely SF and SB, as also SX and SK (§ 95) so that the involution pencil formed by the conjugate lines at *S^ is an orthogonal one. .•. S and its polar XF are focus and directrix for the curve (§ 94). Similarly S' and its polar X' F' are focus and directrix. The eccentricity = SB : BF = GA : GX which is less than unity. Note too that as GS . GX = GA'-, the eccentricity = 6'>S' : GA. Note. We see now how a circle may be regarded as the limiting case of an ellipse whose two foci S and S' coincide with its centre, and its directrix is the line at infinity (see Note to § 94). 100. Prop. A hyperbola (or projection of a circle which is cut by the vanishing line) has two axes of symmetry mutually at i^ight angles, only one of which cuts the curve. THE CONIC SECTIONS 103 Let the vanishing line cut the circle in w and co' , and let c be the pole of the line. Let pp be any chord the line of which passes through c. Then j^P' is divided harmonically at c and its intersection with WW. Using corresponding capital letters in the projection, we shall have that the chord FP' through C is divided harmonically at G and its intersection with the polar of C which is the line at infinity. .-. FC = CP'. Thus every chord through C in the curve of projection is bisected at G, which is therefore called the centre of the curve, and the chords through G are called diameters. Let it be observed that not every line through G meets the curve, since in the plane of the circle there are lines through c which do not meet it. Of each pair of conjugate lines through c only one will meet the circle, for cut and cw' are the double lines of the involution formed by these conjugate lines. Further the involution pencil formed by the conjugate lines through G cannot be an orthogonal one, since it has real double lines, namely the projection of c&> and cw. Thus there will be one and only one pair of conjugate lines through G mutually at right angles' (§ 86). 104 THE CONIC SECTIONS Let this pair be CA and GB, of which the former is the one that meets the curve, namely in A and A'. Note that the curve of projection will have two tangents, from G whose points of contact fl and iV the projections of co and co' are at infinity. These tangents are called asymptotes. Since CH and GDf are the double lines of the involution pencil formed by the conjugate lines through G G {mr, AB)=--l (§82). .•. Since GA and GB are at right angles, they are the bisectors of the angles between Of! and GVL' (§ 72). To prove that the curve is symmetrical about GA and GB we draw chords PQ, PR perpendicular to them and cutting them in N and M. Let Z and Z' be the points at infinity along the lines GA and GB. Then since Z' is the pole of AA' and PQ passes through Z', (PQ, NZ') = -1. .-. PN = NQ. Similarly PM = MR. Thus the curve has two axes of symmetry mutually at right angles, one of which meets the curve, and the other not. AA' which meets the curve is called the transverse axis and GB is called the conjugate axis. Note. At present B is not a definite point on the line GB. We shall find it convenient later on to make it definite ; the point to be emphasised is that the transverse axis does not cut the curve, and we cannot determine points B and B' on it as these points are determined in the case of the ellipse. 101. Prop. A hyperbola (or curve of projection of a circle cut by the vanishing line) has the focus and directiHx property, and the eccentricity is g7-eater than unity. Using the notation of the preceding article, we describe a circle with centre G, and radius GA cutting GH in K and L', and Gfl' in K' and L, as in the figure. The lines KL and K'L' will be perpendicular to the trans- verse axis, since, as we have seen, GA bisects the angle OCH'. THE COXIC SECTIONS 105 .•. the poles of these lines, which we will denote by S and S\ will lie on the line of the transverse axis (§ 95). We will now shew that S and its polar KL are focus and directrix, as are also S' and K'L'. Let KL and K'L' cut AA' in X and A". Then by the harmonic property of the pole and polar {AA', SX) = - 1. .-. CS.CX = CA'=CK\ .'. CKS is a right angle. Now the polar of K must go through S, since that of S goes through K. Moreover the polar of K goes through H, since KCl is a tangent at O. .•. Sn is the polar of A", that is >S7i and SD. are conjugate lines. But n being at infinity SCI is parallel to KQ., that is Sfl is perpendicular to SK. Thus we have two pairs of conjugate lines through S, mutually at right angles, namely Sfl and SK, and >S'6' and the line through S at right angles to SC (§ 95). Hence the pencil formed by the pairs of conjugate lines through S is an orthogonal one, and therefore *S' and its polar KX are focus and directrix for the curve. 106 THE CONIC SECTIONS Similarly S' and K'L' are focus and directrix. The eccentricity is the ratio Sn : perpendicular from II on KL = KVl : the same = GK : OX = CA : CX which is greater than unity. Note too that as C^. CX = CA'^, the eccentricity also = C'>S^: C'J.. We might have obtained the eccentricity thus : It is the ratio SA : AX = CS- CA : CA - CX = CH . CX - CA . CX : CX {CA - CX) = CA (CA - CX) : CX (CA - CX) = CA : CX. 102. Central and non-central conies. Diameters. We have seen that the ellipse and hyperbola have each a centre, that is a point such that ever}^ chord passing through the same is bisected by it. Ellipses and hyperbolas then are classified together as central conies. The parabola has no centre and is called non-central. We have proved in § 95 that the locus of the middle points of a system of parallel chords is a straight line. Clearly in the case of the central conies this line must go through the centre, for the diameter parallel to the chords is bisected at that point. In the case of the parabola, the line which is the locus of the middle points of a system of parallel chords is parallel to the axis. For such a system is the projection of chords of the circle con- current at a point r on the vanishing line; and the polar of r, which projects into the locus of the middle points of the system of chords, passes through ro the point of contact of the circle with the vanishing line. Thus the locus of the middle points of the system of parallel chords of the parabola passes through fl, that is, the line is parallel to the axis. All lines then in the plane of a parabola and parallel to its axis will bisect each a system of parallel chords. These lines are conveniently called diameters of the parabola. They are not diameters in the same sense in which the diameters of a THE CONIC SECTIONS 107 central conic are, for they are not limited in length and bisected at a definite" point. 103. Ordinates of diameters. Def . The parallel chords of a conic bisected by a particular diameter are called double ordinates of that diameter, and the half chord is called an ordinate of the diameter. The ordinates of a diameter are as we have seen parallel to the tangents at the point or points in which the diameter meets the curve. The ordinates of an axis of a conic are perpendicular to that axis. The ordinates of the axis of a parabola, of the major axis of an ellipse, and of the transverse axis of a hyperbola are often called simply ' ordinates ' without specifying that whereto they are ordinates. Thus the ordinate of a point P on a parabola, ellipse or hyperbola must be understood to mean the per- pendicular PN on the axis, the major axis, or the transverse axis, as the case may be. Note. When we speak of the axis of a conic there can be no ambiguity in the case of a parabola, but in the case of the ellipse and hyperbola, which have two axes of symmetry, there would be ambiguity unless we determined beforehand which axis was meant. Let it then be understood that by the axis of a conic will be meant that one on which the foci lie. 104. The contents of the present chapter are of great importance for a right understanding of the conic sections. The student should now have a good general idea of the form of the curves, and, as it were, see them whole, realising that they have been obtained by projecting a circle from one plane on to another. We shall in the next chapter set forth properties which all conies have in common, and in subsequent chapters treat of the parabola, ellipse and hyperbola separately, shewing the special properties which each curve has. 108 CHAPTER X PROPERTIES COMMON TO ALL CONICS 105. Proposition. If the line {produced if necessary) joining two points P and Q of a, conic meet a directrix in F, and S be the corresponding focus, SF will bisect one of the angles between SP and SQ. Fig. 1. For, drawing PM and QR perpendicular to the directrix we have, if e be the eccentricity, SP:PM=e = SQ:QR, .-. SP:SQ = PM:QR = FP:FQ (by similar As FQR, FPM). PROPERTIES COMMON TO ALL CONICS 109 .-. in figs 1 and 3, *S'^ bisects the exterior angle of PSQ, and in fig. 2 it bisects the angle PSQ itself. We see then that SF bisects the exterior angle of PSQ, if P and Q be on the same branch of the curve, and the angle P»SQ itself if P and Q be on opposite branches. 106. Prop. //" the tangent to a conic at a point P meet a directrix in Z and S he the corresiDonding focus, ZSP is a ricfht angle. This is easily seen from the following considerations : The focus and directrix are ' pole and polar ' for the conic, therefore the tangents at the extremities of a focal chord PSQ will meet at Z in the directrix, and Z will be the pole of PQ. Thus SZ and SP lies on SZ. conjugate lines, since the pole of SP 110 PROPERTIES COMMON TO ALL CONICS But the pairs of conjugate lines through a focus are at right angles. Therefore ZSP is a right angle. But as we are going to prove in the next article that every plane curve having the focus and directrix property is the projection of some circle, we will give another proof of the proposition dependent only on this property. Regard the tangent at P as the limiting case of the line of the chord PP' when P' is very close to P. Now if PP' meet the directrix in F, SF bisects the exterior angle of PSP' (§ 105) for P and P' are on the same branch. And the nearer P' approaches P, the more does this exterior angle approximate to two right angles. Thus Z ZSP = the limit of FSP' when P' approaches P = a right angle. It should be observed that this second proof yields also the result that tangents at the ex- tremities of a focal chord intersect in the directrix, since the tangent at either end of the chord PSQ is determined by drawing SZ at right angles to the chord lo meet the directrix in Z; then ZP, ZQ are the tanoents. 107. In the preceding chapter we defined the conic sections as the curves of projection of a circle and showed that they have the focus and directrix property. We shall now establish the converse proposition. Prop. Every 2)lane curve having the focus and directrix property is the projection of some circle. For we have shown in the second jjart of § 106 that a curve having the focus and directrix property is such that tangents at the extremities of any chord through S, the focus, intersect on the directrix on a line through 8 perpendicular to the chord. PROPERTIES COMMON TO ALL CONICS 111 Now project so that the directrix is the vanishing line ; and so that the orthogonal involution at S projects into another orthogonal involution (§ 87). Then the curve of projection has the property that the tangents at the extremities of every chord through s, the pro- jection of S, meet at infinity on a line through s perpendicular to the chord. Hence the tangent at each point of the curve is at right angles to the radius joining the point to s, and therefore the curve is a circle Avith s as centre. 108. It follows of course from what we have established in the preceding chapter that if the curve having the focus and directrix property had its eccentricity unity then the circle into which it has been projected must touch the vanishing line in the plane <jf the circle ; if the eccentricity be less than unity then the circle does not meet the vanishing line ; if the eccen- tricity be greater than unity the circle is cut by the vanishing line. ' 109. Pair of tangents. Prop. J/ a pair of ttDiyents TF, TQ be drawn to a conic from a point P and S be a focus, then SP and i^Q make equal angles with ST, and if PQ meet the corresponding directrix in F Z TSF is a right angle. Let Tti meet FQ in R. 112 TROPERTIES COMMON TO ALL CONICS Since PQ is the polar of T, and this goes through F, the polar of F must go through T. But since F is on the directrix the polar of F must go through 8. Thus 8T is the polar of F. Hence 8F and 8T are conjugate lines, and as they are through a focus, they must be at right angles. Further (PQ, FR) = -l .-. S(PQ,FR) = -1. Thus SR and SF are the bisectors of the angles between SP and SQ (§ 72). Note. It will be seen that if the points of contact of the tangents from T lie on the same branch of the curve ST bisects the angle PSQ, but if they are on different branches then ST bisects the exterior angle of PSQ. The figures given do not shew the case v/here TP and TQ both touch the branch remote from S. The student can easily represent this in a figure of his own. 110. The above proposition gives a simple construction for drawing two tangents to a conic from an external point T. PROPERTIES COMMON TO ALL CONICS 113 Join ST and let it meet the conic in K and K'. The figure of the preceding article can be utilised. Take R in KK' such that {TR, KK') = - 1 (§ 70, Cor. 2). Draw SF at right angles to ST to meet the directrix in F. Draw the line FR and let it cut the conic in P and Q. Then TP and TQ are the tangents. For as ST and SF are at right angles and through a focus, they are conjugate lines. .-. the pole of /SfT lies on SF. But the pole of SI' is on the directrix. . ■. i^ is the pole of ST, that is, the polar of F goes through T. . ■ . the polar of T goes through F. But the polar of T goes through R since (TR, KK') = - 1. Thus the line FR is the polar of T, that is, PQ is the chord of contact of the tangents from T. 111. The Normal. Def. The line through the point of contact of a tangent and at right angles to it is called the normal at that point. Prop. If the normal at P to any conic meet the aocis in G, and S be a focus of the conic then SG = e . SP, where e is the eccentricity. Let the tangent at P meet the directrix corresponding to SinZ. Draw PM perpendicular to the directrix. Then since PM is'Jparallel to the axis, Z MPS = Z PSG. A. G. 8 114 PROPERTIES COMMON TO ALL CONICS Also since PMZ and i^S'Z are right angles, PSZM is cyclic and ^^ SMP = Z SZP = complement of z SPZ = z SPO. Thus the A s ISPG, PMS are similar and SG:SP = PS:PM = e- .-.80 = 6. SP. The student can make for himself a figure shewing the case where P is on the branch of a hyperbola remote from S. In this case it will be found ZSMP = 180°- ZSZP = 90°+ ZSPZ= ZSPG. So that the As SPG and PMS are still similar. 112. The latus rectum. Def. The focal chord perpendicular to the axis on which the focus lies is called the latus rectum of the conic. Thus the latus rectum is the double ordinate through the focus to the axis (§ 103), Prop. The semi-latus rectum of a conic is aharmonic mean between the segments of any focal chord. Let aS7> be the semi-latus rectum, and PSQ any focal chord. Draw PM and QR perpendicular to the directrix, and PN and QK perpendicular to the axis. Then . SP:P3f = e = SL:SX = 8Q : QR. And by similar triangles SP ■.SQ = SN: KS = XN - XS : XS - XK (Fig. 1 ) = MP - XS : A^S' - RQ = e{MP-XS):e(XS-RQ) = SP- SL : SL - SQ. .■. SP, SL and SQ are in harmonic progression and _L J_ - A SP'^SQ'~SL' This proposition requires some modification if P and Q are on opposite branches. We ijow have PROPERTIES COMMON TO ALL CONICS 115 SP : SQ = SN : KS = XS - XR :KX + XS (Fig. 2) = e{XS-3IP):e(QR + XS) = SL-SP:SQ + SL Fig. 1. Fig. 2. .-. SP {SQ + SL) = SQ (SL - SP) . ■ . SQ . SL - SP . SL = 2.ST . SQ 1 __L_ ^ '' SP SQ'SL' Thus in this case it is SP, SL and — SQ that are m h.p. Cor. IVie rectangle contained by tJie segments of any focal chord varies as the length of the chord. ^ SP - SQ SL ' according as P and Q be on the same or opposite branches {P being on the branch adjacent to S), and in both cases we have PQ ^ _2_ SP . SQ SL .-. SP.SQ = '^^'xPQ that is SP.SQozPQ. If SP and SQ are in opposite directions P and Q lie on the same branch of the curve, and if they are in the same direction, P and Q lie on opposite branches. 113. Prop. Any conic can be projected into a circle loith any point in the plane of the conic projected into the centre of the circle. 8—2 116 PROPERTIES COMMON TO ALL CONICS For let P be any point in the plane of the conic. Take the jiolar of P for the vanishing line and project so> that the involution pencil formed by the pairs of conjugate lines through P projects into an orthogonal involution (§ 87); then exactly as in § 98 we can prove that the curve of projection is a circle. Note. In order that the projection may be a real one, the tangents from P to the conic must not be real (Note to § 87), that is P must lie within the conic. Carnot's theorem. 114. Prop. If a conic cut the sides of a triangle ABC in A^, A^; B,, B^; C\, C^; then AB, . AB^ . CA, . CA, . BC, . BC, = AC\.Aa. BA, . BA, . CB, . CB,. Project the conic into a circle ; and denote the points in the projection by corresponding small letters. Then since abi . ah = ac^ . ac2, C«i . COa = c6i . cb.2 , bcy . bc2 = buy. ba.2,, .' . ahi . ah, . ca-i . ca., • bci . bc.2 = aCj . ac.^ . bui . ba^ . cb^ . 063. .•. the triangle formed by the lines ttj^a, b^Co, c^a^ is in per- spective with the triangle abc (§ 68). PROPERTIES COMMON TO ALL CONICS 117 .•. the triangle formed by the lines A^Bo, B^Co, G^Ao is in perspective with the triangle ABC. .-.by §68 AB, . AB, . CA, . CA.^ . BC, . BC, = AC\.ACo_. BA, . BA, . CB, . CB.,. Newton's theorem. 115. Prop. If be a variable jjoint in the plane of a conic, and PQ, RS be chords in fixed directions through 0, then OP.OQ OB. OS IS constant. Let 0' be [any other point and through 0' draw the chords P'Q', R'S' parallel respectively to PQ and RS. Let QP, Q'P' meet in &> at infinity and SR, S'R' in CI. Let P'Q' and RS meet in T. Now apply Carnot's theorem to the triangle coOT and get ayP . oyQ . OR . OS . TP' . TQ' but coP' . o)Q' . TR . TS . OP . OQ -1. 1, (oP , , &>( ~p;= 1 and ~^. loP o)Q 118 PROPERTIES COMMON TO ALL CONICS ■ ■ TR. TS ~ OK.OIS' Next apply Carnot's theorem to the triangle D.TO' and get n R.nS. TF. TQ ' . O'R ■ O'S' nw . ns' . o'P' . O'Q' . tr . Ts ~ TP'.TQ' ^ O'F' .O'Q' • • TR. TS ~ O'R' . O'S' • Hence OP . OQ _0'P' .O'Q OR.OS~ O'R'.O'S" ,L . OP.OQ . that IS, 77^5 — 7T-< IS constant. UK . (Jo This proposition is known as Newton's theorem. Note. In applying Newton's theorem it must be remembered that the lines OP, OQ, &c. have sign as well as magnitude. If OP and OQ are opposite in direction, they have opposite sign, and so for OR and OS. 116. Newton's theorem is of great importance, as we shall see in later chapters, where considerable use will be made of it. We give some propositions illustrating its use. Prop. If two chords of a conic PP' and QQ', intersect in the ratio OP . OP' : OQ . OQ' is equal to that of the lengths of the focal chords parallel to PP' and QQ'. Let the focal chords parallel to PP' and QQ' be pSp' and qSq'. Then by Newton's theorem OP . OF : OQ . OQ' = Sp . Sp : Sq . Sq' = pp':qq (§ 112 Cor.). In the special case where is the centre of the conic we have OP' = - OP and OQ' = - OQ. .-. OP"-:OQ^-=pp':qq'. Note. We have already explained that in using Newton's theorem, the signs of the segments of the line are to be considered. If OP . OP' and OQ . OQ' have opposite signs so also will *S^ . Sj^ and Sq . Sq' have opposite signs. This only happens in PROPERTIES COMMON TO ALL CONICS 119 the case of the hyperbola, and when one of the four points p, p' and q, q lies on the opposite branch to the other three. Then one of the focal chords pp', qq will join two points on opposite branches and the other will join two points on the same branch. If we make the convention that a negative value be attached to the length of a focal chord if it joins two points on opposite branches otherwise it is to count positive the relation OP.OP:OQ.OQ'^pp:qq' is algebraically as well as numerically correct. So also it is true, with the same convention as to sign, that if CP, CQ be the semidiameters parallel to the focal chords, pp', qq' pp' : qq' = CP' : CQ\ And from this we see that of the two diameters parallel to two focal chords, one of which joins two points on the same branch and the other two points on opposite branches, only one can meet the curve in real points for the ratio CP- : CQ- has now a negative value. 117. Prop. //■ OP and OQ be two tangents to a conic then OP' : OQ- is equal to the ratio of the focal chords parallel respectiveli/ to OP and OQ. Let the focal chords be pSp', qSq'. Then regarding OP as meeting the curve in two coincident points P, and OQ similarly, we have by Newton's theorem OP.OP= OQ .OQ = Sp. Sp : Sq . Sq', .-. OP\:OQ^=pp:qq'. Whence we see that the focal chords have the same sign. It is clear too that the ratio of the tangents from a point to a central conic is equal to that of the diameters parallel to them. 118. Prop. //" a circle cut a conic in four points the chords joining their points of intersection in pairs are equally inclined to the axis. Let the conic and circle intersect in the four points P, Q, P', Q'. Let PP' and QQ' intersect in 0. Draw focal chords pSp', qSq parallel to PP' and QQ'. 120 PROPERTIES COMMON TO ALL CONICS Then, by Newton's theorem OP.OP'-.OQ. OQ' = Sp . Sp' : Sq . Sq' = pp':qq'. But from the circle op.OF = OQ.oq, • ' • PP = 'Vl' ^^d ^P • ^P' = ^9 • ^'j'- Thus the parallel focal chords have the same sign, their lengths are ecjual and the rectangles contained by their segments are equal. These choi'ds must then be symmetrically placed and make equal angles with the axis. Thus PP' and QQ', parallel to them, make equal angles with the axis. CoR. If a circle touch a conic at one point and cut it at two others, the tangent at the point of contact and the chord joining the two points of intersection make equal angles with the axis. Circle of curvature. 119. An infinite number of circles can be drawn to touch a conic at a given point P, such circles having their centres along the normal at the point. These circles will in general cut the conic in two other points, but in the special case where one of these other two points coincides with the point of contact P the circle is called the circle of curvature at P. This circle may be regarded as the limiting case of the circle passing through P and through two points on the conic consecutive to P, so that the conic and the circle have two consecutive tangents in common. They have then the same rate of curvature at that point. The subject of curvature properly belongs to the Differential Calculus, but it seems desirable to give here the principal properties of the circles of curvature of conies* Accordingly we shall at the end of the chapters on the para- bola, ellipse, and hyperbola add a proposition relating to the circles of curvature for these curves. It is clear from § 118 that if the circle of curvature at a point P of a conic cut the conic again in Q then PQ and the tangent at P are equally inclined to the axis. For the tangent at P and the chord PQ are the common chords of the circle and the conic. PROPERTIES COMMON TO ALL COXICS 121 The following figure illustrates the circle of curvature at a poipt of a conic. Self-Polar Triangle. 119a. Prop. If a conic pass through the four i^oints of a quadrangle, the diagonal or harmonic triangle is self-polar with regard to the conic — that is, each vertex is the pole of the opposite Mde. Let ABCD be the quadrangle; PQR the diagonal or harmonic triangle. Let PQ cut AD and BC in X and F. 122 PROPERTIES COMMON TO ALL CONICS Then (AD,XR)=-1, .'. the polar of R goes thi'ough A'" (§ 92 (5)), and {BC\YR) = -1, .'. the polar of J? goes through Y. .'. PQ is the polar of R. Similarly QR is the polar of P, and PR of Q. Thus the proposition is proved. Another way of stating this proposition would be to say that the diagonal points when taken in pairs are conjugate for the conic. The triangle PQR is also called self-conjugate with regard to the conic. EXERCISES 1. Given n conic and a focus and corresponding directrix of it, shew how to draw the tangent at any point. 2. Given two points on a conic and a directrix, shew that the locus of the corresponding focus is a circle. 3. POP' and QOQ' are two chords of a conic intersecting in 0, prove that PQ and P'Q' meet on the polar of 0. [Project the conic into a circle and into the centre.] 4. If the tangent at the end of a latus rectum LSL' meet the tangent at the nearer vertex A in T then TA =AS. 5. If the tangent at any point /* of a conic meet a directrix in F, and the latus rectum through the corresponding focus in D then SD : SP = the eccentricity. 6. If the normal at P to a conic meet the axis in G, and GL be perpendicular to the focal i-adius SP, tlien PL = the semi-latus rectum. 7. If PSP' be a focal chord and Q any point on the conic and if PQ and PQ meet the directrix corresponding to the focus S in F and F', FSF' is a right angle. PROPERTIES COMMON TO ALL CONICS 123 8. If a conic touch the sides opposite to A, £, C oi a. triangle ABO in I), U, F respectively then AD, BE, CF are concurrent. [Use §114.] 9. By means of Newton's theorem prove that if PX be the ordinate of a point P on a parabola whose vertex is A, then PN'^ : AJ^ is independent of the position of P on the curve. 10. Conies are drawn through two fixed points I> and E, and are such that DE subtends a constant angle at a focus of them ; shew that the line joining this focus to the pole of DE passes through a fixed point. 11. The polar of any point with respect to a conic meets a directrix on the diameter which bisects the focal chord drawn through the point and the corresponding focus. 12. Pro\'e that the line joining a focus of a conic to that point in the correspcmding directrix at which a diameter bisecting a system of parallel chords meets it is perpendicular to the chords. ■ [Use §§ 95 and 106.] 13. P and Q are two points on a conic, and the diameters bisecting the chords parallel respectively to the tangents at P and Q meet a directrix iu M and N ; shew that MX subtends at the corresponding focus an angle equal to that between the tangents at P and Q. 14. Given a focus and the corresponding directrix of a variable conic, shew that the polar of a given point passes through a fixed point. 15. Given a focus and two points of a variable conic, prove that the corresponding directrix must pass through one or other of two fixed points. 16. If two conies have a common focus, a chord common to the two conies will pass through the point of intersection of the corre- sponding directrices. 17. If P be any point on the tangent at a point P of a conic of which >S is a focus, and if I'M be the perpendicular to SP, and Ty the perpendicular on the directi'ix corresponding to 6', then S2I : TJV - e. (Adams' theorem.) 18. Given a focus of a conic and a chord through that focus, prove that the locus of the extremities of the coi'responding latus rectum is a circle. 124 PROPERTIES COMMON TO ALL CONICS 19. If TP and TQ be two tangents to a conic prove that the portion of a tangent parallel to PQ intercepted between TP and TQ is bisected at the point of contact. 20. A diameter of a conic meets the curve in P and bisects the chord QR which is a normal at Q, shew that the diameter through Q bisects the chord through P which is a normal at P. 21. PQ is a chord of a conic cutting the axis in K, and T is the pole of PQ ; the diameter bisecting PQ meets a directrix in Z and S is the corresponding focus ; prove that TS is parallel to ZK. 22. If AA\ BB\ CC be chords of a conic concurrent at 0, and P any point on the conic, then the points of intersection of the straight lines BC, PA', of CA, PB', and of AB, PC" lie on a straight line through 0. [Project to infinity the line joining to the point of intersection of AB, PC and the conic into a circle.] 23. A, B, C, D are four points on a conic ; AB, CD meet in E; AC and BD in F ; and the tangents at A and D in G ; prove that U, F, G are col linear. [Project AD and BG into parallel lines and the conic into a circle.] 24. If a conic be inscribed in a quadrilateral, the line joining two of the points of contact will pass through one of the angular points of the triangle formed by the diagonals of the quadrilateral. 25. Prove Pascal's theorem, that if a hexagon be inscribed in a conic the pairs of opposite sides meet in three collinear points. [Project the conic into a circle so that the line joining the points of intersection of two pairs of opposite sides is projected to infinity.] 26. ^ is a fixed point in the plane of a conic, and P any point on the polar of A. The tangents from P to the conic meet a given line in Q and R. Shew that AR, PQ, and AQ, PR intersect on a fixed line. [Project the conic into a circle having the j^rojection of A for centime.] 27. A system of conies touch AB and .^C at ^ and C. D is a fixed point, and BD, CD meet one of the conies in P, Q. Shew that PQ meets BC in a fixed point. 28. If a conic pass through the points A, B, C, D, the points of intersection of AG and BD, of xlZ/and CD, of the tangents at B and C, and of the tangents at A and D are collinear. PROPERTIES COMMON TO ALL CONICS 125 29. Til rough a fixed point A on a conic two fixed straight lines AI, AT are drawn, »S' and *S" are two fixed points and P a variable point on the conic ; PS, PS' meet AI, AT in Q, Q' respectively, shew that QQ' passes through a fixed point. 30. If a conic cut the sides BC, CA, AB of a triangle ABC in A-^A.2, B^B^, C1C2, and AA^, BB^, CC^ are concurrent, then will AA.2, BB.2, GC^ be concurrent. 31. When a triangle is self-conjugate for a conic, two and onl}^ two of its sides cut the curve in real points. 32. Prove that of two conjugate diameters of a hyperbola, one and only one can cut the curve in real points. [Two conjugate diameters and the line at infinity form a self- conjugate triangle.] 33. Given four points .S', A, B, C, shew that in general four conies can be drawn through A, B, C having ^S" as focus ; and that three of the conies are hyperbolas with A, B, C not on the same branch, while the remaining conic may be an ellipse, a parabola, or a hyperbola having A, B, C on the same branch. 34. Prove that a circle can be projected into a pai\abola with any given point within the circle projected into the focus. 35. Prove that a circle can be projected into an ellipse with two given points within the circle projected into the centre and a focus of the ellipse. 36. Prove that a circle can be projected into a hyperbola with a given point P within the circle and another given point Q without it projected respectively into a focus and the centre of the hyperbola. 126 CHAPTER XI THE PARABOLA 120. The form of the parabola has ah-eady been indicated in §§ 96 and 97. In this chapter we shall develop the special properties of the curve. Throughout A will stand for the vertex, S for the focus, X for the intersection of the directrix with the axis, and Q for the point at infinity along the axis, at which point, as we have seen, the parabola touches the line at infinity. Prop. The latus rectum = 4<AS. Let LSL' be the latus rectum. Draw LJ\l perpendicular to the directrix. Then LL' = 2LS = 2LM = 'ISX = 4>AS. 121. Prop. If PN be the ordinate of the point P, then PN' = ^AS.AN. Let PN meet the parabola again in P', and let LSL' be the latus rectum. Then by Newton's theorem (§ 115), NP . NP' : SL . SL ' = NA . Na : SA . Sn ^NA:SA since H is at infinity. THE PARABOLA PN'-AAS' = AN':AS; .-. PN"- = 4.AS.AA'. 127 This prop(jsiti(ni will later on be seen to be only a special case of a more general theorem. 122. The preceding proposition shews that a parabola is the locus of a point in a plane such that the square of its distance from a line I varies as its distance from a perpendicular line I'. The line I is the axis, I' the tangent at the vertex, and the constant of variation is the length of the latus rectum. To determine the parabola we ought to know on which side of the line // the point lies. If it may lie on either side then the locus is tivo parabolas, each of which is got from the other by rotating the figure about the tangent at the vertex through two right angles. 123. Tangent and Normal. Prop. If the tangent and normal at P meet the axis in T and G respectively, and PN be the ordinate of P, (1) TA=AN (2) NG = 2AS. The first of these properties has been already proved in § 97, We have seen that if PJ^ meet the curve again in P', the tangents at P and P' meet on the line of the axis, that is, they 128 THE PARABOLA intersect in T. Then by the harmonic property of the pole and polar {TN, Aa) = -l; .-. TA=AN. Also since TPG is a right angle PN'- = TN . NG = ^AN . NG. But PN'-=4.AS. AN {ll^l), .-. NG = 2AS. Def. NG is called the suhnormal of the point P. Thus in a parabola the subnormal is constant. 124. Prop. The tangent at any point of a 'parabola makes equal angles luith the axis and the focal distance of the M Z ?^ <7\ T Is G Let the tangent at P meet the directrix in Z, and the axis in T. Draw PM perpendicular to the directrix. THE PARABOLA 129 Now since SP = PM, and PZ is common to the As SPZ, MPZ which have the angles at M and 8 right angles, .-. A SPZ = A MPZ and Z SPT = Z TPM =zSTP. Cor. If the normal at P meet the axis G then SG = SP = ST. That ,ST = SP follows from the equality of the angles SPT and ST P. Further the complements of these angles must be equal ; .-. zSPG = ^8GP; .-. SG = SP. We note that the equality of SG and SP in a parabola follows from the fact that for any conic SG = e . SP (§ 111). 125. Prop. The foot of the perpendicular F from the focus on to the tangent at any point P of a parabola lies on the tangent at the vertex and SY- = SA . SP. The first part of this proposition is implicitly proved in § 97. We can also prove it thus : Let the tangent at P meet the axis in T. -Since ST=SP, /SF will bisect TP. But if PX be the ordinate of P, TA = AN. .'. ^P^is parallel to XP, that is J. F is the tangent at the vertex. A. G. 9 130 THE PARABOLA Further as SY2^ is a right angle and YA perpendicular to ST, SY' = SA . ST = SA . SF. Cor. 1. zSPY=zSYA. Cor. 2. If the locus of the foot of the perpendicular from a fixed point on a variable line be a straight line, then the variable line touches a parabola having its focus at the fixed point. Def. When a line moves in a plane so as always to touch a certain curve, the curve is called the envelope of the line. Pair of Tangents. 126. Prop. Tangents to a parabola at tJte extremities of a focal chord intersect at right angles in the directrix. That they intersect in the directrix we know already, since the focus and directrix are pole and polar. Let PSQ be a focal chord, and let the tangents meet in the directrix in Z. Draw Pif and QF perpendicular to the directrix. Then as we have seen in § 124 ASFZ^AMFZ, .-. z.SZF = aMZF. Si milarly Z SZQ = Z FZQ. Thus Z QZF = i of two right Z s = a right angle. THE PARABOLA 131 127. Prop. If TP and TQ be two tangents to a parabola, the triangles SPT, STQ are similar. We know already that the angles PST, TSQ are equal (§ 109). Let the tangents at P and Q meet the tangent at the vertex in Y and Z, then S7T and SZT are right angles (§ 125). Then Z>^PV=Z8YA (§125) = Z STZ since SZYT is cyclic. Thus ZSPT = ZSTQ; .-. the remaining angle STP = Z. >S'Qrand the triangles tiPT and STQ are similar. Cor. 1. ST' = SP.SQ fur SP : ST = ST : SQ. Cor. 2. TP' : TQ' = SP : SQ i for TP' : 7Y,)'^ = ASPT : A,STQ = SP.ST:ST.SQ since ZPST = Z.TSQ = SP:SQ. 128. Prop. TAe exterior angle between ttvo tangents to a parabola is equal to half the angle luhich their chord of contact subtends at the focus. 132 THE PARABOLA Let the tangents at P and Q meet in T, and let them meet the axis in F and K respectively. Then Z FTK = Z SKQ - Z 8FP = zSQK-zSPT = 2 right angles - Z SQT - Z SPT = 2 right angles - Z STP - Z ^PT = ZT.SP = 1 zPSQ. 129. Parabola escribed to a triangle. When the sides of a triangle are tangents to a parabola, the triangle is said to circumscribe the parabola. But it must be clearly understood that the triangle does not enclose the parabola, for no finite triangle can enclose a parabola, which is infinite in extent. When a triangle circumscribes a parabola, the parabola is really escribed to it, that is, it touches one side of the triangle and the other two sides produced. Only triangles which have the line at infinity for one of their sides can enclose the parabola, and in the strict sense of the word be said to circumscribe it. It is convenient however to extend the meaning of the word ' circumscribe ' and to understand by a triangle circumscribing a conic a triangle whose sides touch the conic whether the triangle encloses the conic or not. 130. Prop. The circumcircle of the triangle formed by three tangents to a j^arabola passes througli the focus. THE PARABOLA 133 Understanding the word ' circumscribe ' as explained in § 129 we may state this proposition thus : If a triangle circumscribe a parabola its circumcircle goes through the focus. This can be seen from the fact that the feet of the perpen- diculars from the focus on the three sides of the triangle which touch the parabola are collinear, lying as they do on the tangent at A. .'. S lies on the circumcircle of the triangle (§ 7). Or we may prove the proposition in another way : Let the tangents at P, Q, R form the triangle TLM as in the figure. Then as ASPL is similar to A SLR, z SLR = zSPL. And as ASPT is similar to ASTQ, Z STQ = z SPT. .-. zSLM= zSTM, that is, SLTM is cyclic, or S lies on the circle through T, L and M. Cor. The orthocentre of a triangle circumscribing a j'jora- bola lies on the directrix. For if 'TLM be the triangle, the line joining S to the ortho- centre is bisected by the tangent at the vertex, which is, as we have seen, the pedal line of *S (§ 8). .•. the orthocentre must lie on the directrix. 134 THE PARABOLA 131. Prop. //' the tangents at P and Q to a parabola meet in T, and a third tangent at R cut them in L and M, the triangle SLM is similar to the triangles SPT and STQ, and PL.LT= TM : MQ = LR : RM. By the preceding proposition 8 lies on the circumcircle of TLM. .-. zSML= Z.STL and ZSLM = ^ SPT from the similar As SPL, SLR. .-. A SLM is similar to A SPT and therefore also to A STQ. Further A SLR is similar to ASTM for ZSLM=ZSTM, and Z SRL = Z SLP = 180° - z SLT = z SMT. .-. LR:TM=SR:SM = MR : MQ (by similar As SRM, SMQ). .-. LR : RM = TM : MQ. Similarly MR ■.RL = TL: LP. Hence PL : LT = TM : MQ = LR : RM. 132. Diameters. We have already explained in § 102 what is meant by diameters of a parabola. Every line parallel to the axis is a diameter, and every diameter bisects a system of parallel chords. It must be remembered too that the tangents at the extremities THE PARABOLA 135 of each of the parallel chords bisected by a diameter intersect on that diameter (§ 95). Prop. // TQ and TQ' be tangents to a parabola, and TV be the diameter bisecting QQ' in V and cutting the curve in P then TP = PV. For PV being parallel to the axis goes through O. Thus by the harmonic property of the pole and polar {TV,Pn) = -l, .-. TP = PV. Note that TA = AX (§ 123) is only a special case of this. 133. Prop. The length of any focal chord of a parabola is four times the distance of the focus from the point where the diameter bisecting the chord meets the curve. Let RSR' be any focal chord, PV the diameter which bisects it in V. Let the tangents at R and R' meet in Z. Then Z is both on the directrix and on the line of the diameter PV. Also ZP=PF(§ 132). But ZSV is a right angle (§ 106). .-. SP = PV = ZP. Now draw RM and R'M' perpendicular to the directrix. 136 THE PARABOLA Then 2 VZ=RM + R'M' since V is the middle point of RR' = RS + 8R' = RR'; .-. RR' = 4.PV=4.SP. A focal chord bisected by a particular diameter is called the pcwameter of the diameter which bisects the chord. Thus RR' is the parameter of the diameter PV, and we have proved it equal to ^SP. In particular the latus rectum is the parameter of the axis, and we proved it equal to 4>S^^. 134. Prop. // QV he an ordinate of a diameter PV then QV = 4aSP . P V. Produce the ordinate QF to meet the curve again in Q'. .-. QV=VQ'. Draw the focal chord RSR' parallel to the chord QQ'. RR' will be bisected by PV in tl (say). The diameter PV meets the curve again in D, at an infinity, thus by Newton's theorem VQ . VQ' ■.VP.Vn=UR.UR':UP. Un ; .-. VQ. VQ' : UR . UR'= VP. Vn.UP.Un=VP: UP. .-. QV':RU^-=PV:PU: THE PARABOLA 137 FV PU SP = 4SP. .-. QV"- = ^SP.PV. (§ 133) It will be seen that the property FN- = 4 AS. AS of § 121 is only a special case of the general proposition just proved. 135. The preceding proposition shews that a parabola may be regarded as the locus of a point in a plane such that the square of its distance from a fixed line I varies as its distance from another fixed line I' not necessarily at right angles to /. The line I is a diameter of the parabola and I' is a tangent at the point where / and I' intersect. If a be the angle which QV makes with the axis in § 134 QF=perp. from*Q on PF x cosec a. and PF= perp. from Q on tangent at P x cosec a; .*. (Perp. from Q on PV)- x cosec- a = 4<SP X perp. fi'om Q on tangent at P x cosec a. (Perp. from Q on PF)- , cd • i ot- rerp. from y on tangent at P where »SF is the perpendicular firom *S^ on the tangent at P. 138 THE PARABOLA Thus if a point move in a plane so that the square of its distance from a line I is k times its distance from another line i', k being constant, the locus of the point is a parabola having its axis parallel to I. The focus lies in a line parallel to I' and distant -r from it, and it also lies in a line through the intersec- tion of I and I' and making with I' the same angle that I makes with it. The line I' is the tangent to the parabola at its point of intersection with I. As already explained in ^122 if the lines I and V are at right angles / is the axis itself and I' the tangent at the vertex. 136. Circle of curvature. We have explained in § 119 what we mean by the circle of curvature at any point of a conic. We shall now shew how the circle of curvature at any point of a parabola can be determined. The centre of the circle of curvature at P lies on the nonnal at P, if then we can find the length of the chord through P of this circle in any direction, the length of the diameter of the circle can be found by drawing a line through the other extremity of this chord and perpendicular to it to meet the normal in D, then PD will be the diameter of the circle of curvature. Prop. The chords of the circle of curvature parallel to the axis and through the focus at any point P of a parabola = 4>SP. It is clear that these two chords must be equal for they make equal angles with the tangent at P. Now consider a circle touching the parabola at P and cutting it again at a near point Q. Draw the diameter of the parabola through Q and let it meet the circle again in K and the tangent at P in R. Draw QV the ordinate to the diameter PV. Then from the circle we have RQ.RK = RP"'; RP"' 07^ THE PARABOLA 139 Now in the limit when Q moves up to and ultimately coin- cides with P, RK becomes the chord of the circle of curvature through P parallel to the axis. Hence this chord is of length 4<SP. AiSP- CoR. The diameter of the circle of curvature at P = -o-r>-, ox where <SFis the perpendicular on the tangent. For -r^. = cos ( Z between normal and axis) Diameter = sin ( Z between tangent and axis) = sin { Z between tangent and SP) _SY ~ sp- Note. The diameter of the circle of curvature is commonly called the diameter of curvature, and the chords of the circle of curvature through P are called simply chords of curvature. 140 THE PARABOLA EXERCISES 1. PSP' is a focal chord of a paralwla, Pi^' and P'N' ordinates to the axis, prove PN. P'N' = 4. AS'' = iAN. AN'. 2. A series of parabolas touch a given line and have a common focus, prove that their vertices lie on a circle. 3. If a straight line rotate about a point in a plane containing the line the directions of motion of each point in it at a given moment are tangents to a parabola. 4. The ordinates of points on a parabola are divided in a given ratio, prove that the locus of the points dividing them is another parabola. 5. The locus of the middle points of focal chords of a parabola is also a parabola, whose latus rectum is half that of the original parabola. 6. li QV be an ordinate of the diameter PV and QD be perpendicular to the diameter QD^ = 4:AS . PV. 7. If tlie normal at P to a parabola meet the axis in G then PG"^iAS.SP. 8. PQ, PR are two chords of a parabola; PQ meets the diameter through R in the point F and PE meets the diameter through Q in E ; prove that EF is parallel to the tangent at P. [Project the parabola into a circle with E projected into the centre.] 9. If a circle touch a parabola at P and cut it at Q and R, the diameters through Q and A' will meet the circle again in points on a line parallel to the tangent at P. [Use§ 116 Cor.] 10. If TP and TQ be two fixed tangents to a parabola, and a variable tangent cut them in L and M respectively the ratios PL : TM and TL : QM are equal and^constant. 11. If two tangents TP and TQ to a parabola be cutl)y a third tangent in L and M respectively then TL TM_ TP "^ TO ~ THE PARABOLA 141 12. If the normal at P to a parabola meet the curve again in Q and PN be the ordinate of P, and T the pole of PQ^ Pq:PT = PN:AN. 13. If the tangent at P meet the directrix in Z and the latus rectum produced in D, then SD ~ SZ. 14. TP and TQ are tangents to a parabola, and the diameters through Pand Q meet any line drawn through Tin J/ and N ; prove TM- = TX- = TP . TQ. 15. If The any point on the tangent at P to a parabola and the diameter through T meet the curve in Q, then nP- = 4:SP.RQ. IG. If the chord PQ which is normal at P to a parabola subtends a right angle at aS', then SQ = 2.S'P. 17. If PN and P'X' be two ordiuates of a parabola such that the circle on XX' as diameter touches the parabola, then XP + X'P' = XX'. IS. 7'P and TQ are two tangents to a parabola, and PK and QL are drawn perpendicular respectively to TQ and Tl', prove that the triangles STK and STL are equal. 19. PP and PP' are tangents to a paral)ola, and the diameter through T cuts the curve in Q. If PQ, P'Q cut TP', TP respec- tively in R and A", and the diameters through li and R' cut the curve in V, V respectively, prove that PV, P'V intersect on I'Q. [Project the parabola into a circle and the line through 7' parallel to PP' to infinity.] 20. Pi"Ove that no circle described on a chord of a parabola as diameter can meet the directrix unless the chord be a focal chord, and then the circle touches the directrix. 21. If 7'P and TQ be a pair of tangents to a parabola and the chord PQ be normal at P, then TP is bisected by the directrix. 22. The triangle ABC circumscribes a parabola having S as focus, prove tliat the lines through A, B, C perpendicular to 8A, SB, SC respectively are concurrent. 23. The tangents to a parabola at P and P' intersect in Q ; the circles circumscribing the triangles SPQ, SPQ' meet the axis again in R and R' . Prove that PR and QR' are parallel. 142 THE PARABOLA 24. A given triangle ABC moves in a plane, with one side AB passing through a fixed point, and with the vertex ^4 on a given straight line. Shew that the side AG will envelop a parabola. 25. Shew that the envelope of a line which moves so as to intercept equal chords on two given circles is a parabola having the radical axis of the two given circles as the tangent at its vertex. 2G. A line meets a parabola in P and p on the same side of the axis. xiQ is drawn parallel to Pp to meet the curve again in Q. Prove that the ordinate of Q is equal to the sum of the ordinates of P and J)- 27. The distance of the point of intersection of two tangents to a parabola from the axis is half the sum of the ordinates of their points of contact. 28. If LL' be the latus rectum of a parabola and the tangent at any point P meet that at L in F, then SL . LP = VL . VL'. 29. A cii'cle touches a parabola at a point P, and passes through the focus S. Shew that the parabola meets the circle again or not accoi'ding as the latus rectum is or is not less than SP. 30. QQ' is the normal at (^ to a parabola meeting the parabola again in Q', QP is equally inclined to the axis with the normal and meets the curve again in P ; T is the middle point of QQ' , and PV meets the axis in R ; shew that QSPP lie on a circle, *S' being the focus. 31. Two paralwlas with equal latus rectum are on the same axis, and are such that the part of any tangent to one which is cut off by the other is equal to the perpendicular upon this tangent from the focus of the first parabola. Shew that the latus rectum of each is sixty-four times the distance between the vertices. 32. A circle touches a parabola at both ends of a double ordinate PP' to the axis. The normal at P meets the circle in R and the parabola in Q. The diameter of the parabola through Q meets PP' in U. Prove that the circle Q RU iowches PP' at U. 33. EF is a double ordinate of the axis of a parabola, R any point on it, and the diameter through R meets the curve in P ; the tangent at P intersects in M and N the diameters through E and F. Prove that PR is a mean proportional between EM and FN. 34. If a parabola roll on another equal parabola, the vertices being originally in contact, its focus will trace out the directrix of the fixed parabola. THE PARABOLA 143 35. If P be any point on a parabola whose vertex is A, and PR perpendicular to AP meet the axis R, a circle whose centre is R and radius RP will pass through the ends of the ordinate to the parabola through R. Also if common tangents be drawn to the circle and parabola the ordinates at the points where they touch the parabola will be tangents to the circle. 36. Through any point on a parabola two chords are drawn equally inclined to the tangent there. Shew that their lengths are proportional to the portions of their diameters bisected between them and the curve. 37. TQ, TR, tangents to a parabola, meet the tangent at P in Y and Z, and TU is drawn parallel to the axis, meeting the parabola in U. Prove that the tangent at U passes through the middle point of YZ, and that, if S be the focus," YZ- = ^HP.TV. 38. PQ is a normal chord of a parabola meeting the axis in G. Prove that the distance of G from the vertex, the ordinates of P and Q, and the latus rectum are four proportionals. 39. Lines are drawn through the focus of a parabola to cut the tangents to it at a constant angle. Prove that the locus of their intersection is a straight line. •40. The radius of curvature at an extremity of the latus rectum of a parabola is equal to twice the normal. 41. The diameter at either extremity of the latus rectum of a parabola passes through the centre of curvature at its other extremity. 42. If the tangent at any point /* of a parabola meet the axis in T, and the circle of curvature meet the curve in (?, then PQ = iPT. 43. If R he the middle point of the radius of curvature at P on a parabola, PR subtends a right angle at the focus. 44. The tangent from any point of a parabola to the circle of curvature at its vertex is equal to the abscissa of the point. 45. The chord of curvature through the vertex A at any point iPY- Pof a i^arabola is , Y being the foot of the perpendicular from the focus on the tangent at /-•. 144 CHAPTER XII THE ELLIPSE 137. We have already in ^§ 98 and 99 indicated the general form of the ellipse, shewing that it has two axes of symmetry at right angles to one another and intersecting in C the centre. On the major axis AA' are two foci S and ^*', at a distance equal to CA from B and B' the ends of the minor axis, and to these foci correspond directrices at right angles to A A' and cutting it externally in X and X' so that GS : CA = CA : CX = e, the eccentricity. It is convenient to call A and A', the ends of the major axis, the vertices of the ellipse. It will be understood that the smaller is the ratio CS : CA the more does the ellipse approximate to circular form, and the greater C>S' : CA is without reaching unity the more does the ellipse flatten out. We have always CS'=CA'-CB' so that, keeping CA constant, CB diminishes as GS increases, and vice versa. And we have already explained that a circle may be regarded as the limiting case of an ellipse whose two foci coincide with the centre. We now proceed to establish the chief geometrical properties which all ellipses have in common. 138. Sum of focal distances constant. Prop. The sum of the focal distances of any point on an ellipse is constant and equal to AA'. THE ELLIPSE 145 Let P be any point on an ellipse, MPM' the perpendicular through P to the directrices as in the figure. Then >S'P = e . PM and S'P = e . PM' ; . • . SP + S'P = e (MP + PM) = e . XX ' = 2e.CX = '2CA = AA'. Cor. Two confocal ellipses (that is, which have both foci in conunon) cannot intersect. 139. The proposition just proved shews that an ellipse may be regarded as the locus in a plane of a point the sum of whose distances from two fixed points in the plane is constant. And we learn that an ellipse can be drawn by tying the ends of a piece of string to two pins stuck in the paper so that the string is not tight, and then holding the string tight by means of the pencil pressed against it, and allowing the point of the pencil to make its mark in all possible positions thus determined. By keeping the string the same length and changing the distance between the pins we can draw ellipses all having the same major axis but having different eccentricities. It will be seen that the nearer the pins are together the more does the ellipse approximate to circular form. 140. Tangent and Normal. Prop. The tangent and normal at any point of an ellipse bisect respectively the exterior and interior angles between the focal distances of the point. A. G. 10 146 THE ELLIPSE Let the tangent and normal at P meet the major axis in T and G respectively. Then hy^lll SG = e.SP, and S'G = e . 8' P. .-. SG:GS' = SP:PS'; .-. PG bisects the angle SPS' ; .-. PT which is at right angles to PG must bisect the exterior angle of SPS'. Cor. CG.CT=CS-' for since PG and PT are the bisectors of the angles between SP and S'P, (GT, SS') = - 1. Prop. 141. If SY, S'Y' be the perpendiculars from the foci on the tangent at any point P of an ellipse, Y and Y' lie on the circle described on the major axis A A' as diameter, and SY.S'Y' = BC\ Produce SY to meet S'P in K. Then ASPY= AKPY for ^SPY=ZKPY (§140) Z SYP = 4 KYP being right angles and PY is common. .-. PK^SPsiUclKY^^SY, . • . KS'= KP + PS' = SP + PS' = AA' (§ 138). Now since Y and C are the middle points of SK and SS', CY is parallel to S'K and CY = ^S'K= CA. THE ELLIPSE 147 Thus Y (and similarly Y') lies on the circle on AA' as diameter. Moreover as Y' YS is a right angle, YS will meet the circle again in a point Z such that Y'Z will be a diameter, that is, Y'Z goes through C. And ACSZ = ACS'Y' so that SZ=S'Y'. .-. SY.S'Y'^SY.SZ = AS . SA' = CA' - CS'^ = BC\ Cor. 1. The diameter parallel to the tangent at P will meet SP and S'P in points £'aild E' such that PE = PE'= AC. For PYGE' is a parallelogram. .-. PE'==CY = AC and similarly PE = AC. Cor. 2. The envelope of a line such that the toot of the perpendicular on it from a fixed point S lies on a fixed circle, which has S within it, is an ellipse having *S' for a focus. Cor. 3. The envelope of a line such that the product of the perpendiculars on it ft-om two fixed points, lying on the same side of it, is constant is an ellipse having the fixed points for its foci. Def. The circk on the major axis of an ellipse as diameter is called the auxiliary circle. 142. Prop. If TQ and TQ' be a pair of tangents to an ellipse wJiose centre is C, and CT meet QQ' in V and the curve inP,CV.CT=CP\ 10—2 148 THE ELLIPSE For let PC meet the ellipse again in F'. Then as T and QQ' are pole and polar, {TV, PF) = - 1. .-.as C is the middle point ofPP', GV . CT=GP\ 143. The preceding proposition is an important one and includes the following as a special case : If the tangent at P meet the major and minor axes in T and t, and PN, PM he the ordinates to these axes then GN .CT^GA' GM.Gt=GB\ p t B 4 " \ "^ N C 1 For T is the intersection of the tangents at P and at the point Avhere PN again meets the curve, and t is the intersection of the tangents at P and at the point where PM again meets the curve. THE ELLIPSE 149 144. Prop. If the normal at any point P on an ellipse meet the major and minor axes in G and g, and the diameter jmrallel to the tangent at F in F, then FF.FG=BC- and FF.Fg = AC\ Draw the ordinates FX and FM to the axes and let these meet the diameter parallel to the tangent at F in K and L as in the figure. t F y B '/^^^ ^--^^ L y^ I N M ^\^^ T aI ^ C Let the tangent at F meet the major and minor axes in T and t. Then NKFG is cyclic since N and F are right angles. . •. FF.FG = FN . FK = CM . Ct = BC-. Also gFML is cyclic since F and M are right angles. .-. FF.Fg = FM.FL = CN.CT^CA\ 144a. Prop. //' the normal at F on an ellipse meet the major axis in G, and FN be the ordinate to this axis, CG = e- . ON. Let the tangent at F meet the major axis in 2\ We have already proved in § 140 Cor. that GG.CT=CS\ But CN.CT=CA'; .-. CG:CN=CS-':CA' = e'; .-. CG^eKCN. Cor. NG:NC = BG':AC-. For as CG ■.GN=CS' : CA"-, .-. CN- CG : CN = CA' - CS' : CA'^ .= BC'-:AC\ 150 THE ELLIPSE 144b. Pair of tangents. Prop. The two tangents chawn from an external point to an ellipse make equal angles with the focal distances of the point Let TP and TQ be the tangents ; it is required to prove /.PTS = ^S'TQ. Draw 8Y, S' Y' perpendicular to TP, and 8Z and S'Z' per- pendicular to TQ. Then SY. S'Y' = BC'^SZ. S'Z' (| 141) ; .-. SY:8Z = S'Z':S'Y'. Also Z F5f^ = supplement of Z YTZ (since SYTZ is cyclic) = z Y'S'Z' (since Y'TZ'S' is cyclic). . •. the A s SYZand S'Z' F'are similar.and Z SZY= Z S'Y'Z', But Z ;SfZF= Z *STF in the same segment and Z S'T'Z' = Z >S^TZ' in the same segment. .-. ZSTY=S'TZ'. 145. Director Circle. Prop. The locus of points, from which the tangents to an ellipse are at 7'ight angles is a ciixle {called the director circle of the ellipse). Let TP and TQ be two tangents at right angles. Draw SY perpendicular to TP to meet S'P in K. Then by § 141, SY= YK and S'K = AA'. THE ELLIPSE 151 Also AaSFT= A A^FT, for SY= KY and YT is common and the angles at Y are right angles. . •. >ST = KT and z KTP = z STP = z QT8' (§ 144 b). .-. Z ATra' = Z PTQ = a right angle. T Now 2CT= + 26'»S'-^ = .ST- + ^T- (§ 10) = 4C^^ . •. CT' = 2CA' - CS' = 2CA ' - {CA' - CB') = CA' + GB\ Thus the locus of T is a circle round C, the stjuare of whose radius is CA- + CB'-. 146. Conjugate Diameters. The student is already familiar with the idea of conjugate lines of a conic, two lines being called conjugate when each contains the pole of the other. When a pair of conjugate lines meet in the centre of an ellipse, each being a diameter it is convenient to call them conjugate diameters. It is clear that they are such that the tangents at the points where either meets the curve are parallel to the other. Moreover all the chords which are parallel to one of two conjugate diameters are bisected by the other (§ 95), and these chords are double ordinates of the diameter which thus bisects them. The axes of the ellipse are that particular pair of conjugate diameters which are mutually at right angles. 152 THE ELLIPSE 147. Prop. If QV he an ordinate of the diameter PGP' of an ellipse, and BCD' be the diameter conjugate to CP, QV'-:PV.VP'=CIP:CP\ Fur, producing QV to meet the ellipse again in Q', by Newton's theorem we have VQ . VQ' : VP. VP' = CD . CD' : CP . CP' . But VQ' = - VQ, CD'=- CD, CP' = - CP; .-. QV':PV.VP' = CD':CP\ 148. Special cases of the preceding proposition are these : If PN and PM he ordinates of the major and minor axes of an ellipse then PN' -.AN. NA' = BC : A C' PM-':BM.MB' = AC-':BC\ B P,--""^ — T7^~^\ / M ^\ aI ^ J C j 2BC'- CoR. The latus rectum = ---, , AC THE ELLIPSE 153 For if SL be the semi-latus rectum SD : AS . SA' - BG^ :AC' and AS . SA' = BC\ 149. These properties in § 148 shew that an ellipse may be regarded as the locus of a point in a plane such that the square of its distance from a fixed line I in the plane bears a constant ratio to the product of its distances from two other fixed lines I' and I", perpendicular to the former and on opposite sides of the point. The line I is one of the axes of the ellipse, and the lines I' and I" are the tangents at the ends of the other axis. The property established in § 147 shews that an ellipse ma}- also be regarded as the locus of a point in a plane such that the square of its distance from a fixed line I in the plane, bears a constant ratio to the product of its distances from two other fixed lines I' and I" which are parallel to each other (but not necessarily perpendicular to I), and on opposite sides of the point. The line I is a diameter of the ellipse and the lines l' and I" are the tangents at the points where I meets the curve. For the student can easily prove for himself that in the notation of § 147 QV^ :PV. VP' = square of perpendicular from Q on PP' : product of perpendiculars from Q on tangents at Q and Q' . Auxiliary Circle. 150. Prop. If P he any point on an ellipse and PN the ordinate of the major axis, and if NP meet the auxiliary circle in p, then NP:Np = BC:AC. For by § 148 PN':AN.NA' = BC-':AC-' and as Z ApA' being in a semicircle is a right angle pN^- = AN.NA'] .-. PN:pN=BG:AC. P and p are said to be corresponding points on the ellipse and the auxiliary circle. The tangents at two corresponding 164 THE ELLIPSE points will meet the line of the major axis in the same point. For let the tangent at P meet it in T, then CN'.CT=CA- (§ 143). .'. 7' is the pole of i^N for the circle, that is, the tangent at p goes through T. The student can prove for himself by the same method that if an ordinate FM to the minor axis meet the circle on BB' as diameter in p' then PM:p'M=AC:BG. From all this it follows that if the ordinates of a diameter of a circle be all divided in the same ratio, the points of division trace out an ellipse having the diameter of the circle as one of its axes. 151. Prop. If GP and CD he a pair of conjugate semi- diameters of an ellipse, and p, d the points on the auxiliary circle corresponding to P and D, then pGd is a right angle. Let the tangents at P and p meet the major axis in T. Draw the ordinates PN, DM. Now since CD is parallel to TP, /\PNT is similar to ADMC. .-. PN:DM = NT:MC. But as PN :pN = BG:AG = DM: dM, .-. PN:DM = pN:dM. THE ELLIPSE 155 .-. pN:dM==NT:MC, that is pN : NT = dM : MC. .'. as the As pNT unci dMC have the angles at ilf and i\^ equal, they are similar. .-. zMCd^zXTp. .'. Cd is parallel to Tp. . • . Z dCp = Z CpT = a right angle. Cor. pX = CM and dM = CN for A CNp = A dMC. Whence also we have PN:CM = BC:AC, DM:CN = BC.AC. 152. Prop. If CP and CD be conjurjate semidiameters of an ellipse CP-'+CD"- = CA-'+CB\ For using the figure of the last proposition cp^ = GX-' + PX' = av-^ + ~, . px-' 7?r" RC- and CD' = CM-' + DM' = i^^' + 1^ ^^' " ^^' "^ I^ ^'^'• .-. CP' + CD' = (l+ 1^) (pX' + CX^) = (i+^)ac-' = ac' + bc"^. 15G THE ELLIPSE Thus the sum of the squares of two conjugate diameters of an ellipse is constant and = AA"^ + BB'". Or we may prove the proposition thus : In § 151 we proved pN = CM, . • . CM' + CN-' = pN'- + CN-' = Cp' = A C\ In exactly the same way by drawing ordinates to the minor axis and working with the circle on BB' as diameter we have FN' + DM'' = BC\ Whence by addition. GP' + CD- = AC + BG\ 152a. Equiconjugate Diameters. There is one pair of conj ugate diameters of an ellipse which are equal to one another namely those which lie along the diagonals of the rectangle formed by the tangents at the extremities of the major and minor axes. X y\ V y^ \ J That the diameters along these lines are equal is clear from the symmetry of the curve, and that they are conjugate diameters is seen from the fact that in the figure here presented since KG bisects AB, which is parallel to LL', GK and GL' are conjugate lines. 152b. Prop diameters of an If CP and CD he a pair of conjugate senii- SP . S'P = GD\ THE ELLIPSE 157 Since is the middle point of SS' = (ST + spy - 2SP . S'P = 4^CA'-2SP.S'P. .-. SP . S'P = 2CA' - CP' - CS' = CA"- + CB' - CP' = CD' (by §152). 153. Prop. // P be amj point on an ellipse, and the twrmal at P meet BCD', the diameter conjugate to CP, in F, then PF,CD = AC.BC. Draw the tangent at Pand (h-opthe perpendiculars .S'Fand S'Y' from the foci (jn it. Join SP and S'P, and let S'P cut DD' in E. Then the As SPY, S'PY' are similar. 158 THE ELLIPSE .-. SY:8P==S'Y':S'F .-. SV. S'Y' : SP . S'P = SY' : SP' = PF' : PE' since the As SYP, PFE are similar. .-. BC'-:CD-' = PF':Aa' that is, PF.CD = AG.BC. Cor. The area of the parallelogram formed by the tangents at the extremities of a pair of conjugate diameters is constant = 4.AC.Ba For the area = 4 area of parallelogram PD^ = ^PF.GD = 4:AG.BG. Circle of Curvature. 154. Prop. The chord of the circle of curvature at any 'point P of an ellipse and through the centre of the ellipse is 2G]> GP • Let Q be a point on the ellipse near to P and QV the ordinate of the diameter PGP'. Consider the circle touching the ellipse at P and cutting it in Q. Let QK be the chord of this circle parallel to GP. Let QK meet the tangent at P in R. THE ELLIPSE 159 Then from the circle RQ.RK = RP\ Thus the chord of the circle of curvature through the centre being the limit of RK when Q approaches F X Limit VF' 2CD' ^x2Crp=^^. Cor. The diameter of the circle of curvature 2CD"- FF F being the point in which the normal meets CD, for Diameter 2011 CF = sec ( Z between normal and OF) Diameter CF : FF. 2CJy FF 2CD' ACTBC' 160 THE ELLIPSE EXERCISES 1. Prove that in the notation of this chapter as : ex = CS'^ : CA^ 2. If SL be tlie semi-latus rectum of an ellipse then SL = e . SX ; prove from this that Obtain also the length of the latus rectum by using the fact (§§ 116, 117) that the lengths of two focal chords are in the ratio of the squares of the diameters parallel to then). 3. If Y, Z be the feet of the perpendiculars fi'om the foci of an ellipse on the tangent at P, of which PN is the ordinate ; prove that the circle circumscribing YNZ passes through the centre of the ellipse. \. If P be any point on an ellipse whose foci are S and S' the circle circumscribing SPS' will cut the minor axis in the points where it is met by the tangent and normal at /^ 5. If two circles touch internally the locus of the centres of circles touching them both is an ellipse, whose foci are the centres of the given circles. 6. If the tangent at P to an ellipse meet the major axis in 'l\ and NG be the subnormal, CT .NG=BC\ 7. If PN be the ordinate of any point P of an ellipse and Y, Y' the feet of the perpendiculars from the foci on the tangent at P, then PN bisects the angle YNY'. 8. If the normal at P to an ellipse meet the minor axis in </, and the tangent at P meet the tangent at the vertex A in F, shew t\i-A.t Sij ■ SG = PV : VA. 9. If the normal at P meet the major axis in G, PG is a harmonic mean between the perpendiculars from tlie foci on the tangent at P. 10. If an ellipse inscribed in a triangle have one focus at the orthocentre, the other focus will be at the circumcentre. THE ELLIPSE 161 11. If an ellipse slide between two straight lines at right angles, the locus of its centre is a circle. 12. Lines are di'awn through a focus of an ellipse to meet the tangents to the ellipse at a constant angle, prove that the locus of the points in which they meet the tangents is a circle. 13. The locus of the incentre of the ti'iangle whose vertices are the foci of an ellipse and any point on the curve is an ellipse. 14. The opposite sides of a quadrilateral described about an ellipse subtend supplemeutaiy angles at either focus. 15. Prove that the foci of an ellipse and the points where any tangent to it meets the tangents at its vertices are concyclic. 16. If CQ be a semidiameter of an ellipse conjugate to a chord which is normal to the curve at P, then CP is conjugate to the normal at Q. 17. If P be any point on an ellipse, foci .S' and .S", and .4 be a vertex, then the bisectors of the angles PSA, PS' A meet on the tangent at P. 18. In an ellipse whose centre is C and fuci S and S', GL is drawn perpendicular to C P, and CJ/is drawn parallel to S' P meeting PG in M. Prove that the triangles CLM, CMP are similar. 19. A circle is drawn touching an ellipse at two points, and Q is any point on the ellipse. Prove that if ^7' be a tangent to the circle from Q, and QL perpendicular to the common chord, then QT=^e.QL. 20. If a parabola have its focus coincident with one of the foci of an ellipse, and touch its minor axis, a common tangent to the ellipse and parabola will subtend a right angle at the focus. 21. Shew how to determine the magnitude and position of the axes of an ellipse, having given two conjugate diameters in magni- tude and position. 22. Construct an ellipse when the position of its centre and a self -conjugate triangle are given. 23. If P be any point on an ellipse whose vertices are A and A ', and AP, A' P meet a directrix in E and F, then ^i*^ subtends a right angle at the corresponding focus. 24. Deduce from Ex. 23 the property that PX'^ : AN . NA' is constant. A. G. 11 162 THE ELLIPSE 25. Prove that chords joining any point on an ellipse to the ends of a diameter are parallel to a pair of conjugate diameters. [Two such chords are called supplemental chords.] 26. If a circle touch a fixed ellipse at P, and intersect it at the ends of a diameter QQ', then PQ and PQ' are fixed in direction. 27. Ellipses have a common fixed focus and touch two fixed straight lines, prove that their director circles are coaxal. 28. SY is the perpendicular from the focus S of an ellipse on a tangent, and K the point in 5 F produced such that SY = YK. Prove that the square of the tangent from A' to the director circle is double the square on SY. 29. The circle of curvature at an extremity of one of the equal conjugate diameters of an ellipse meets the ellipse again at the extremity of that diameter. .30. If PN be the ordinate of a point P on an ellipse, the chord 26'2)2 of curvature in the direction of PN = x PN. 31. If S and S' be the foci of an ellipse and B an extremity of the minor axis, the circle SS'B will cut the minor axis in the centre of curvature at B. 32. The circle of curvature at a point P of an ellipse passes through the focus S, and SE is drawn parallel to the tangent at P to meet in E the diameter through P ; shew that it divides the diameter in the ratio 3:1. 33. The circle of curvatui-e at P on an ellipse cuts the curve ao'ain in Q ; the tangent at P meets the other common tangent, which touches the ellipse and circle at E and F, in ; prove {TO, EF) = -\. 34. The tangent at P to an ellipse meets the equiconjugates ia Q and Q' ; shew that CP is a symmedian of the triangle QCQ'. 163 CHAPTER XIII THE HYPERBOLA 155. We have eeen in |§ 100, 101 that a hyperbola, which is the projection of a circle cut by the vanishing line, has two axes of symmetry at right angles, one of which, named the transverse axis, meets the curve in what are called the vertices A and A', while the other called the conjugate axis does not meet the curve. These two axes meet in C the centre of the curve, and there are two tangents from C to the curve having their points of contact at infinity. These tangents are called the asymptotes and they make equal angles with the axes. The curve has two foci S and S' lying on the line of the transverse axis, and such that the feet of the perpendiculars from them on the asymptotes lie on the circle on A A' as diameter. The directrices which are the polars of the foci are at right angles to the transverse axis, and pass through the feet of these perpendiculars. If A' and X' be the points in which the directrices cut the transverse axis and C be the centre, then the eccentricity (e) = CS : CA = CA : CX. 156. In this chapter we shall set forth the principal properties that all hyperbolas have in common. Some of these are the same as those of the ellipse and can be established in much the same way. But the fact that the hyperbola has a pair of asymptotes, that is, tangents whose points of contact are at infinity, gives the curve a character and properties of its own. 11-2 164 THE HYPERBOLA 157. Difference of focal distances constant. Prop. Tlie difference of the fucal distances of any point on a hyperbola is constant, and equal to the length A A' of the transverse axis. Let P be any point on the hyperbola. Draw PMM' as in the figure perpendicular to the directrices, then S'P=e.PM' and SP = e.PM. .-. S'Pr^SP = e.XX'^AA'. For points on the one branch we have S'P — SP =AA' and for points on the other SP — S'P = A A'. Cor. Two confocal hyperbolas cannot intersect. Tangent and Normal. 158. Prop. The tangent and normal at any point of a hyperbola bisect respectively the interior and exterior angles of the focal radii of the point. Let the tangent and normal at P meet the transverse axis in T and G. Then by § 111, SG = e . SP, S'G = e . S'P. .-. SG:SP = S'G:S'P. THE HYPEEBOLA 165 .•. PG is the bisector of the exterior angle of SPS', and PT, perpendicular to PG, must therefore bisect the interior angle. Cor. 1. CG.CT= CS\ for {SS', TG) = - 1. Cor. 2. If an ellipse and hyperbola are confocal their tangents at the points of intersection of the curves are at right angles, or, in other words, the curves cut at right angles. 159. Prop. //' SY, S'Y' he the perpendiculars from the foci on the tangent to a hyperbola at any point P, Y and Y' will lie on the circle on A A' as diameter {called the auxiliary circle), and 8Y. S'Y' luill he constant. Let SY meet S'P in K. Then since ^ SPY = A KPY (§ 158) and ^SYP=ZKYP and PF is common, .-. A5rPF= A/fPF and SY^YK, PK = SP. And since F and C are the middle points of *S^ and SS', CY is parallel to S'K and GY= ^S'K = i {S'P - KP) = ^{S'P-SP) = ^AA' = CA. Thus Y (and similarly F') lies on the circle on AA'. 166 THE HYPERBOLA If S'Y' meet the circle again in Z, then since YY'Z is a right angle, YZ must be a diameter and pass through C. Also ^SCY= AS'CZ and S'Z=SY. .-. SY.S'Y'=S'Z.S'Y' = S'A'.S'A^CS"--CA' which is constant. Cor. 1. The diameter parallel to the tangent at P will meet SP and ST in points E and E' such that PE= PE' = GA. Cor. 2. The envelope of a line such that the foot of the perpendicular on it from a fixed point S lies on a fixed circle which has S outside it, is a hyperbola having >S' for a focus. Cor. 3. The envelope of a line such that the product of the perpendiculars on it from two fixed points, lying on opposite sides of it, is constant is a hyperbola having the fixed points for its foci. Compare § 141, Corr. 2 and 3. 160. On the length of the conjugate axis. We have seen that the conjugate axis of a hyperbola does not meet the curve, so that we cannot say it has a length in the same way that the minor axis of an ellipse has for its length that portion of it intercepted by the curve. It is convenient, and this will be understood better as wo THE HYPERBOLA 16^ proceed, to measure off a length BB' on the conjugate axis such that B and B' are equidistant from G, and BC' - GS-^ - GA"- = ^,S' . A'S. This will make SY.S'Y' in the preceding proposition equal to BG\ (Compare § 141.) The length BB' thus defined will for convenience be called the length of the conjugate axis, but it must be clearl}^ under- stood that BB' is not a diameter length of the hyperbola, for B and B' do not lie on the curve. 161. It will easily be seen that if a rectangle be drawn having a pair of opposite sides along the tangents at A and A', and having its diagonals along the asymptotes, then the portion of the conjugate axis intercepted in this rectangle will be this length BB' which we have marked off as explained above. For if the tangent at A meet the asymptote CO in 0, and the directrix corresponding to S meet CO in K, we have, since GK8 is a right angle and GK=GA (§ 101), AGKS= A GAG. Hence A0' = SIC-= GS' - GIO = GS' - GA\ 168 THE HYPERBOLA Pair of tangents. 162. Prop. TJte two tangents drawn from a point to a hyperbola make equal or supplementary angles with the focal distances of the point. Fig. 1. Let TP, TQ be the tangents, it is required to prove that the angles STP, S'TQ are equal or supplementary. Draw SY, S'Y' perp. to TP, and SZ, S'Z' to TQ. Then SY. S' Y' = BO' = SZ . S'Z'. .-. SY:SZ=S'Z':S'Y'. Also Z YSZ = /.Z'S'Y' these being the supplements of the THE HYPERBOLA 169 equal angles ZTY and Z'TY' in fig. 1, and in fig. 2 each being equal to YTZ, since SZTY and 8'Y'TZ' are cyclic. Hence the As SYZ and S'Z'Y' are similar and z >S'rP = z >SZF= z aS"F'Z' = z >S'TZ' = supplement of Z >S"TQ in fig. 1, while in fig. 2 it = Z S'TQ. Thus the two tangents from an external point make equal or supplementary angles with the focal distances of the point according as the tangents belong to opposite branches or the same branch of the curve. 163. Director Circle. Prop. The locus of points tangents from tvliich to a hyperbola are at right angles is a circle (called the director circle of the hyperbola). Let TP and TQ be two tangents at right angles. Draw 8Y perp. to TP to meet S'P in K. Then by § 159, SY= YK and S'K=AA'. Also A8YT= £^KYT. Thus ST= KT and z KTY = Z ST Y= Z. QTS' (§ 162). . •. Z KTS' = Z PTQ = a right angle. 170 THE HYPERBOLA Now ICT^ + 2GS- = ST' + 8'T (§ 10) = Kr~ + S'T' = KS"'==AA'-' = WA\ ♦ .-. CT = 2GA^-GS'' = GA' - {GS' - GA^) = GA' - GB\ Thus the locus of T is a circle whose centre is G and the square of whose radius is GA" — GB~. (Cf. § 145.) Cor. 1. U GA = GB, GT== 0, that is the tangents from G, or the asymptotes, are at right angles. Cor. 2. If GA < GB, there are no points from which tangents at right angles can be drawn. The Conjugate Hyperbola. 164. A hyperbola is completely determined when we know the length and position of its transverse and conjugate axes ; for when AA' and BB' are fixed, S and S' the foci on the line A A' are determined by GS' = GS'' = GA' + GB\ Also the eccentricity e = GS:GA, and the directrices are determined, being the lines perp. to A A' through points X and X' on it such that GA:GX = e=^GA':GX. 165. We are going now to take a new hyperbola having for its transverse and conjugate axes respectively the conjugate and transverse axes of the original hyperbola. This new hyper- bola will clearly have the same asymptotes as the original hyperbola (§ 161) and it will occupy space in what Ave may call the exterior angle between those asymptotes, as shown in the figure. This new hyperbola is called the conjugate hyperbola in relation to the original hyperbola, and it follows that the original hyperbola is the conjugate hyperbola of the new one. THE HYPERBOLA 171 The hyperbola and its conjugate are two distinct curves, each having its foci and directrices, nor will they in general have the same eccentricity. The foci S and S' of the original hyperbola lie on the line of A A' and are such that CS- = GA^ + GB-, and the eccentricity is CS:GA. The foci 2 and S' of the conjugate hyperbola lie on the line of BB' and are such that CI- = GA^ + GB\ Thus 02 = G8, but the eccentricity is (7S : GB, which is only the same as that of the original hyperbola if GA = GB. In this special case the asymptotes are the diagonals of a square (§ IGl) and are therefore at right angles. When the asymptotes are at i-ight angles the hyperbola is said to be rectangular. In the next chapter we shall investigate the special properties of the rectangular hyperbola. The conjugate hyperbola is, as we shall see, a very useful adjunct to the hyperbola and considerable use will be made of it in what follows. Asymptotic properties. 166. Prop. // R be any point on an asymptote of a hyper- bola, and RN^ perpendicular to the transverse axis meet the hyperbola in P and 'p then RP . Rp = BG^. 172 THE HYPERBOLA Let the tangent at A meet the asymptote on Avhich R lies in E. Then fl being the point of contact of the asymptote with the curve at infinity we have by Newton's theorem RP . Rp : Rn^' = EA- : En--. .'. RP.Rp = EA' = BG\ This can also be written RN' - PN' = BC\ It will presently be seen that this proposition is only a special case of a more general theorem. 167. Prop. // PN he the ordinate to the transverse axis of any point P of a Ityperhola PN"~:AN.A'N^BC':AG\ Using the figure of § 166 we have RN-' - PN' = BC\ .-. PN^ = RN'-BG\ But RN^ : BG"- - RN' : EA' = GN' : GA\ .-. RN"' - BG' : BG' = GN-' - GA' : GA\ .-. PN"-:BG' = AN.A'N:CA'. .-. PN':AN.A'N = BG'':AG\ THE HYPERBOLA 173 or we may write this BC':AC\ This too will be found to be but a special case of a more general theorem. Comparing this property with the corresponding one in the ellipse (§ 148) we see that it was not possible to establish the property for the hyperbola in the same way as for the ellipse, because the conjugate axis does not meet the hyperbola. 168. Prop. If from any poixt R in an asymptote of a hyperbola RPN, RDM he drawn perpendicular to the transverse and conjugate axes to cut the hyperbola and its conjugate respectively in P and D, then PD is parallel to the other asymptote, and CP, CD are conjugate lines for both the hypei^hola and the conjugate hyperbola. Let n and DJ be the points of contact of the hyperbola and its asymptotes at infinity. We first observe that AB is parallel to CVl'. For drawing the lines through A and B perpendicular to 174 THE HYPERBOLA the axes to meet the asymptotes, as indicated in the figure, in E, e, E\ we have EB-.BE' =EA -.Ae. Also 3IN is parallel to A B, for GA : CB = GA:AE^ CN : RN = CN : CM. .'. GA : CN = CB : CM. Now RN"^-PN' = BC' and RM'^ - DM' = AC ' ' ^^^' .-. RN"- - PA''' : RM' - DM' = CM' : CN' ^RN'-.RM'. .-. RN':RM' = PN':DM'. Thus PD is parallel to MN and therefore to CD.'. Thus PD will be bisected by CO in the point T (say). Now DP will meet Gil' at Df, and we have (DP, Tn') = -i. .'. GP and CD will belong to the involution of which CO and CVl' are the double lines. But Cfl and Cfl' being tangents from C to both the hyper- bola and its conjugate are the double lines of the involution pencil formed by the pairs of conjugate lines through C. .-. GP and CD are conjugate lines for both the hyperbola and its conjugate. For this it follows that the tangent at P to the hyperbola is parallel to CD, and the tangent at D to the conjugate hyperbola is parallel to GP. 169. On the term conjugate diameters. ' If the lines PC, DC in the figure of § 168 meet the hyper- bola and its conjugate again in the points P' and D' respectively, then PGP' and DCD' are called conjugate diameters for each of the hyperbolas. But it must be clearly understood that PGP' is a diameter of the original hyperbola, whereas DCD' is not, but it is a diameter of the conjugate hyperbola. THE HYPERBOLA 175 Of two so-called conjugate diameters one is a diameter of the hyperbola and the other of the conjugate hyperbola. The line BCD' is a diameter even for the original hyperbola in so far as it is a line through the centre and it Avill bisect a system of parallel chords, but it is not a diameter in the sense that it represents a length intercepted by the curve on the line, for D and D' are not on the hyperbola. DGD' does not meet the hyperbola in real points, though of course as the student acquainted with Analytical Geometry will know it meets the curve in imaginary points, that is, points whose coordinates involve the imaginary quantity V— 1. 170. Prop. The tangents at the extremities of a pair of conjugate diameters form a parallelogram whose diagonals lie along the asymptotes. Let PGP' and BCD' be the conjugate diameters, as in the figure. We have already proved (§ 168) that PD is bisected by CO. The tangents at P and D are respectively parallel to CD and GP. These tangents then form with GP and GD a parallelogram having one diagonal along CO. 176 THE HYPERBOLA Similarly the tangents at P' and D' meet on Cfl, and those at P', i) and P, D' on CiY. Coil. The portion of the tangent at any point intercepted between the asymptotes is bisected at the point of contact. For LP = L)C=GD' = Pl. 171. The property given in the Corollary ot § 170 can be independently established by projecting into a circle, and we may use the same letters in the projection without confusion. Let the tangent at P to the circle meet the vanishing line nn' in K. C L a The polar of K goes through G, since that at G goes through K, and the polar of K goes through P. .-. GP is the polar of K. Let GP meet HH' in F. .-. (/iP, an') = - 1. .-. C(ifp, aa') = -i. .-. {KP,Ll) = -\. Thus in the hyperbola L and / are harmonically conjugate with P and the point at infinity along LI. .-. LP = Pl. 172. Prop. // through any point R on an asymptote of a hyperbola a line he drawn cutting the same branch of the hyper- bola in Q and q, then RQ . Rq is equal to the square of the semi- diameter of the conjugate hyperbola parallel to RQq. THE HYPERBOLA 177 Let V be the middle point of Qg. Let CV cut the hj^erbola in P. Then the tangent at P is parallel to Qq. Let it meet the asymptotes in L and I. Let CD be the semi-diameter of the conjugate hyperbola parallel to Qq. n By Newton's theorem we have RQ.Rq: Rn^- = LP' : LQ.\ .-. RQ.Rq = LP^ = CD\ Thus if the line RQq be always drawn in a fixed direction the rectangle RQ . Rq is independent of the position of the point R on the asymptote. We may ^vl•ite the above relation RV'-QV'^CD'. And if RQq meet the other asymptote in r we have rV2-qV^=CI)\ ... rV - qV = RV' - QV\ .-. RV= Vr and .'. RQ = q>: Hence any chord of a hyperbola and the length of its line intercepted between the asjanptotes have the same middle point. A. G. 12 178 THE HYPERBOLA Wc thus have the following remarkable property of the hyperbola : If Rr joining any tivo points on the asymptotes of a hyperbola cut the curve in Q and q then EQ = qr. 173. Prop. // a line he drawn through a point R on an asymptote of a hyperbola to meet opposite branches of the curve in Q and q then qR. RQ = CP'- where GP is the semi- diameter of the liyperbola parallel to Qq. For by Newton's theorem RQ.Rq: RVt"- = GP . GP' : CHl .-. RQ.Rq = -GP\ .-. qR.RQ = GP\ As in the preceding article we can shew that Qq and the portion of it intercepted between the asymptotes have the same middle point. 174. Prop. If QV he an ordinate of the diameter PGP' and DGD' the diameter conjugate to PP' then QV-':PV.P'V = GD':GP\ Let Q V meet the asymptote Gfl in R and the curve again in Q'. Through R draw the chord qq' of the hyperbola parallel to PGP'. THE HYPERBOLA Then by Newton's theorem VQ . VQ' : VP. VP' = RQ. RQ' : Rq . Rq' .-. - QV : VP . VP' = CD' : - CP\ that is QV':PV.P'V=CD':CP"-. 179 This is the general theorem of which that of § 107 is a special case. We may write the relation as QV:CV'-CP'=CD':CP\ 175. From §§ 1G7, 174 we can see that a hyperbola may be regarded as the locus of a point in a plane such that the ratio of the scjuare of its distance froni a fixed line I varies as the product of its distances from two other fixed lines l' and I" parallel to one another and such that the point is on the same side of both of them. If I' and l" be perpendicular to I then I is the transverse axis of the hyperbola and l', I" the tangents at its vertices. If r and /" are not perpendicular to /, then I is a diameter of the hyperbola, and l', I" are the tangents at the points where it meets the hyperbola. 12—2 180 THE HYPERBOLA 176. Prop. If QQ', RR' he chords of a hyperbola inter- secting in then the ratio OQ . OQ' : OR . OR' is equal to that of tJie squares of tlie diameters -parallel to the respective chords. Let OQQ' meet an asymptote in L. Through L draw irr' parallel to ORR' to meet the curve in r and r'. Then by Newton's theorem OQ . OQ' lOR.OR' =LQ.LQ':Lr. Lr' = sq. of diameter parallel to QQ' : sq. of diameter parallel to RR\ 177. This proposition holds equally well if the ends of either or both of the chords lie on opposite branches of the hyperbola, provided that OQ . OQ' and OR . OR' be regarded simply as positive magnitudes. For suppose that Q, Q' lie on opposite branches and R, R' on the same branch, then as LQ and LQ' are in opposite directions we must for the application of Newton's theorem say LQ . LQ' = — QL . LQ' = — sq. of diameter parallel to QQ', so that, asOQ.OQ' = -QO.OQ' we have QO . OQ' : OR . OR' = sq. of diameter parallel to QQ' : sq. of diameter parallel to RR'. THE HYPERBOLA 181 178. It may perhaps seem imnecessarj^ to make a separate proof for the hyperbola of the proposition proved generally for the central conies in § 117. But our purpose has been to bring- out the fact that in § 117 the diameters must be length diameters of the curve itself, and for these the proposition is true. But as diameters of the hyperbola do not all meet the curve in real points, we wanted to shew how the diameters of the conjugate diameter may be used instead. Whenever the signs of OQ . OQ' and OR . OR' in the notation of §§ 176, 177 are different, this means that the diameters parallel to QQ', RR' are such that only one of them meets the hyperbola. The other meets the conjugate hyperbola. 179. We can see now that if DCD' be a diameter of the conjugate hyperbola, the imaginary points 8, S' in which it meets the original hyperbola, are given by (78^ = Ch'-' = - CD\ and this to the student acquainted with Analytical Geometry is also clear from the following : The equation of the hyperbola is ., ,., = l (1), a- b- ^ and of the conjugate hyperbola 1 (2). Thus Ci^Tcsponding to every point {x, y) on (2) there is a point {ix, ii/) on (1) and vice versa. And if a line through the centre meet the conjugate hyper- bola (2) in (x, y) it will meet the original hyperbola in {ix, iy). 180. Prop. If CP and CD be conjugate semi-diameters of a ItyperboUt CP' - CD' = CA' - CB\ Draw the ordinates PN and DM to the transverse and conjugate axes. 182 THE HYPER150LA These intersect in a point R on an asymptote (| 168), and we have CT- = C'iV- + PN- = CR' - {RN^ - PN^) ^CR'-BC^ (§166) ^ M D./ ^ / B ^ P C A\ N \ \ and CD'~ ^C]\P' + DM-' = GR'- (R]\P - DM'') = GR'-GA\ .'. CP' - CD-' ^ CA' - CB\ Cor. If CA = CB, that is if the hyperbola be a rectangular one, then CP = CD. 181. Prop. If P he any poiyit on a hyperbola 8P.S'P = CD-' where CD is the semi-diameter conjugate to CP. Since C is the middle point of SS' SP-' + ST^' = 2CS"- + 2CP-' (§ 10). . •. (S'P - spy- + 2SP . S'P = 2CS-' + 2CP\ that is 4>CA'- + 2SP. S'P = 2CS-' + 2CP\ '. SP . S'P = CP' + CS' - 2CA' = CP' + CB' + CA'-2CA'^ = CP' - {CA' - CB') = CD' (§180). THE HYPERBOLA 183 182. Prop. If GP and CD he conjugate semi-diameters of a hyperbola, and the normal at P meet CD in F, then PF.CB = AC.BC. Draw the perpendiculars SY and S'Y' from the foci on the tangent at P. Then the As SPY, S'PY being similar we have SY S'Y' S'Y'-SY -IPF PF SP S'P that is S'P- SP 2 AC SY.S'Y ' PF^ SP.S'P ~ AG'' BC-' pjr. CD' AC-'' AC- PF.CD = AC.BC. Cor. The area of the parallelogram formed by the tangents at the ends of a pair of conjugate diameters is constant = AA'.BB'. 183. Prop. The area of the triangle formed hy tlie asymptotes and any tangent to a hyperbola is constant. Let the tangent at P meet the asymptotes in L and I. (Use fig. of § 170.) Let P' be the other end of the diameter through P and let DCD' be the diameter conjugate to PP'. Then L and I are 184 THE HYPERBOLA angular points of the parallelogram formed by the tangents at P,P',D,D' (§170). Moreover /\CLl is one quarter of the area of the parallelo- gram formed by these tangents, that is (§ 182), A GLl = CA . GB, which is constant Cor. The envelope of a line which forms with two fixed lines a triangle of constant area is a hyperbola having the fixed lines for its asymptotes, and the point of contact of the line with its envelope will be the middle point of the portion intercepted between the fixed lines. 184. Prop. //■ TQ and TQ' he tangents to the same branch of a hyperbola, and GT meet the cu7've in P and QQ' in V, then GV.GT = GP\ This follows at once from the harmonic property of the pole and polar, for we have {PP\ TV) = - 1. .-. GV.GT = GP\ 185. Prop. If TQ and TQ' be tangents to o-p-^ositebranches of a hyperbola, and GT meet QQ' in V and the conjugate hyper- bola in P then VG.CT=GP\ THE HYPERBOLA 185 This can be surmised from the preceding proposition, for if CT meet the original hyperbohi, in the imaginary point p, we have (§ 179) Cp'=-CP\ .'. CV.GT^Cj)' (§184) = - CP\ .-. VG.CT = CP\ We give however the foHowing i)urely geometrical proof, which does not introduce imaginary points. Let BCD' be the diameter conjugate to PP' and meeting the hyperbola in D, D' , and TQ in t. Draw the ordinate QW to the diameter DI)', that is, QW is el to PP'. Then by similar As tWQ., tCT TC:WQ=Gt:tM\ .'. TC. ]VQ:WQ'=Ct.CW:GW.tW ^Gt.GW-.GW'-Gt.GW. Butby §184, C'<.(71^=Ci)^ .-. TG. WQ : WQ' = GD'^ : CW'^ - GD\ 186 THE HYPERBOLA But GP' : WQ'=CD' : GW - GD'^ (§ 174). .-. TG.WQ = GP\ .-. VG.GT=GP\ 186. The following arc special cases of the two preceding propositions. If the tangent at P to a hyperbola meet the transverse and conjugate axes in T and t respectively and PN, PM he ordiiiates to these axes GT.GN^GA"- MG.Gt=GB\ For the tangents from Twill be TP and TP' where P' is the point in which PN again meets the hyperbola ; and the tangents from t will be tP, tQ where Q is the point in which PM again meets the hyperbola. 187. Prop. If the normal at P to a hyperbola meet the transverse and conjugate axes in G and g, and the diameter parallel to the tangent at P in F, then FP.PG = BG% PF.Pg = AG'. THE HYPERBOLA 18^ This proposition can be established exactly like the corresponding one in the ellipse (§ 144). 9 M ':A K X A/f c / / \ ^ G 188. Prop. If the iwn/ial at F to a Injperbola meet the transverse axis in G, and PN be the ordinate CG^e\GN and NG:CN = BG::AC\ This is proved in the same way as in § 144 a. Circle of curvature. 189. Prop. The chord of the circle of curvature at any point P of a hyperbola and through the centre of the hyperbola is'-^^.a,uUkedia,neteroftkecircleis'-'^^. This is proved in the same way as for the ellip.se. EXERCISES 1. If a line be drawn through a focus S of a hyperbola parallel to one of the asymptotes and a perpendicular S'K be drawn from the other focus *S'' on to this line aS^A''= AA'. [Use § 157. Take P at O.] 2. Find the locus of the centre of a circle touching two fixed circles externally. 188 THE HYPERBOLA 3. If the tangent at any point /* of a hyperbola cut an asymptote in T and SP cut the same asymptote iu (^) then SQ = Q'f. \^TP and TO. subtend equal angles at S.'\ 4. Shew that when a pair of conjugate diameters of a hyper- bola are given in magnitude and position the asymptotes are completely determined. Hence shew that there are only two hyperbolas liaving a given pair of conjugate diameters. 5. If two hyperbolas have the same asymptotes a chord of one touching the other is bisected at the point of contact. 6. If PH, PK be drawn parallel to the asymptotes CO, CO' of a hyperbola to meet CO' and CO in // and A', then Pll.PK^\CH\ [Use § 183.] 7. The tangent to a hyperbola at P meets an asymptote in T and TQ is drawn parallel to the other asymptote to meet the curve in Q. PQ meets the asymptote in L and M. Prove that LM is trisected at P and Q. 8. From any point R on an asymptote of a hyperbola IIPN is drawn perpendicular to the transverse axis to cut the curve in P ; RK is drawn at right angles to CR to meet the transverse axis in K. Prove that PK is the normal at P. [Prove that CN= e'- . OK. % 188.] 9. Prove that in any central conic if the normal at P meet the axes in G and g then PG . Py - CD" where CD is conjugate to CP. 10. If the tangent at a point P of a hyperbola meet the asymptotes in L and I, and the normal at P meet the axes in G and g, then L, I, G, g lie on a circle which passes through the centre of the hyperbola. 11. The intercept of any tangent to a hyperbola between the asymptotes subtends at the further focus an angle equal to half the angle between them. 12. Given a focus of an ellipse and two points on the curve shew that the other focus describes a hyperbola. 13. If P bfi any point on a central conic whose foci are S and S', the circles on SP, S'P as diameters touch the auxiliary circle and have for their radical axis the ordinate of P. 14. The pole of the tangent at any point /* of a central conic with respect to the auxiliary circle lies on the ordinate of P. THE HYPERBOLA 189 15. If PP' and DD' be conjugate diameters of a hyperbola and Q any point on the curve then QP- + QP'- exceeds QD'~ + QD'- by a constant quantity. 16. Given two points of a parabola and the direction of its axis, prove that the locus of its focus is a hyperbola. 17. If two tangents be drawn to a hyperbola the lines joining their intersections with the asymptotes will be parallel. 18. If from a point /* in a hyperbola PK be drawn parallel to an asymptote to meet a directrix in K, and S be the corresponding focus, then PK = SP. 19. If the tangent and normal at a point P of a central conic meet the axis in T and G and PN be the ordinate, XG . CT= BC-. 20. The base of a triangle being given and also the point of contact with the base of the inscribed circle, the locus of the vertex is a hyperbola. 21. If tangents be drawn to a series of confocal hyperbolas the normals at their points of contact will all pass through a fi.xecl point, and the points of contact will lie on a circle. 22. A hyperbola is described touching the principal axes of a hyperbola at one of their extremities ; prove tha<^ one asymptote is parallel to the axis of the parabola and that the other is parallel to the chords of the parabola bisected by the first. 23. If an ellipse and a hyperbola confocal with it intersect in P, the asymptotes of the h3'perbola pass through the points of intersection of the ordinate of P with the auxiliary circle of the elli]>so. 24. Prove that the central distance of the point where a tangent to a liyperbola meets one asymptote varies as the distance, parallel to the transverse axis, of the point of contact from the other asymptote. 25. Tangents PPR', TQT' are drawn to a hyperbola, P, T being on one asymptote and R\ T' on the other; shew that the circles on RT' and R'T as diameters are coaxal with the director circle. 26. From any point P on a given diameter of a hyperbola, two straight lines are drawn parallel to the asymptotes, and meeting the hyperbola in Q, Q' ; prove that PQ, PQ' are to one another in a constant ratio. 190 THE HYPERBOLA 27. The asymptotes and one point on a hyperbola being given, determine the points in which a given Hne meets the curve. 28. If PN be^the ordinate and PG the normal of a point P of a hyperbola wliose centre is C, and the tangent at P intersect the asymptotes Jin L and L', then half the sum of CL and CL' is the mean proportional between CN and CO. 29. The tangents to a conic from any point on the director circle are the bisectors of the angles between every pair of conjugate lines through the point. 30. Given a focus, a tangent and the eccentricity of a conic, the locus of the centre is a circle. 31. If P be a point on a central conic such that the lines joining P to the foci are at right angles, CJJ^--2BC-. 32. Find the position and magnitude of the axes of a hyperbola which has a given line for an asymptote, passes through a given point, and touches a given straight line at a given point. 33. If P be any point on a hyperbola whose foci are ^S' and <S" the incentre of the triangle >SP>S' lies on the tangent at one of the vertices of the hyperbola. 34. With two conjugate diameters of an ellipse as asymptotes a pair of conjugate hyperbolas are constructed ; prove that if one hyperbola touch the ellipse the other will do likewise and that the diameters drawn through the points of contact are conjugate to each other. 35. Prove that a circle can be described to touch the four straight lines joining the foci of a hyperbola to any two points on the same branch of the curve. 36. Tangents are drawn to a hyperbola and the portion of each tangent intercepted by the asymptotes is divided in a given ratio ; shew that the locus of the point of section is a hyperbola. 37. From a point A' on an asymptote of a hyperbola PB is drawn touching the hyperbola in P, and P2\ PVuve drawn through P parallel to the asymptotes, cutting a diameter in T and V ; RV is joined, cutting the hyperbola in P, }) ; shew that TP and Tj) touch the hyperbola. [Project the hyperbola into a circle and V into the centre.] 38. CP and CD are conjugate semi-diameters of a hyperbola, and the tangent at P meets an asymptote in L; prove that if PD meet the transverse axis in F, LFC is a right angle. THE HYPERBOLA 191 39. From a given point on a hyperbola draw a straight line such that the segment between the other intersection with the hyperbola and a given asymptote shall be equal to a given line. When does the problem become impossible? 40. If P and Q be two points on two circles S^ and *S'o belonging to a coaxal system of which L is one of the limiting points, such that the angle PLQ is a right angle, prove that the foot of the perpendicular from L on PQ lies on one of the circles of the system, and thus shew that the envelope of PQ is a conic having a focus at L. 41. If a conic touch the sides of a triangle at the feet of the perpendiculars from the vertices on the opposite sides, the centre of the conic must be at the symmedian point of the triangle. 192 CHAPTER XIV THE RECTANGULAR HYPERBOLA 190. A rectangular hyperbola as we have already explained is one which has its asymptotes at right angles and its trans- verse and conjugate axes equal. The eccentricity of a rectangular hyperbola = ^2, for e = CS : GA, and CS' = CA' + CB' = WA\ We will now set forth a series of propositions giving the chief properties of the curve. 191. Prop. 1)1 a rectangular hyperbola conjugate diameter's are equal, and if QV he an ordinate of a diameter POP', QV"-=PV.P'V.' For we have CP' - CD' = GA ' - GB' (§ 1 80) = and . QV-':PV.PV=GD':GP' (§174) = 1. 192. Prop. Gonjugate diameters of a rectangular hyper- bola are equally inclined to each of the asymptotes. For the asymptotes are the double lines of the involution pencil formed by the pairs of conjugate lines through G, and therefore the asymptotes are harmonically conjugate with any pair of conjugate diameters. Hence as the asymptotes are at right angles they must be the bisectors of the angles between each pair of conjugate diameters (§ 72). Cor. 1. Any diameter of a rectangular hyperbola and the tangents at its extremities are equally inclined to each of the asymptotes. THE RECTANGULAR HYPERBOLA 193 Cor. 2. Any chord of a rectangular hyperbola and the diameter bisecting it are equally inclined to each asymptote. 193. Prop. Any diameter of a rectangular hyperbola is equal to the diameter perpendicular to it of the conjugate hyperbola. This is obvious when we consider that the conjugate hyper- bola is in our special case equal to the original hyperbola and can be obtained by rotating the whole figure of the hyperbola through a right angle about an axis through its centre perpen- dicular to its plane. 194. If a hyperbola have two perpendicular diameters equal to one another, the one belonging to the hyperbola itself and the other to its conjugate, the hyperbola must be a rectangular one. Let CP and CQ be the semi-diameters at right angles to one another and equal, P being on the hyperbola, and Q on the conjugate. Q M B /^ \ P C A N Draw PN and QM perpendicular conjugate axes respectively. Then ACXP^ACMQ. to the transverse and 13 194 THE RECTANGULAR HYPERBOLA Now PN^':CN'-CA' = BG-':AG' (§167) and QM^- : CM"- -BC^ = AC^: BC\ ON"- FN' , whence we get vtv> — tvtt: = 1 & ^C- BC GAP QM'- and = 1. BG^' AG- Subtract and use CN-'=GM" and PN''=QM-. J 1 BG-' AG- since GN-' + PN' + 0, J.C' = J5a ^^'U^-fil?^, PN'' 195. Prop. // a i-ectangular hyperbola pass through the vertices of a triangle it passes also through the ortliocentre. Let ABG be the triangle, P its orthocentre and AD the perpendicular from A on BG. Let the rectangular hyperbola meet AD again in p. Since the chords Ap and BG are at right angles the diameters parallel to them will meet one the hyperbola and the other the conjugate hyperbola. Thus DB.DG and Dp. DA will have opposite signs (§ 177), and the ratio of their numerical values will be unity since the diameters parallel to them being at right angles are equal. .-. BD.DG=Dp.DA. THE RECTANGULAR HYPERBOLA 195 But BD . DC = AD . DQ where Q is the point in which AD produced meets the circumcircle = -AD.DP (§6). .-. Dp. DA = DA .DP. .: Dp==DP that is ]) coincides with P. Cor. When a rectangular hyperbola circumscribes a tri- angle, if the three vertices lie on the same branch of the curve, the orthocentre will lie on the other branch, but if two of the vertices lie on one branch and the third on the other, the ortho- centre will lie on that branch on which are the two vertices. 196. Prop. //' a conic circumscribing a triangle pass through the orthocentre it must be a rectangular hyperbola. Let ABC be the triangle and AD, BE, C'i^the perpendiculai-s meeting in the orthocentre P. It is clear that the conic must be a hyperbola, since it is impossible for two chords of an ellipse or parabola to intersect at a point external to one of them and not to the other, and the chords A P and BG do so intersect. Now since BD . DC = AD . PD, the diameter parallel to BC= the diameter parallel to AP. And' these diameters must belong the one to the hyperbola and the other to its conjugate since DB . DC and DP . DA have opposite sign (§ 177). Therefore the hyperbola is a rectangular one (§ 194). 197. Prop. If a rectangular hyperbola circumscribe a triangle, its centre lies on the nine points circle of the triangle. Let ABC be the triangle, and D, E, F the middle points of the sides. Let be the centre of the rectangular hyperbola and DLL' an asymptote cutting AB and AC in L and L'. Since OF bisects the chord AB, OF and AB make equal angles with OIL' (§ 192, Cor. 2). .-. zFOL = zFLO. 13—2 19() THE RECTANGULAR HYPERBOLA Similarly z EOL' = z EL'O. .'. zFOE = zALL' + zAL'L = zBAO = ^FI)E. .'. lies on the circle round DEF, which circle is the nine points circle of the triangle. 198. Prop. The angle between any chord PQ of a rect- angular hyperbola and the tangent at P is equal to the angle subtended by PQ at P', the other end of the diameter through P. Let the chord PQ and the tangent at P meet the asymptote on in R and L. Let V be the middle point of PQ. THE RECTANGULAR HYPERBOLA 197 Then Z VRC = Z VCR (§ 192, Cor. 2). and Z PLC = Z PCL (§192, Cor. 1 ). .-. zLPR = ^CLP-zCRV . = zP6'Z-zFCi^ = zF6'P ==ZQP'P (since CV is parallel to QP'). 199. Prop. Art/ chord of a rectangular hyperbola sub- tends at the ends of any diameter angles ivJiich are equal or supplementary. Let QR be a chord, and PCP' a diameter. Let the tangents at P and F meet the asymptotes in L, I and L', I'. In fig. 1, where Q and R lie on the same branch and PP' cuts QR internally, zQPL = zQP'P (§198) and Z RPl = Z RP'P. .-. Z QPR = supplement of sum of QPL and RPl = supplement of Z QP'R. Fiff. 1. Fig. 2. In fig. 2, where Q and P lie on the same branch and PP' cuts QR externally, ZLPR = ZRFP and ^LPQ = ZQP'P. .-. ^RPQ = zQFP-zRFP = zQFR. IDS thp: rectangular hyperbola In fig. 3, where Q and R lie on opposite branches and PP' cuts QR internally, z QPR = z QPL + z LPP' + z P'PR = z QP'P + z PP7' + z PP7' = zQP'R. In fig. 4, where Q and P lie on opposite branches and PP' cuts QR externally, z QPP = z QPP + z PPP = z QP'P + z LPR and Z QP'P = Z QP'X' + z P7^'P = z QP'P' + z RPP'. .-. zQPR + zQP'R' = z.L'P'P + zLPP' = 2 right Z s. EXERCISES 1. The portion of any tangent to a rectangular hyperbola in- tercepted between its asymptotes is double the distance of its point of contact from the centre. 2. If PJV be the ordinate of any point P on a rectangular hyperbola, and PG the normal at P, prove GJ^= NG, and the tangent from N to the auxiliary circle = PN. 3. If CK be the perpendicular from the centre on the tangent at any point P of a rectangular hyperbola the triangles PC A, CAK are similar. THE RECTANGULAR HYPERBOLA 199 4. PQR is a triangle inscribed in a rectangular hyperbola, and the angle at P is a right angle; prove that the tangent at P is per- j^endicular to QR. 5. If PP' and QQ' be perpendicular chords of a rectangular hyperbola then PQ', QP' will be at i-ight angles, as also PQ and P'Q'. 6. PP' is any chord of a rectangular hyperbola, and a diameter perpendicular to it meets the hyperbola in Q ; prove that the circle PQP' touches the hyperbola at Q. 7. If from the extremities of any diameter of a rectangular hyperbola lines be drawn to any point on the curve, they wull be equally inclined to each asymptote. 8. Focal chords of a rectangular hyperbola which are at right angles to one another are equal. [See§§ 116, 117.] 9. The distance of any point on a rectangular hyperbola from the centre is the geometric mean between its distances from tlie foci. 10. If PP' be a double ordinate to the transverse axis of a rectangular hyperbola whose centre is C, then C P' is perpendicular to the tangent at P. 11. The centre of the inscribed circle of a triangle lies on any rectangular hyperbola circumscribing the triangle whose vertices are the e centres. 12. Focal chords parallel to conjugate diameters of a rectangular hyperbola are equal. 13. If the tangent at any point P of a rectangular hyperbola, centre C, meet a pair of conjugate diameters in E and F, PC touches the circle CEF. 14. Two tangents are drawn to the same branch of a rectangular hyperbola. Prove tliat the angles which these tangents subtend at the centre are respectively equal to the angles which they make with the chord of contact. 15. A circle and a rectangular hyperbola intersect in four points and one of their common chords is a diameter of the hyper- bola ; shew that the other common chord is a diameter of the circle. 16. Ellipses are described in a given parallelogram ; shew that their foci lie on a rectangular hyperbola. 200 THE RECTANGULAR HYPERBOLA 17. If from any point Q in the conjugate axis of a rectangular hyperbola QA be drawn to the vertex, and QR parallel to the trans- verse to meet the curve, QR = AQ. 18. The lines joining the extremities of conjugate diameters of a rectangular hyperbola are perpendicular to the asymptotes. 19. The base of a triangle and the difference of its base angles being given the locus of its vertex is a rectangular hyperbola. 20. The circles described on parallel chords of a rectangular hyperbola are coaxal. 21. If a rectangular hyperbola circumscribe a triangle, the pedal triangle is a self-conjugate one. 22. At any point P of a rectangular hyperbola the radius of curvature varies as C/*", and the diameter of the curve is equal to the central chord of curvature. 23. At an}' point of a rectangular hyperbola the normal chord is equal to the diameter of curvature. 24. FN is drawn perpendicular to an asymptote of a rectangular hyperbola from any point P on it, the chord of curvature along PN IS equal to -p^ - 201 CHAPTER XV ORTHOGONAL PROJECTION 200. When the vertex of projection by means of which a figure is projected from one plane p on to another plane vr is at a very great distance from these planes, the lines joining corre- sponding points in the original figure and its projection come near to being parallel. What we may call cylindrical projection is the case in which points on the p plane are projected on to the tt plane by lines which are all drawn parallel to each other. We regard this as the limiting case of conical projection when the vertex V is at infinity. In the particular case where the lines joining corresponding points are perpendicular to the tt plane on to which the figure on the p plane is projected, the resulting figure on the tt plane is said to be the orthogonal projection of the original figure. Points in space which are not necessarily in a plane can be orthogonally projected on to a plane by drawing perpendiculars from them to the plane. The foot of each perpendicular is the projection of the point from which it is drawn. Thus all points in space which lie on the same line perpendicular to the plane on to which the projection is made will have the same projection. In the present chapter it will be shewn how certain pro- perties of the ellipse can be obtained from those of the circle, for, as we shall see, every ellipse is the orthogonal projection of a circle. It is first necessary to establish certain properties of orthogonal projection. 201. It may be observed at the outset that in orthogonal projection we have no vanishing line as in conical projection. 202 (ORTHOGONAL PROJECTION The line at infinity in the ji plane projects into the line at infinity in the plane ir. This is clear from the fact that the perpendiculars to the ir plane from points in it on the line at infinity meet the p plane at infinity. It follows that the orthogonal projection of a parabola will be another parabola, and of a hyperbola another hyperbola, while the orthogonal projection of an ellipse Avill be another ellipse or in particular cases a circle. 202. The following propositions relating to orthogonal pro- jection are important : Prop. Tlie projection of a straight line is another straight line. This is obvious from the fact that orthogonal is only a limiting case of conical projection. It is clear that the line in the TT plane which will be the projection of a line I will be that in which the plane through I and perjjendicular to the plane tt cuts this TT plane. 203. Prop. Parallel straight lines p7'oject into 2Ja'>'allel straigJit lines, and in the same ratio as regards their length. Let J^i) and CI) be tAvo lines in space parallel to one another. y\ B y^ y D A ^ F C G / c ( 7 7 £ I b / Let ah, cd be their orthogonal projections on to the plane tt. ORTHOGONAL PROJECTION 203 Then ab and cd must be parallel, for if they were to meet in a point p, p Would be the projection of a point common to AB and CD. Now draw AF and CG parallel respectively to ab and cd to meet Bb and Dd in F and G. Then AabF is a parallelogram so that AF= ab, and similarly CG = cd. Now since AB is parallel to CD, and ^i^ to CG (for these are respectively parallel to ab and cd which we have proved to be parallel), the angle FAB = the angle GCD ; and the angles at F and G are right angles. .'. A ^i^5 is similar to A CG^D. .-. AF : CG = AB : CD. .-. ab:cd = AB:CD. Cor. Lengths of line lying along the same line are projected in the same ratio. 204. Prop. // I be a limited line in the p plane parallel to p's intersection with the ir plane, the orthogonal projectdon of I OH IT will be a line parallel to and of the same length as I. Let AB be the limited line I, and ab its orthogonal projection. 204 ORTHO(;OJN'AL PROJECTION Draw J.Cand BD perpendicular to the line of intersection of p and TT. Then ACDB is a parallelogram. Also since Aa and Bb are perpendicular to tt, Ga and Dh are perpendicular to CD and therefore they are parallel to each other. Further AACa = ABDb for AC^BD, zAaC = Z BbD and /.AGa= ^ BDb for AO and Ca are parallel to BD and Db. .-. Ca = Db. .". as C'a and Db are parallel, (7i)&a is a parallelogram. .-. ab = CD = AB. ' 205. Prop. A limited line in the p plane perpendicular to the line of intersection of p and ir will project into a line also perpendicular to this line of intersection and whose length will ■bear to the original line a ratio equal to the cosine of the angle between the planes. Let AB be perpendicular to the intersection of p and tt, and let its line meet it in C. Let ab be the orthogonal projection of AB. Then ab and AB meet in C, and ab : AB = ac : AC = cos aCA = cos ( Z between p and tt). ORTHOGONAL PROJECTION 205 206. Prop. A closed figure on the p plane ivill project into a closed figure luhose area will bear to that of the original figure a ratio equal to the cosine of the angle betiveen the planes. For we may suppose the figure to be made up of an infinite number of" narrow rectangular strips the length of which runs parallel to the intersection of p and tt. The lengths of the slips are unaltered by projections, and the breadths are diminished in the ratio of the cosine of the angle between the planes. 207. The ellipse as the orthogonal projection of a circle. We have seen (§ 150) that corresponding ordinates of an ellipse and its auxiliary circle bear a constant ratio to one another, viz., BC:AC. Now let the auxiliary circle be turned about its major axis AA' until it comes into a plane making with that of the ellipse an angle whose cosine is BG:AC. 206 ORTHOGONAL PROJECTION It is clear that the lines joining each point on the ellipse to the new position of the point corresponding to it on the auxiliary circle will be perpendicular to the plane of the ellipse. Thus the ellipse is the orthogonal projection of the circle in its new position. Certain properties of the ellipse then can be deduced from those of the circle by orthogonal projection. We proceed to some illustrations. 208. Prop. If CP and CD he a pair of conjugate semi- diameters of an ellipse, and CP', CD' another such pair, and the ordinates P'M, D'N he drawn to CP, tlien P'M : CN = D'N: CAl = CD : CP. For let the corresponding points in the auxiliary circle BC adjusted to make an angle cos~^ -r- -, with the plane of the ellipse, be denoted by small letters. Then Cp and Cd are perpendicular radii as are also Cp and Cd', and p'm, d'n being parallel to Cd will be perpendicular to Cp, and we have A Cmp =Ad 'nC. . • . p'm : Cd = Cn : Cp and d'n : Cd = Cm : Cp. by § 203 P'M:GD=CX:CP D'N:CD=CM:CP. PM : CN - CD : CP = D'N : CM. ORTHOGONAL PROJECTION 207 209. Prop. If the tangent at a point P of an ellipse meet any pair of conjugate diameters in T and T' and CD he conjugate to CP, then rP . PT' = CD\ For in the corresponding figure of the circle Ct and Ct' are at right angles, and Cp is perpendicular to tt'. t' J' .'. tp . pt' = Cp- = Cd '". .-. tp:Cd = Cd:pt. TP:CD = CD:PT'. TP.PT' = CD\ 210. Prop. The area of an ellipse ichose semi-axes are CA and CB is IT. CA.cn. For the ellipse is the orthogonal projection of its auxiliary BC circle tilted to make an angle cos'^jy; with that of the ellipse. . •. Area of ellipse : Area of auxiliary circle = BC : A C (§ 206 ). .-. Area of ellipse = tt . BC .AC. 211. Prop. The orthogonal projection of a circle from a plane p on to another plane tt is an ellipse whose major axis is jKirallel to the intersection of p and tt, and equal to the diameter of the circle. Let xiA' be that diameter of the circle which is parallel to the p and tt planes. Let A A' project into aa equal to it (§ 20-1). 208 ORTHOGONAL PROJECTION Let PN be an ordinate to the diameter A A' and let pn be its projection. .•. p/i = PN cos a where a is the Z between p and ir, and pn is perpendicuhxr to cm' (§ 205), Now p7i- : CD) . na = PN" cos^ a : ^iV . NA' = cos- a : 1. Hence the locus of p is an ellipse having aa for its major axis, and its minor axis = aa x cos a. The eccentricity is easily seen to be sin a. CoR. 1. Two circles in the same plane project orthogonally into similar and similarly situated ellipses. For their eccentricities will be equal and the major axis of the one is parallel to the major axis of the other, each being parallel to the line of intersection of the planes. Cor. 2. Two similar and similarly situated ellipses are the simultaneous orthogonal projections of two circles. EXERCISES 1. The locus of the middle points of chords of an ellipse which pass through a fixed point is a similar and similarly situated ellipse. 2. If a parallelogram be inscribed in an ellipse its sides are parallel to conjugate diameters, and the greatest area of such a parallelogram is BC.A<'. 3. If PQ be any chord of an ellipse meeting the diameter con- jugate to CP in 7\ then PQ . PT --=2CR- where CR is the semi- diameter parallel to PQ. 4. If a variable chord of an ellipse bear a constant ratio to the diameter parallel to it, it will touch anoth(,M- similar ellipse having its axes along those of the original ellipse. 5. The greatest triangle which can be inscribed in an ellipse has one of its sides bisected by a diameter of the ellipse and the others cut in points of trisection by the conjugate diameter. ORTHOGONAL PROJECTION 209 6. If a straight line meet two concentric, similar and similarly situated ellipses, the portions intercepted between the curves are equal. 7. The locus of the points of intersection of the tangents at the extremities of pairs of conjugate diameter is a concentric, similar, and similarly situated ellipse. 8. If CP, CD be conjugate serai-diameters of an ellipse, and BP, BD be joined, and AD, A'P intersect in 0, the figure BDOP will be a parallelogram. 9. Two ellipses whose axes are at right angles to one another intersect in four points. Shew that any pair of common chords will make equal angles with an axis. 10. Shew that a circle of curvature for an ellipse and the ellipse itself can be projected orthogonally into an ellipse and one of its circles of curvature. 14 210 CHAPTER XVI CROSS-RATIO PROPERTIES OF CONICS 212. Prop. If A, B, G, D he four fixed points on a conic, and P a variable point on the conic, P (ABCD) is constant and equal to the corresponding cross-ratio of the four points in which the tangents at A, B, 0, D meet that at P. Project the conic into a circle and use corresponding small letters in the projection. Then P{ABGD)=p{ahcd). But p (abed) is constant since the angles apb, bpc, cpd are constant or change to their supplements as p moves on the circle; therefore P {ABCD) is constant. Let the tangents at a, b, c, d cut that in p in a^, b^, Ci, d^ and let be the centre of the circle. CROSS- RATIO PROPERTIES OF CONICS 211 Then Oa-^, Ob^, Oc^, Od^ are perpendicular to pa, ph, pc, pd. .•. p {abed) = (ciibiCidi) = (ajbiCidi). .-. P{ABCD) = {A^B,C,D,). Cor. If J.' be a point on the conic near to A, we have A'{ABGD) = P{ABGD). .-. if J. T be the tangent at A, A (TBCD) = P{ABCD). Note. In the special case where the pencil formed by joining any point P on the conic to the four fixed points A, B, C, D is harmonic, we speak of the points on the conic as harmonic. Thus if P {ABCD) = -\, we say that A and C are harmonic conjugates to B and D. 213. Prop. If A, B, C, D be four fixed non-collinear points in a plane and P a point such that P (ABCD) is constant, the locus of P is a conic. Let Q be a point such that Q {ABCD) = P (ABCD). % Then if the conic through the points A, B, C, D, P does not pass through Q, let it cut QA in Q'. .-. P {ABGD) = Q' (ABCD) by § 212. .-. Q' (ABCD) = Q (ABCD). Thus the pencils Q (A, B, C, D) and Q (A, B, G, D) are homographic and have a common ray QQ'. 14—2 212 CROSS- RATIO PROPERTIES OF CONICS Therefore (§ 64) they are coaxally in perspective ; that is, A, B, C, D are collinear. But this is contrary to hypothesis. Therefore the conic through A, B, C, D, P goes through Q. Thus our proposition is proved. We see from the above that we may regard a conic through the five points A,B, G,D, E as the locus of a point P such that P {ABCD) = E {ABGD). 214. Prop. TJie envelope of a line wJiich cuts four non- concurrent coplanar fixed straight lines in four points forming a range of constant cross-ratio is a conic touching the four lines. This proposition will be seen, when we come to the next chapter, to follow by Reciprocation directly from the proposition of the last paragraph. The following is an independent proof Let the line p cut the four non-concurrent lines a, b, c, d in the points A, B, G, D such that {ABGD) = the given constant. CROSS- RATIO PROPERTIES OF CONICS 213 Let the line q cut the same four lines in A', B', C, D' such that {A'B'C'D') = (A BCD). Then if q be not a tangent to the conic touching a, h, c, d,p, from A' in q draw q' a tangent to the conic. Let b, c, d cut q' in B", C", D". .-. {A'B"C"D") = {ABCD) by § 212 = {A' BCD'). The ranges A'B'C'D" and A'B'C'D' are therefore homo- graphic and they have a common corresponding point. Therefore they are in perspective (§ 60), which is contrary to our hypothesis that a, b, c, d are non-concurrent. Thus q touches the same conic as that which touches a, b, c, d, p. And our proposition is established. 215. Prop. //' F (A, B, C, D) be a pencil in the plane of a conic S, and A,, B„ C„ D, the poles of FA, FB, FC, FD with respect to S, then P{ABCD) = (A,B,CM. We need only prove this in the case of a circle, into which as we have seen, a conic can be projected. 214 CROSS-RATIO PROPERTIES OF CONICS Let be the centre of the circle. Then OA^, OB^, 00^, OD^ are perpendicular respectively to PA, PB, PC, PR .. P (ABCD) = (A,B,CM = (A.BAD,). This proposition is of the greatest importance for the pur- poses of Reciprocation. We had already seen that the polars of a range of points form a pencil ; we now see that the pencil is homographic with the range. 216. PascaPs theorem. If a conic pass through six points A, B,0, D, E, F, the opposite pairs of sides of each of the sixty different hexagons (st^-sided figures) that can be formed with these points intersect in collinear points. This theorem may be proved by projection (see Ex. 25, Chap. X). Or we may proceed thus : Consider the hexagon or six-sided figure formed with the sides AB, BG, CD, DE, EF, FA. CROSS-RATIO PROPERTIES OF CONICS 215 The pairs of sides which are called opposite are AB and DE ; BG and EF; CD and FA. Let these meet in X, Y, Z respectively. Let CD meet EF in H, and DE meet FA in G. Then since A (BDEF) = C (BDEF), .-. (XDEG) = (YHEF). These homographic ranges XDEG and YHEF have a common corresponding point ^. .-. XY,DH and FG= are concurrent (§ 60), that is, Z, the intersection of DH and FG, lies on XF. Thus the proposition is proved. The student should satisfy himself that there are sixty different hexagons that can be formed with the six given vertices. 217. Brianchon's theorem. If a conic be inscribed in a hexagon the lines joining opposite vertices are concurrent. This can be proved after a similar method to that of § 216, and may be left as an exercise to the student. We shall content ourselves with deriving this theorem from Pascal's by Recipro- cation. To the principles of this important development of modem Geometry we shall come in the chapter immediately following this. 218. Prop. I'he locus of the centres of conies through four fixed points is a conic. Let be the centre of one of the conies passing through the four points A, B, C, D. 516 CROSS-RATIO PROPERTIES OF CONICS Let M,,M,, M,, M, be the middle points of AB, BC, CD, DA respectively. Draw Oil//, OM.:, OM,', OM,' parallel to AB, BG, CD, DA respectively. Then Oifi, OM,'; OM,, OM,' ; OM,, OM; ■ 0M„ OM,' are pairs of conjugate diameters. Therefore they form an involution pencil. .-. {M,M,M,M,) = {M,'M.^M;m:). But the right-hand side is constant since OM,, OM, &c. are in fixed directions. .-. (MJl^MsM,) is constant. .*. the locus of is a conic through M,, M„ M^, M,. Cor. 1. The conic on which lies passes through ifg, M^ the middle points of the other two sides of the quadrangle. For if Oi, 0.^, O3, O4, O5 be five positions of 0, these five points lie on a conic through M^, M„ M,, M, and also on a conic through ilfi, M„ M„ M^. But only one conic can be drawn through five points. Therefore J/j, M„ M.., M^, M^, M, all lie on one conic, which is the locus of 0. Cor. 2. The locus of also passes through P, Q, R the diagonal points of the quadrangle. For one of the conies through the four points is the pair of lines AB, CD ; and the centre of this conic is P. So for Q and R. 219. Prop. If [AA', BB', CC'\ he an involution pencil and if a conic he draivn through to cut the rays in A, A', B, B', C, C, then the chords A A', BB' , GC are concurj-ent. Let A A' and BB' intersect in P. Project the conic into a circle with the projection of P for its centre. Using small letters in the projection, we see that aoa, hoh' are right angles, being in a semicircle. CROSS-RATIO PROPERTIES OF CONICS Hence they determine an orthogonal involution. .•. coc is a right angle ; that is, cc goes through p. /. CC goes through P. 217 It will be understood that the points AA' , BB', CC when joined to any other point on the conic give an involution pencil ; for this follows at once by the application of § 212. A system ^f points such as these on a conic is called an involution range on the conic. The point P where the corresponding chords intersect is called tlie j)ole of the involution. EXERCISES 1. If {P, F), (Q, Q') be pairs of harmonic points on a conic (see Note on § 212), prove that the tangent at P and PP' are harmonic conjugates to PQ and PQ'. Hence shew that if PP' be normal at P, PQ and PQ' make equal angles with PP. 2. The straight line PP' cuts a conic at Pand P' and is normal at P. The straight lines PQ and PQ' are equally inclined to PP' and cut the conic again in Q and Q'. Prove that PQ and P'Q' are harmonic conjugates to PP and the tangent at P. 3. Shew that if the pencil formed by joining any point on a conic to four fixed points on the same be harmonic, two sides of the quadrangle formed by the four fixed points are conjugate to each other with respect to the conic. 218 CROSS-RATIO PROPERTIES OF CONICS 4. The tangent at any point P of a hyperbola intersects the asymptotes in il/j and vl/, and the tangents at the vertices in L^ and Lr,, prove that PAf,- = PL^.rL,. 5. Deduce from Pascal's theorem that if a conic pass through the vertices of a triangle the tangents at these points meet the opposite sides in collinear points. [Take a hexagon AA'BB'CC in the conic so that A', B' , C are near to ^, B, C] 6. Given three points of a hyperbola and the directions of both asymptotes, find the point of intersection of the curve with a given straight line drawn parallel to one of the asymptotes. 7. Through a fixed point on a conic a line is drawn cutting the conic again in P, and the sides of a given inscribed triangle in A\ B\ C". Shew that [PA'B'C) is constant. 8. A, B, C, D are any four points on a hyperbola ; CK parallel to one asymptote meets AD \n A", and DL parallel to the other asymptote meets CB in //. Prove that KL is parallel to AB. 9. The sixty Pascal lines corresponding to six points on a conic intersect three by three. 10. Any two points D and E are taken on a hyperbola of which the asymptotes are GA and CB ; the parallels to CA and CB through D and U respectively meet in Q ; the tangent at D meets CB in P, and the tangent at PJ meets CA in T. Prove that T, Q, R are collinear, lying on a line parallel to DE. 1 1. The lines CA and CB are tangents to a conic at A and B, and D and E are two other points on the conic. The line CD cuts ABinG, AE in H, and BE in K. Prove that CD^ : GD'^CII. CK : GH . GK. 12. Through a fixed point A on a conic two fixed straight lines AI, AI' are drawn, iS'and S' are two fixed points and P a variable point on the conic ; PS, PS' meet AI, A I' in Q, Q' respectively, shew that QQ' passes through a fixed point, 13. If two triangles be in perspective, the six points of inter- section of their non-corresponding sides lie on a conic, and the axis of perspective is one of the Pascal lines of the six points. CROSS-RATIO PROPERTIES OF CONICS 219 14. If two chords PQ, PQ' of a conic through a fixed point P are equally inclined to the tangent at P, the chord QQ' passes through a fixed point. 15. If the lines AB, BC, CD, DA touch a conic at P, Q, R, S respectively, shew that conies can be inscribed in the hexagons APQCRS and BQRDSP. 16. The tangent at P to an ellipse meets the auxiliary circle in Y and Y'. ASS' A' is the major axis and SY, S'Y' the perpendiculars from the foci. Prove that the points A, Y, Y', A' subtend at any point on the circle a pencil whose ci-oss-ratio is independent of the position of P. 17. li A, B he given points on a circle, and CD be a given diameter, shew how to find a point P on the circle such that PA and PB shall cut CD in points equidistant from the centre. 220 CHAPTEE XVII RECIPROCATION 220. I f we have a number of points P, Q, R, fcc. in a plan e _an d take t he_p olars p, q, r, &c. o^ _these points with respect to a__conie F in the plane , thenj bhe^m^^jommg^a^y;!^^^ ^goin ts F and Q is . a,g y e- Ji aye jtlgeady_ §een. the_polai^ with rg_spect to r of . th e i nterse ction of th, e_c orrespondinff J ines p and q. It will be convenient to represent the intersection of the lines p and q by the symbol (pq), and the line joining the points P and Q by (PQ). The p oint P corre sponds with the line p^ in t he sense I hat PJsjb he pole of j ), j^cLthfi^line^ C^^ correspond s with th e point ( pq) in the sense that (PQ) is _th e polar o £Xj2i^- Thus if we have a fi gure P cojisisti ng of an aggregate of points and lines, t hea^jcorrespondinp' toJ.t^ w e h av e a GignreJ^' consistmg^f_lin£Sjand-^>oints — Two snch fignrea^Xand^^re called in relation to one another Recipr ocal Ji^ reS;_JThe medium o ftheir_ Recipro cit y is the_ conic^J\_ Using § 215 we see that a range of points in jP corresponds to a pencil of lines, homographic with the range? in F'. RECIPROCATION 221 221. "Rj^mPflTis nf t,V.p prinniple. of correspondence enun- ciated in the last paragraph ^ve. are able frOT nji k n ov^^^ ,property of a figure consisting of p oints and lines to infer another property of a fig ure consi st ing of lin as nnd points. The o ne property is called the Reciprocal of the other, and the process of passin g f rom the one to the other is known as Reciprocation. We will now give examples. 222. We know that if the vertices of two triangles ABC, A'B'C be in perspective, the pairs of corresponding sides {BG) (BV), (GA) (C'A'), (AB) (A'B') intersect in collinear points X, Y, Z. Fig. F. Now if we draw the reciprocal figure, corresponding to the vertices of the triangle ABG, we have three lines a, h, c forming a triangle whose vertices will be (6c), {ca), {ah). And similarly for A'B'C'. Corresponding to the concurrency of {AA'), {BB'), (CC) in the figure F, we have the collinearity of (aa), (bb'), (cc) in the figure F'. Corresponding to the collinearity of the intersections of (BC) {B'C'), {CA) {C'A'), {AB) {A'B') in figure F, we have the concurrency of the lines formed by joining the pairs of points (6c) (6'c'), {ca) {ca), {ah) {a'b') in the figure F'. 222 RECIPROCATION Thus the theorem of the figure F reciprocates into the following : If two triangles whose sides are abc, a'h'c respectively be such that the three intersections of the corresponding sides are collinear, then the lines joining corresponding vertices, viz. {ah) and {ah'), {he) and (6'c'), (ca) and {ca), are concurrent. (ab) Fig.F'. The two reciprocal, theorems placed side by side may be stated thus : Triangles in perspective are I Coaxal triangles are in per- coaxal. I spective. The student will of course have realised that a triangle regarded as three lines does not reciprocate into another triangle regarded as three lines, but into one regarded as three points ; and vice versa. 223. Let us now connect together by reciprocation the harmonic property of the quadrilateral and that of the quadrangle. Let a, h, c, d be the lines of the quadrilateral ; A, B, C, D the corresponding points of the quadrangle. RECIPROCATION 223 (ac) Let the line joining (ub) and (cd) be p, „ „ „ {ac) and {hd) be q, „ „ „ (ad) and (he) be r. (bd) Fig. F'. 224 RECIPROCATION The harmonic property of the quadrilateral is expressed symbolically thus : {{ab){cd), {pr)(pq)] = -l, {(ad) (be), (pr){qr)}=-l, {{ac){hd), (pq){qr)}=-l. The reciprocation gives {{AB)iCD), (FR){PQ)} = -1, [{AD){BC\ (PR){QR)} = -1, {(AC){BD), iPQ)(QR)] = -l. If these be interpreted on the figure we have the harmonic property of the quadrangle, viz. that the two sides of the diagonal triangle at each vertex are harmonic conjugates with the two sides of the quadrangle which pass through that vertex. The student sees now that the ' diagonal points ' of a quadrangle are the reciprocals of the diagonal lines of the quadrilateral from which it is derived. Hence the term ' diagonal points.' 224. Prop. A71 involution range reciprocates with respect- to a. conic into an involution pencil. For let the involution range be A, A,; B,B,; C,G,kc. on a line p. The pencil obtained by reciprocation will be a, a^ ; h, h^; c, Ci &c. through a point P. Also {ahca,) = {ABGA,) and {aAc.a) = (A,B,C,A) by § 215. But (ABCA,) = {A,B,C, A) by § 78. .". {ahca^) = (a^biCia). Thus the pencil is in involution. 225. Involution property of the quadrangle and quadrilateral. Prop. Ally transversal cuts the pairs of opposite sides of • a quadrangle in pairs of points which are in involution. RECIPROCATION Let ABCD be the quadrangle (§ 76). A, 225 Let a transversal ^ cut the opposite pairs of sides AB, CD in E, E^, AG, BD in F,F„ AD, BG in G, G,. Let AD and BG meet in P. Then {GEFG,) = A{GEFG,) = (FBCG,) = D{PBGG,) = {GF,E,G,) = ( Gt] £", Fi Or) by interchanging the letters in pairs. (aB) ad I Hence E, £", ; F, F^ ; G, Gi belong to the same involution. A. G. 15 226 RECIPROCATION We havL' only to reciprocate the above theorem to obtain tliis other: The lines joining any point to the 2mirs of opposite vertices of a complete quadrilateral form a j^encil in involution. Thus in our figure T, which corresponds to the transversal t, joined to the opposite pairs of vertices (ac), (bd) ; (ad), (be) ; (ab), (cd) gives an involution pencil. 226. Prop. IVie circles described on the three diagonals of a complete quadrilateral are coaxal. Let AB, BC, CD, DA be the four sides of the quadrilateral. The diagonals are AC, BD, EF. Let P be a point of intersection of the circles on AG and BD as diameters. .-. APC and BPD are right angles. But PA, PC; PB, PD; PE, PF are in involution (§ 225). i .-. by § 86 Z EPF is a ri^ht angle. I .•. the circle on EF as diameter goes through P. ' RECIPROCATIOX 22/ Similarly the circle on EF goes through the other point of intersection of the circles on BD and AC. That is, the three circles are coaxal. Cor. The middle points of the three diagonals of a quadri- lateral are collinear. This important and well-known property follows at once, since these middle points are the centres of three coaxal circles. The line containing these middle points is sometimes called the diameter of the quadrilateral. 227. Desargues' theorem. Conies through four gicen points are cut by any transversal in iniirs of points belonging to the same involution. Let a transversal t cut a conic through the four })oints A, B, G,D in P and P,. Let the same transversal cut the two pairs of opposite sides AB, CD; AC, BD of the quadianglo in E, E, : F, F,. We now have {PEF1\) = A{PEFI\) = A {PBCP,) = D(PBCP,)hy^2l2 = (PF,EJ\) = (PiEiF^P) by interchanging the letters in pairs. .-. P, Pi belong to the involution determined by E, E^ ; F, F,. 15—2 228 RECIPROCATION 'I'hus all the conies through ABCD will cut the transversal t in pairs of points belonging to the same involution. Note that the proposition of § 225 is only a special case of Desargues' theorem, if the two lines AD, BC be regarded as one of the conies through the four points. 228. As we shall presently see, the reciprocal of Desargues' theorem is the following : If conies touch four given lines the jKiirs of tangents to them from any point in their plane belong to the same involution pencil, namely that determined by the lines joining the point to the pairs of opposite vertices of the quadrilateral formed by the four lines. Reciprocation applied to conies. 229. We are now going (jn to explain how the principle of Reciprocation is applied to conies. Suppose the point P describes a curve S in the plane of the C(Hiic r, the line p, which is the polar of F with regard to r> will envelope some curve which we will denote by S'. Tangents to S' then correspond to points on S ; but we must observe further that tangents to S correspond to points on >S". For let P and P' be two near points on S, and let p and p' be the corresponding lines. Then the point (pp) corresponds to the line (PP') Now as P' moves up to P, (PP') becomes the tangent to >S' at P and at the same time {p}^') becomes the point of contact of p with its envelope. Hence to tangents of *S' correspond points on S'. Each of the curves S and S' is called the polar reciprocal of the other with respect to the conic T. 230. Prop. If S be a conic then S' is another conic. Let A, B, C, D he four fixed points on S, and P any other point on S. Then P {ABCD) is constant. RECIPROCATION 229 But P (ABOD) = {(pa) (pb) (pc) (pd)} by § 215. .-. {(pa) (pb) (pc) (pel)] is constant. .•. the envelope of j3 is a conic touching the lines a, b, c, d (§ 214). Hence S' is a conic. This important proposition might have been proved as follows. <S, 'being a conic, is a curve of the second order, that is, straight lines in its plane cut it in two and only two points, real or imaginary. Therefore S' must be a curve of the second class, that is a curve such that from each point in its plane two and only two tangents can be drawn to it ; that is, S' is a conic. 231. Prop. If S and S' be two conies reciprocal to each other tvith respect to a conic F, then pole and polar of S corre- spond to polar and pole of S' and vice versa. Let F and TUhe pole and polar of S. [It is most important that the student should understand that TU is the polar of P with respect to S, not to F. The polar of P with respect to F is the line we denote by p.] Let QR be any chord of S which passes through P : then the tangents at Q and R meet in the line TU, at T say. Therefore in the reciprocal figure p and (tii) are so related that if any point (qt) be taken on p, the chord of contact t of tangents from it to S' passes through (tu). .'. p and (tu) are polar and pole with respect to S\ 230 RECIPROCATION Cor. 1. Conjugate jxiints of S reciprocate into conjugate lines of S' an(i vice versa. Cor. 2. A self-conjugate triangle of S will reciprocate into a self-conjugate triangle of S'. 232. We will now set forth some reciprocal theorems in lei columns. If a conic be circnmscribed to a triangle (a three-point figure), tho intersections of the sides of the triangle with the tangents at the opposite vertices are collinear. 1. If a conic be inscribed in a triangle (i.e. a three-side figure), the joining lines of the vertices of the triangle and the points of contact of the conic with the opposite sides are concurrent. (acj (ab) 2. The six points of inter- section with the sides of a triangle of the lines joining the opposite vertices to two fixed points lie on a conic. 3. The three points of inter- section of the opposite sides of each of the six-side figures formed by joining six points on a conic are collinear. — PascaPs theorem. 4. If a conic circumscribe a quadrangle, the triangle formed by its diagonal points is self- conjugate for the conic. The six lines joining the vertices of a tx'iangle to the points of intersection of the opposite sides and two fixed lines envelope a conic. The three lines joining the opposite vertices of each of the six-point figures formed by the intersections of lines touching a conic are concurrent. — Brianchon's theorern. If a conic be inscribed in a quadrilateral, the triangle formed by its diagonals is self-conjugate for the conic. RECIPROCATION 231 233. Prop. The conic S' is an ellipse, parabola or hyper- bola, according as the centre of F is within, on, or tvithout S. For the centre of F reciprocates into the line at infinity, and lines through the centre of F into points on the line at infinity. Hence tangents to S from the centre of F will reciprocate into points at infinity on S', and the points of contact of these tangents to S will reciprocate into the asymptotes of S'. Hence if the centre of F be outside S, S' has two real asymptotes and therefore is a hyperbola. If the centre of F be on S, S' has one asymptote, viz. the line at infinity, that is, <S' is a parabola. If the centre of F be within *S', S' has no real asymptote and is therefore an el 234. Case where F is a circle. If the auxiliary ov base conic F be a circle (in which case we will denote it by C and its centre by 0) a further relation exists between the two figures F and F' which does not otherwise obtain. The polar of a point P with respect to G being perpendicular to OP, we see that all the lines of the figure F or F' are perpendicular to the lines joining to the corresponding points of the figure F' or F. And thus the angle between any two lines in the one figure is equal to the angle subtended at by the line joining the corresponding points in the other. In particular it may be noticed that if the tangents from to S are at right angles, then S' is a rectangular hyperbola. For if OP and OQ are the tangents to S, the asymptotes of S' are the polars of P and Q with respect to C, and these are at right angles since POQ is a right angle. If then a parabola be reciprocated with respect to a circle whose centre is on the directrix, or a central conic be recipro- cated with respect to a circle with its centre on the director circle, a rectangular hyperbola is always obtained. Further let it be noticed that a triangle whose orthocentre 232 RECIPROCATION is at will reciprocate into another triangle also having its orthocentre at 0. This the student can easily verify for himself. 235. It can now be seen that the two following propositions are connected by reciprocation : 1. The orthocentre of a triangle circumscribing a parabola 'lies on the directrix. 2. The orthocentre of a triangle inscribed in a rectangular hyperbola lies on the curve. These two propositions have been proved independently (i 95, 130). Let us now see how the second can be derived from the first by reciprocation. Let the truth of (1) be assumed. Reciprocate with respect to a circle G having its centre at the orthocentre of the triangle. Now the parabola touches the line at infinity, therefore the pole of the line at infinity with respect to C, viz. 0, lies on the reciprocal curve. And the reciprocal curve is a rectangular hyperbola because is on the directrix of the parabola. Further is also the orthocentre of the reciprocal of the triangle circumscribing the parabola. Thus we see that if a rectangular hyperbola be circum- scribed to a triangle, the orthocentre lies on the curve. It is also clear that no conies but rectangnlar hyperbolas can through the vertices of a triangle and its orthocentre. 236. Prop. If 8 be a circle and we reciprocate luith respect to a circle G luhose centre is 0, S' ivill be a conic having for a focus. Let A be the centre of S. Let p) be any tangent to S, Q its point of contact. Let P be the pole of p, and a the polar of A with respect to G. Draw P3I perpendicular to a. RECIPROCATIOX 233 Then since AQ is perpendicular to jh "e have by Salmon's theorem (§17) qp_PM 0A~ AQ- • • TTiT= the constant ^rr, • FM A Q Thus the locus of P which is a point on the reciprocal curve is a conic whose focus is 0, and corresponding directrix the polar of the centre of S. Since the eccentricity of S' OA AQ that >S' is an ellipse, parabola, or hyperbola according as is within, on, or without >S'. This is in agreement with § 233. Cor. The polar reciprocal of a coilic with respect to a circle having its centre at a focus of the conic is a circle, whose centre is the reciprocal of the corresponding directrix. 237. Let us now reciprocate with respect to a circle the theorem that the angle in a semicircle is a right angle. Let A be the centre of S, KL any diameter, Q any point on the circumference. In the reciprocal figure we have corresponding to A the directrix a, and a point {kl) on it corresponds to {KL). 234 RECIPROCATION /,; and I are tangents from {kl) to >S" which correspond- to K and L, and q is tlie tangent to *S'' corresponding to Q. Now {QK) and \QL) are at right angles. Therefore the line joining {qk) and {ql) subtends a right angle at the focus of /S". Hence the reciprocal theorem is that the intercept on anij tangent to a conic between tiuo tangents which intersect in the directrix subtends a right angle at the focus. 238. Prop. A system, of non-intersecting coaxal circles can be reciprocated into confocal conies. Let L and L' be the limiting points of the system of circles. Reciprocate with respect to a circle C whose centre is at L. Then all the circles will reciprocate into conies having L for one focus. Moreover the centre of the reciprocal conic of any one of the circles is the reciprocal of the polar of L with respect to that circle. RECIPROCATION 235 But the polar of L for all the circles is the same, viz, the line through L' perpendicular to the line of centres (§ 22). Therefore all the reciprocal conies have a common centre as well as a common focus. Therefore they all have a second common focus, that is, they are confocal. 239. We know that if ^ be; a common tangent to two circles of the coaxal system touching them at P and Q, PQ subtends a right angle at L. Now reciprocate this with regard to a circle with its centre L. The two circles of the system reciprocate into confocal conies, the common tangent t reciprocates into a common point of the confocals, and the points P and Q into the tangents to the confocals at the common point. Hence confocal conies cut at right angles. This fact is of course known and easily proved otherwise. We are here merely illustrating the principles of reciprocation. .240. Again it is known (see Ex. 40 of Chap. XIII) that if Si and S.2 be two circles, L one of the limiting points, and P and Q points on S^ and ^2 respectively such that PLQ is a right angle, the envelope of PQ is a conic having a focus at L. Now reciprocate this property with respect to a circle having its centre at L. >S\ and S., reciprocate into confocals with 236 RECIPROCATION L as one focus ; the points F and Q reciprocate.' into tangents to /S/ and SJ, viz. p and q, which will be at right angles ; and the line {PQ) reciprocates into the point (pq). As the envelope of {PQ) is a conic with a focus at L, it follows that the locus of {jiq) is a circle. Hence we have the theorem : If tivo tangents from a point T, one to each of two confocals, he at right angles, the locus of T is a circle. This also is a well-known property of confocals. 241. We will conclude this chapter by proving two theorems, the one having reference to two triangles which are self-conjugate for a conic, the other to two triangles reciprocal for a conic. Prop. //' tivo triangles be self-conjugate to the same conic their six vertices lie on a conic and their six sides touch a conic. Let ABC, A'B'C be the two triangles self-conjugate with respect to a conic S. Project 8 into a circle with A projected into the centre ; then (using small letters for the projections) ah, ac are conjugate diameters and are therefore at right angles, and h and c lie on the line at infinity. Further a'h'c' is a triangle self-conjugate for the circle. .•. a the centre of the circle is the orthocentre of this triangle. RECIPROCATION '237 Let a conic be placed through the five points a, U, c, a and h. This must be a rectangular hyperbola, since as we have seen no conies but rectangular hyperbolas can pass through the vertices of a triangle and its orthocentre. .•. c also lies on the conic through the five points named above, since the line joining the two points at infinity on a rectangular hyperbola must subtend a right angle at any point. Hence the six points a, h, c, a, h', c all lie on a conic. ,•. the six points A, B, C, A' , B', C also lie on a conic. The second part of the proposition follows at once by reciprocating this which we have just proved. 242. Prop. If two triangles are reciprocal for a conic, they are in perspective. Let ABC, A'B'C be two triangles which are reciprocal for the conic S; that is to say, A is the pole of B'C, B the pole of G'A', C the pole of A'B ; and consequently also A' is the pole of BC, B' of CA, and C" of AB. Project >S' into a circle with the projection of A for its centre. .•. B' and 6" are projected to infinity. 288 RECIPROCATION Using small letters for the projection, we see that, since a' is the pole of he, aa' is perpendicular to he. Also since h' is the pole of ac, ah' is perpendicular to ac : hh' which is parallel to ah' is perpendicular to ac. Similarly cc is perpendicular to ah. .'. aa', hh', cc meet in the orthocentre of the triangle ahc. .'. A A', BB', CC are concurrent. EXERCISES 1. If tlTe conies *S' and aS" be reciprocal polars with respect to the conic V, the centre of *S" corresponds to the polar of the centre of r with respect to aS'. 2. Parallel lines reciprocate into points collinear with the centi-e of the base conic V. 3. Shew that a quadrangle can be reciprocated into a parallelogram. RECIPROCATION 239 4. Reciprocate with respect to any conic the theorem : The locus of the poles of a given line with respect to conies passing- through four fixed points is a conic. Reciprocate with respect to a circle the theorems contained in Exx. 5 — 12 inclusive. 5. The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 6. The tangent to a circle is perpendicular to the radius thi-ough the point of contact. 7. Angles in the same segment of a circle are equal. 8. The opposite angles of a quadrilateral inscribed in a circle are together equal to two right angles. 9. The angle between the tangent at any point of a circle and a chord through that point is equal to the angle in the alternate .segment of the circle. 10. The polar of a point with respect to a circle is perpen- dicular to the line joining the point to the centre of the circle. 11. The locus of the intersection of tangents to a circle which cut at a given angle is a concentric circle. 12. Chords of a circle which subtend a constant angle at the centre envelope a concentric circle. 13. Two conies having double contact will reciprocate into conies having double contact. 14. A circle .S' is reciprocated by means of a circle C into a conic *S". Prove that the radius of C is the geometric mean between the radius of IS and the semi-latus rectum of 6". 15. Prove that with a given point as focus four conies can be drawn circumscribing a given triangle, and that the sum of the latera recta of three of them will equal the latus rectum of the fourth. 1 6. Conies have a focus and a pair of tangents common ; prove that the corresponding directrices will pass through a fixed point, and all the centres lie on the same straight line. 17. Prove, by reciprocating with respect to a circle with its centre at *S', the theorem : If a triangle ABC circumscribe a parabola 240 KECIPROCATION whose focus is S, the lines through A, B, C perpendicular respectively to SA, SB, SC are concurrent. 18. Conies are described with one of their foci at a fixed point ^S", so that each of the four tangents from two fixed points subtends the same angle of given magnitude at S. Prove that the directrices corresponding to the focus «S' pass through a fixed point. 19. If be any point on the common tangent to two parabolas with a common focus, prove that the angle between the other tangents from to the parabolas is equal to the angle between the axes of the parabolas. 20. A conic circumscribes the triangle ABC, and has one focus at 0, the orthocentre ; shew that the corresponding directi-ix is perpen- dicular to 10 and meets it in a point -V such that 10 . 0X= AO . OD, where / is the centre of the inscribed circle of the triangle, and D is the foot of the perpendicular from A on BC. Shew also how to find the centre of the conic. 21. Prove that chords of a conic which subtend a constant angle at a given point on the conic will envelope a conic. [Reciprocate into a parabola by means of a circle having its centre at the fixed point.] 22. If a triangle be i-eciprocated with respect to a circle having its centre on the circumcircle of the triangle, the point will also lie on the circumcircle of the reciprocal triangle. 23. Prove tlie following and obtain from it by reciprocation a theorem applicable to coaxal circles : If from any point pairs of tangents p, p ; q, q, be drawn to two confocals S-^ and S.;,, the angle between p and q is equal to the angle between p and q. 24. Prove and reciprocate with respect to any conic the following : If ABC be a triangle, and if the polars oi A, B, C with respect to anj'^ conic meet the opposite sides in P, Q, R, then P, Q, R are collinear. 25. A fixed point in the plane of a given circle is joined to the extremities A and B of any diameter, and OA, OB meet the circle again in P and Q. Shew that the tangents at P and Q intersect on a fixed line parallel to the polar of 0. 26. All conies through four fixed points can be projected into rectangular hyperbolas. RECIPROCATION 241 27. If two triangles be reciprocal for a conic (§ 242) their centre of perspective is the pole of the axis of perspective with regard to the conic. 28. Prove that the envelope of chords of an ellipse which subtend a right angle at the centre is a concentric circle. [Recipi'ocate with respect to a circle having its centre at the centre of the ellipse.] 29. ABC is a triangle, /its incentre ; A^, L\, C\ the points of contact of the incircle with the sides. Prove that the line joining / to the point of concurrency of AA^, BBy, CCi is perpendicular to the line of collinearity of the intersections of BiC\, BC ; C^A^, CA ; A^By, AB. [Use Ex. 27.] 16 242 CHAPTER XVIII CIRCULAR POINTS. FOCI OF CONICS 243. We have seen that pairs of concurrent lines which are conjugate for a conic form an involution, of which the tangents from the point of concurrency are the double lines. Thus conjugate diameters of a conic are in involution, and the double lines of the involution are its asymptotes. Now the conjugate diameters of a circle are orthogonal. Thus the asymptotes of a circle are the imaginary double lines of the orthogonal involution at its centre. But clearly the double lines of the orthogonal involution at one point must be parallel to the double lines of the orthogonal involution at another, seeing that we may hy a motion of trans- lation, without rotation, move one into the position of the other. Hence the asymptotes of one circle are, each to each, parallel to the asymptotes of any other circle in its plane. Let a, h be the asymptotes of one circle G, a, h' of another C, then a, a being parallel meet on the line at infinity, and h, h' being parallel meet on the line at infinity. But a and a meet C and C" on the line at infinity, and h and h' „ C and C „ „ „ Therefore Q and C go through the same two imaginary points on the line at infinity. Our conclusion then is that all circles in a plane go through the same two imaginary points on the lirte at infinity. These two points are called the circular points at infinity or, simply, the circular points. CIRCULAR POINTS. FOCI OF CONICS 243 . The circular lines at any point are the lines joining that point to the circular points at infinity ; and they are the imagi- nary double lines of the orthogonal involution at that point. 244. Analytical point of view. It may help the student to think of the circular lines at any point if we digress for a moment to touch upon the Analytical aspect of them. The equation of a circle refeired to its centre is of the form X- -f y- = a-. The asymptotes of this circle are X- + ?/2 = 0, that is the pair of imaginary lines y = ix and y = — ix. These two lines are the circular lines at the centre of the circle. The points where they meet the line at infinity are the circular points. If we rotate the axes of coordinates at the centre of a circle through any angle, keeping them still rectangular, the equation of the circle does not alter in form, so that the asymptotes will make angles tan~'(i) and tan~'(— i) with the new axis of x as well as with the old. This at first sight is paradoxical. But the paradox is ex- plained by the fact that the line y = ix makes the same angle tan~^ {%) with every line in the plane. For let y = mx be any other line through the origin. Then the angle that y = ix makes with this, measured in the positive sense from y —- mx, is tan~' :; ;- = tan ^ { -, -. — - \ = tan ^ i. W + imj \ 1 +tm ) 245. Prop. // AOB be an angle of constant magnitude and n, n' he the circular points, the cross-ratios of the jJencil (fl, n', A, B) are constant. 16-2 244 CIRCULAR POINTS. FOCI OF CONICS ■n ^ ,r^r^> 4 i,s sin HOCl' sin AOB For (nQ.'AB) = -. — „^., . — tvy7v^> ^ sin nOB sni ^ Of i but the angles nOfl', flOB, AOi^' are all constant since the circular lines make the same angle with every line in the plane, and Z AOB is constant by hypothesis. .-. 0(nn'.4i?) is constant. 246. Prop. All cunics passing throuc/h the circular points (fre circles. Let C be the centre of a conic >S' passing through the circular points, which we will denote by O and Q'. Then CO., CD.' are the asymptotes of >S'. But the asymptotes are the double lines of the involution formed by pairs of conjugate diameters. And the double lines completely determine an involution, that is to say there can be only one involuti<)n with the same double lines. Thus the conjugate diameters of S are all orthogonal. . Hence ^ is a circle. The circular points may he utilised for establishing properties of conies passing through two or more fixed points. For a system of conies all passing through the same twO' points can be projected into circles simultaneously. This is effected by projecting the two points into the circular points on the plane of projection. The projections of the conies will now go through the circular points in the new plane and so they are all circles. The student of course understands that such a projection is an imaginary one. 247. We will now proceed to an illustration of the use of the circular points. It can be seen at once that any transversal is cut by a system of coaxal circles in pairs of points in involution (the centre of this involution being the point of intersection of the line with the axis of the system). CIRCULAR POINTS. FOCI OF CONICS 245 From this follows at once Desargues' theorem (§ 227), namely that conies through four points cut any transversal in pairs of points in involution. For if we project two of the points into the circular points the conies all become circles. Moreover the circles form a coaxal system, for they have two other points in common. Hence Desargues' theorem is seen to follow from the involu- tion property of coaxal circles. The involution property of coaxal circles again is a particular case of Desargues' theorem, for coaxal circles have four points in common, two being the circular points, and two the points in which all the circles are cut by the axis of the system. 248. We will now make use of the circular points to prove the theorem: If a triangle he self -con jugate to a rectangular hyper- bola its circumcircle passes through the centre of the hyperbola. Let be the centre of the rectangular hyperbola, ABC the self-conjugate triangle, O, fl' the circular points. Now observe first that OHO' is a self-conjugate triangle for the rectangular hyperbola. For OD., Ofl' are the double lines of the orthogonal involution at to which the asymptotes, being at right angles, belong. Therefore OH, OH' belong to the involution whose double lines are the asymptotes (§ 82), that is the involution formed by pairs of conjugate lines through 0. .•. OD., on' are conjugate lines, and is the pole of ClQ' which is the line at infinity. ,•. OnO' is a self-conjugate triangle. Also ABC is a self-conjugate triangle. .-. the six points A, B, C, 0, ft, D,' all lie on a conic (§ 241) - and this conic must be a circle as it passes through ft and ft'. .•. A, B, C, are concyclic. Cor. If a rectangular hyperbola circumscribe a triangle, its centre lies on the nine-points circle. This well-known theorem is a particular case of the above proposition, for the pedal triangle is self-conjugate for the rectangular hyperbola. (Ex. 21, Chapter XIV.) 246 CIRCULAR POINTS. FOCI OF CONICS 249. Prop. Concentric circles have double contact at infinity. For if be the centre of the circles, D,, Vl' the circular points at infinity, all the circles touch Ofl and Ofl' at the points n and O'. That is, all the circles touch one another at the points f) and n'. 250. Foci of Conies. Prop. Every conic has four foci, tiuo of which lie on one axis of the conic and are real, and two on the other axis and are imaginary. Since conjugate lines at a focus form an orthogonal involution, and since the tangents from any point are the double lines of the involution formed by the conjugate lines there, it follows that the circular lines through a focus are the tangents to the conic from that point. But the circular lines at any point go through il and H' the circular points. Thus the foci of the conic will be obtained by drawing tangents from H and iV to the conic, and taking their four points of intersection. Hence there are four foci. To help the imagination, construct a figure as if H and H' were real points. Draw tangents from these points to the conic and let S, 8', F, F' be their points of intersection as in the figure ; >S, S' being opposite vertices as also F and F'. Let FF' and SS' intersect in 0. Now the triangle formed by the diagonals FF', SS' and HO' is self-conjugate for the conic, because it touches the sides of the quadrilateral. (Reciprocal of § 119 a.) .•. is the pole of HO', i.e. of the line at infinity. .•. is the centre of the conic. CIRCULAR POINTS. FOCI OF CONICS 247 Further OflCl' is the diagonal, or harmonic, triangle of the quadra^^Ze SS'FF'. .-. 0(nn',FS)==-h (§76) .'. OF and OS are conjugate lines in the inA^olution of which Ofl and on' are the double lines. .'. OF and OS are at right angles. And OF and OS are conjugate lines for the conic since the triangle formed by the diagonals FF', SS', On' is self-conjugate for the conic ; and is, as we have seen, the centre. .•. OF and OS, being orthogonal conjugate diameters, are the axes. Thus we have two pairs of foci, one on one axis and the other on the other axis. Now we know that two of the foci, say S and >S", are real. It follows that the other two, F and F', are imaginary. For if F were real, the line FS would meet the line at infinity in a real point, which is not the case. .•. F and F' must be imaginary. Cor. The lines joining non-corresponding foci are tangents to the conic and the points of contact of these tangents are coney clic. 248 CIRCULAR POINTS. FOCI OF CONICS 251. Prop. A system of conies touching the sides of a quadrilateral can he projected into confocal conies. Let ABCD be the quadrilateral, the pairs of opposite vertices being A, C; B,D; E,F. Project E and F into the circular points at infinity on the plane of projection. .'. A, C and B, D project into the foci of the conies in the projection, by § 250. Cor. Confocal conies form a system of conies touching four lines. 252. We will now make use of the notions of this chapter to prove the following theorem, which is not unimportant. If the sides of two triangles all touch the same conic, the six vertices of the triangles all lie on a conic. Let ABC, A'B'C be the two triangles the sides of which all touch the same conic *S^. Denote the circular points on the tt plane or plane of pro- jection by w, w. Project B and C into twand w'; .-. S projects into a parabola, since the projection of »S^ touches the line at infinity. Further A will project into the focus of the parabola, since the tangents from the focus go through the circular points. CIRCULAR POINTS. FOCI OF CONICS 249 Using corresponding small letters in the projection, we see that, since the circumcircle of a triangle whose sides touch a parabola goes through the focus, a, a , b', c' are concyclic. .•. a, a, h', c', 0), co' lie on a circle. .-. A, B, C, A', B', C lie on a conic. The converse of the above proposition follows at once by reciprocation. 253. We have in the preceding article obtained a proof of the general proposition that if the sides of two triangles touch a conic, their six vertices lie on another conic by the projection of what is a particular case of this proposition, viz. that the circumcircle of a triangle whose sides touch a parabola passes through the focus. This process is known as generalisiiig by projection. We will proceed to give further illustrations of it. 254. Let us denote the circular points in the p plane by n, n', and their projections on the tt plane by w, co'. Then of course co and to' are not the circular points in the tt plane. But by a proper choice of the ir plane and the vertex of projection 0) and Q)' may be any two points we choose, real or imaginary. For if we wish to project H and H' into the points a) and &>' in space, we have only to take as our vertex of projection the point of intersection of the lines wD. and co'il', and as the plane TT some plane passing through o) and to'. 255. The following are the principal properties connecting figures in the ji and tt planes when Ci. and fl' are projected into &> and co' : 1. Circles in the p plane project into conies through the points o) and co' in the tt plane. 2. Parabolas in the p plane project into conies touching the line ww' in the tt plane. 3. Rectangular hyperbolas in the p plane, for which, as we have seen, H and ft' are conjugate points, project into conies having co and co' for conjugate points. 250 CIRCULAR POINTS. FOCI OF CONICS 4. The centre of a conic in the p plane, since it is the pole of fin', projects into the pole of the line cow'. 5. Concentric circles in the p plane project into conies having double contact at w and a>' in the tt plane. 6. A pair of lines OA, OB at right angles in the p plane project into a pair of lines oa, oh harmonically conjugate with Oft), Ota'. This follows from the fact that OH, OH' are the double lines of the involution to which OA, OB belong, and therefore 0{AB, 120') = -! (§ 82); from which it follows that o{ah, &)&)') = — 1. 7. A conic with ^ as focus in the p plane will project into a conic touching the lines sw, sw' in the tt plane. And the two foci S and *S^' of a conic in the j9 plane will project into the vertices of the quadrilateral formed by drawing tangents from ft) and &>' to the projection of the conic in the tt plane. 256. It is of importance that the student should realise that ft) and tu' are not the circular points in the tt plane when they are the projections of O and Xl'. In § 252 we have denoted the circular points in the tt plane by &) and w', but they are not there the projections of the circular points in the p plane. Our practice has been to use small letters to represent the projections of the corresponding capitals. So then we use to and wl for the projection of 12 and fl' respectively. If H and H' are the circular points in the p plane, <w and w are not the circular points in the ir plane ; and if w and co' are the circular points in the tt plane, ft and H' are not the circular points in the p plane. That is to say, only one of the pairs can be circular points at the same time. 257. We will now proceed to some examples of general- isation by projection. Consider the theorem that the radius of a circle to any point A is perpendicular to the tangent at A. CIRCULAR POINTS. FOCI OF CONICS 251 Project the circle into a conic through to and w : the centre C of the circle projects into the pole of &)&>'. The generalised theorem is that if the tangents at two points CO, co' of a conic meet in c, and a be any point on the conic and at the tangent there a{tc, oi(o') = —1. 258. Next consider the theorem that angles in the same segment of a circle are equal. Let AQB be an angle in the segment of which AB is the base. Project the circle into a conic through w and to' and we get the theorem that if q be any point on a fixed conic through the four points a, b, (o, w , q (abuxo) is constant (§ 245). Thus the property of the equality of angles in the same seg- ment of a circle generalises into the constant cross-ratio property of conies. 262 CIRCULAR POINTS. FOCI OF CONICS 259. Again we have the property of the rectangular hyper- bola that if PQR be a triangle inscribed in it and having a right angle at P, the tangent at P is at right angles to QR. Project the rectangular hyperbola into a conic having eo and a>' for conjugate 'points and we get the following property. If p be any iwint on a conic for which w and co are con- jugate j)oints and q, r tivo other jmints on the conic such that ]) {qr, coco') = — 1 and if the tangent at p meet qr in k then k (pq, coco') = — 1. 260. Lastly we will generalise by projection the theorem that chords of a circle which touch a concentric circle subtend a constant angle at the centre. CIRCULAR POINTS. FOCI OF CONICS 253 Let PQ be a chord of the outer circle touching the inner and subtending a constant angle at G the centre. The concentric circles have double contact at the circular points fl and H' and so project into two conies having double contact at w and &)'. The centre C is the pole of Hfl' and so c, the projection of C, is the pole of woi'. The property we obtain by projection is then : If tiuo conies have double contact at two jJoints co aiid lo' and if the tangents at these points meet in c, and if pq be any chord of the outer conic touchinfj the inner conic, then c (pqcoo)') is constant. EXERCISES 1. If be tlie centre of a conic, 12, 12' the circular points at infinity, and if Dim' be a self-conjugate triangle for the conic, the conic must be a rectangular hyperbola. 2. If a variable conic pass through two given points F and F', and touch two given straight lines, shew that the chord which joins the points of contact of these two straight lines will always meet FF' in a fixed point. 3. If three conies have two points in common, the opposite common chords of the conies taken in pairs are concurrent. 4. Two conies **>', and X, circumscribe the quadrangle ABCD. Through A and B lines AEF, BGH&.VQ drawn cutting <S'._> in ^and G, and S^ in F and //. Prove that CD, EG, Fll are concurrent. 5. If a conic pass through two given points, and touch a given conic at a given point, its chord of intersection with the given conic passes through a fixed point. 6. If f2, 12' be the circular points at infinity, the two imaginary foci of a parabola coincide with 12 and 12', and the centre and second real focus of the parabola coincide with the point of contact of 1212' with the parabola. 254 CIRCULAR POINTS. FOCI OF CONICS 7. If a conic be drawn through the four points of intersection of two given conies, and through tlie intersection of one pair of common tangents, it also passes through the intersection of the other pair of common tangents. 8. Prove that, if three conies pass through the same four points, a common tangent to any two of the conies is cut harmonically by the third. 9. Reciprocate the theorem of Ex. 8. 10. If from two points P, P' tangents be drawn to a conic, the four points of contact of the tangents with the conic, and the points P and P' all lie on a conic. [Project P and P' into the circular points.] 11. If out of four pairs of points every combination of three pairs gives six points on a conic, either the four conies thus deter- mined coincide or the four lines determined by the four pairs of points are concurrent. 12. Generalise by projection the theorem that the locus of the centre of a rectangular hyperbola circumscribing a triangle is the nine-points circle of the triangle. 13. Generalise by projection the theorem that the locus of the centre of a rectangular hyperbola with respect to which a given triangle is self-conjugate is the circumcircle. 14. Given that two lines at right angles and the lines to the circular points form a harmonic pencil, find the reciprocals of the circular points with regard to any circle. Deduce that the polar reciprocal of any circle with regard to any point has the lines from to the circular points as tangents, and the reciprocal of the centre of the circle for the corresponding chord of contact. 15. Prove and generalise by projection the following theorem : The centre of the circle circumscribing a triangle which is self- conjugate with regard to a parabola lies on the directrix. 16. P and P' are two points in the plane of a triangle ABC. D is taken in BC such that BC and BA are harmonically conjugate with DP and DP' ; E and F are similarly taken in CA and AB respectively. Prove that AD, BE, OF are concurrent. 17. Generalise by projection the following theorem : The lines perpendicular to the sides of a triangle through the middle points of the sides are concurrent in the circumcentre of the triangle. CIRCULAR POINTS. FOCI OF CONICS 255 IS. Generalise : The feet of the perpendiculars on to the sides of a triangle from any point on the circumcircle are collinear. 19. If two conies have double contact at A and B, and if PQ a chord of one of them touch the other in Ji and meet AB in T, then {PQ, RT)^-\. 20. Generalise by projection the theorem that confocal conies cut at right angles. 21. Prove and generalise that the envelope of the polar of a given point for a system of eonfocals is a parabola touching the axes of the eonfocals and having the given point on its directrix. 22. If a system of conies pass through the four points A, B, C, D, the poles of the line AB with respect to them will lie on a line /. Moreover if this line I meet CD in P, PA and PB are hai-monic conjugates of CD and I. 23. A pair of tangents from a fixed point T to a conic meet a third fixed tangent to the conic in L and L'. P is any point on the conic, and on the tangent at P a point X is taken such tiiat X{PT, LL') = — 1 ; prove that the locus of A' is a straight line. 24. Defining a focus of a conic as a i)oint at which each pair of conjugate lines is orthogonal, prove that the polar reciprocal of a circle with respect to another circle is a conic having the centre of the second circle for a focus. 256 CHAPTER XIX INVERSION 261. We have already in § 13 explained what is meant by- two ' inverse points ' with respect to a circle. being the centre of a circle, P and P' are inverse points if they lie on the same radius and OP. OP' = the square of the radius. P and P' are on the same side of the centre, unless the circle have an imaginary radius, = i/c, where k is real. As P describes a curve >S', the point P' will describe another curve S'. S and 8' are called inverse curves. is called the centre of inversion, and the radius of the circle is called the radius of inversion. If P describe a curve in space, not necessaril}^ a plane curve, then we must consider P' as the inverse of P with respect to a sphere round 0. That is, whether P be confined to a plane or not, if be a fixed point in space and P' be taken on OP such that OP. OP' = A constant k", P' is called the inverse of P, and the curve or surface described by P is called the inverse of that described by P' , and vice versa. It is convenient sometimes to speak of a point P' as inverse to another point P with respect to a j)oint 0. *By this is meant that is the centre of the circle or sphere with respect to which the points are inverse. 262. Prop. 'Phe inverse of a circle with respect to a point in its jilane is a circle or straight line. First let 0, the centre of inversion, lie on the circle. Let k be the radius of inversion. INVERSION Draw the diameter OA, let A' be the inverse of A. Let P be any point on the circle, P' its inverse. Then OP. OP' = Ic'=OA. 0A\ .-. PAA'P' is cyclic. 257 ,*. the angle AA'P' is the supplement of APP', which is a right angle. ,•, J. 'P' is at right angles to A A'. .•. the locus of P' is a straight line perpendicular to the diameter OA, and passing through the inverse of A. Next let not be on the circumference of the circle. Let P be any point on the circle, P' its inverse. Let OP cut the circle again in Q. Let A be the centre of the circle. Then OP. OP' = ^'•^ and OP. OQ = sq, of tangent from to the circle = t" (say). qp;^^ • • OQ ~ t'' OB k' Take B on OA such that and BP' is parallel to AQ. OA f 5 is a fixed point 17 258 INVERSION , ^ BP' OB h^ ^ ^ And .vY = 7=nr = t, > ^^ constant. AQ OA t- .'. P' describes a circle round B. Thus the inverse of the circle is another circle. Cor. 1. The inverse of a straight line is a circle passing through the centre of inversion. Cor. 2. If two circles be inverse each to the other, the centre of inversion is a centre of similitude (§ 25) ; and the radii of the circles are to one another in the ratio of the distances of their centres from 0. The student should observe that, if we call the two circles 8 and S', and if OPQ meet S' again in Q', Q' will be the inverse of Q. Note. The part of the circle *.S which is convex to corresponds to the part of the circle S' which is concave to 0, and vice versa. Two of the common tangents of S and S' go through 0, and the points of contact with the circles of each of these tangents will be inverse points. 263. Prop. The inverse of a sphere with respect to any point is a sphere or a plane. This proposition follows at once from the last by rotating the figures round OA as axis ; in the first figure the circle and line will generate a sphere and plane each of which is the inverse of the other; and in the second figure the two circles will generate spheres each of which will be the inverse of the other. INVERSION 259 264. Prop. The inverse of a circle luith respect to a point 0, not in its plane, is a ciixle. For the circle may be regarded as the intersection of two spheres, neither of which need pass through 0. These spheres will invert into spheres, and their intersection, which is the inverse of the intersection of the other two spheres, that is of the original circle, will be a circle. 265. Prop. A circle will invert into itself with respect to a point in its plane if the radius of inversion he the length of the tangent to the circle from the centre of inversion. This is obvious at once, for if OjT be the tangent from and OPQ cut the circle in P and Q, since OP . OQ = OT- it follows that P and Q are inverse points. That is, the part of the circle concave to inverts into the part which is convex and vice versa. Cor. 1. Any system of coaxal circles can be simultaneously inverted into themselves if the centre of inversion be any point on the axis of the system. Cor. 2. Any three coplanar circles can be simultaneously inverted into themselves. For we have only to take the radical centre of the three circles as the centre of inversion, and the tangent from it as the radius. 17—2 260 INVERSION 266. Prop. Ttuo coplanar curves cut at the same angle as their inverses with respect to any point in their plane. Let P and Q be two near points on a curve 8, P' and Q' their inverses with respect to 0. Then since OP. OP' = k' = OQ . OQ'. .'. QPP'Q' is cyclic. .-. ^OPQ = zOQ'P'. Now let Q move up to P so that PQ becomes the tangent to S at P; then Q' moves up at the same time to P' and P'Q^ becomes the tangent at P' to the inverse curve S'. .-. the tangents at P and P' make equal angles with OPP'. The tangents however are antiparallel, not parallel. Now. if we have two curves »S\ and *S^2 intersecting at P, and PTi, PTo be their tangents there, and if the inverse curves be INVERSION 261 /S/, So intersecting at P', the inverse of P, and PT/, P'T^' be their tangents, it follows at once from the above reasoning that ^t^pt, = at,'P't:. Thus >S'i and S^ intersect at the same angle as their inverses. Cor. If two curves touch at a point P their inverses touch at the inverse of P. 267. Prop. If a circle 8 he inverted into a circle S', and P, Q be inverse points luith respect to 8, then P' and Q', the inverses of P and Q ^respectively, ivill be inverse points luith respect to 8'. Let be the centre of inversion. Since P and Q are inverse points for 8, therefore 8 cuts orthogonally every circle through P and Q, and in particular the circle through 0, P, Q. Therefore the inverse of the circle OPQ will cut 8' orthogonally. But the inverse of the circle OPQ is a line ; since 0, the centre of inversion, lies on the circumference. Therefore P'Q' is the inverse of the circle OPQ. Therefore P'Q' cuts 8' orthogonally, that is, passes through the centre of 8'. Again, since every circle through P and Q cuts 8 orthogonally, it follows that every circle through P' and Q' cuts 8' orthogonally (§ 266). 262 INVERSION Therefore, if ^j be the centre of S', A^P'. J.iQ' = square of radius of S'. Hence P' and Q' are inverse points for the circle S'. 268. Prop. A system of non-intersecting coaxal circles can be inverted into concentric circles. The system being non-intersecting, the limiting points L and L' are real. Invert the system with respect to L. Now L and L' being inverse points with respect to each circle of the system, their inverses will be inverse points for each circle in the inversion. But L being the centre of the circle of inversion, its inverse is at infinity. Therefore L' must invert into the centre of each of the circles. 269. Feuerbach's Theorem. ' The principles of inversion may be illustrated by their application to prove Feuerbach's famous theorem, viz. that the nine-points circle of a triangle touches the inscribed and the three escribed circles. Let ABC be a triangle, I its incentre and /j its ecentre opposite to A. Let M and ilfj be the points of contact of the incircle and this ecircle with BC. Let the line AII^ which bisects the angle A cut BG in R. Draw AL perpendicular to BO. Let 0, P, U be the circum- centre, orthocentre and nine-points centre respectively. Draw OD perpendicular to BC and let it meet the circum- circle in K. Now since BI and BI^ are the internal and external bisectors of angle B, .'. (AR,II,) = -1, .-. L(AR, II,) = -1. .'. since RLA is a right angle, LI and X/j are equally inclined to BC (§ 27, Cor. 2). INVERSION 263 .-. the polars of L with regard to the incircle and the ecircle will be equally inclined to BC. Now the polar of L for the incircle goes through M and that for the ecircle through M^. Let MX be the polar of L for the incircle cutting OD in X. Then since D is the middle point of MM, (§ 12, Cor.) AXM,D = AXMD. .-. aXM,D = zXMD. .". MiX is the polar of L for the ecircle, i.e. L and X are conjugate points for both circles. Let N be the middle point of XL, then the square of the tangent from N to both circles = NX^ = ND^. 264 INVERSION .-. iV is on the radical axis of the two circles ; but so also is i) since DM = DM,. .'. ND is the radical axis, and this is perpendicular to J/j. Now the pedal line of K goes through D, and clearly also, since K is on the bisector of the angle A, the pedal line must be perpendicular to AK. .\ DN is the pedal line of K. But the pedal line of K bisects KP. .'. KNP is a straight line and N its middle point. And since U is the middle point of OP, UN=\OK. .'. JV is a point on the nine-points circle. Now invert the nine-points circle, the incircle and ecircle with respect to the circle whose centre is N and radius ND or NL. The two latter circles will invert into themselves ; and the nine-points circle will invert into the line BG ; for N being on the nine-points circle the inverse of that circle must be a line, and D and L, points on the circle, invert into themselves, .*. DL is the inverse of the nine-points circle. But this line touches both the incircle and ecircle. .*. the nine-points circle touches both the incircle and ecircle. Similarly it touches the other two ecircles. CoK. The point of contact of the nine-points circle with the incircle will be the inverse of M, and with the ecircle the inverse of Mi. EXERCISES 1. Prove that a system of intersecting coaxal circles can be inverted into concurrent straight lines. 2. A sphere is inverted from a point on its surface ; shew that to a system of meridians and parallels on the surface will correspond two systems of coaxal circles in the inverse figure. [See Ex. 16 of Chap. II.] INVERSION 265 3. If A, B, C, D be four collinear points, and A', B', C, D' the four points inverse to them, then AC.BD A'C .B'D' AB.CD~ A'B' . CD'' 4. If P be a point in the plane of a system of coaxal circles, and i*i, P^, Pg jfec. be its inverses with respect to the different circles of the system, Pi, P.-,, Pj &c. are concyclic. 5. If P be a fixed point in the plane of a system of coaxal circles, P' the inverse of P with respect to a circle of the system, P" the inverse of P' with respect to another circle, P'" of P" with respect to another and so on, then P', P", P'" &c. are concyclic. 6. POP', QOQ' are two chords of a circle and is a fixed point. Prove that the locus of the other intersection of the circles POQ, P'OQ' is a second fixed circle. 7. Shew that the result of inverting at any odd number of circles of a coaxal system is equivalent to a single inversion at one circle of the system ; and determine the circle which is so equivalent to three given ones in a given order. 8. Shew that if the circles inverse to two given circles ACD, BCD with respect to a given point P be equal, the circle PCD bisects (internally and externally) the angles of intersection of the two given circles. 9. Three circles cut one another orthogonally at the three pairs of points AA\ BB', CC ; prove that the circles through ABC, AB'C touch at A. 10. Prove that if the nine-points circle and one of the angular points of a triangle be given, tiie locus of the orthocentre is a circle. 1 1 . Prove that the nine-points circle of a triangle touches the inscribed and escribed circles of the three triangles formed by joining the orthocentre to the vertices of the triangle. 12. The figures inverse to a given figure with regard to two circles C'l and Cn are denoted by ^S*! and S.. respectively ; shew that if C\ and Co cut orthogonally, the inverse of S^ with regard to C, is also the inverse of So with regard to Cj. 13. If A, B, C be three collinear points and any other point, shew tliat the centres P, Q, R of the three circles circumscribing the triangles OBC, OCA, OAB are concyclic with 0. Also that if three other circles are drawn through 0, A ; 0, B ; 0, C to cut the circles OBC, OCA, OAB respectively, at right angles, then these circles will meet in a point which lies on the circumcircle of the quadrilateral OPQR. 266 INVERSIOX 14. Shew that if the circle PAR cut orthogonally the circle FCD; and the circle PAC cut orthogonally the circle PBD; then the circle PAD must cut the circle PBG orthogonally. 15. Prove the following construction for obtaining the point of contact of the nine- points circle of a triangle ABC with the in circle : The bisector of the angle A meets BC in H. From H the otlier tangent HY is drawn to the incircle. The line joining the point of contact Y of this tangent and D the middle point of BC cuts the incircle again in the point required. 16. Given the circumcircle and incircle of a triangle, shew that the locus of the centroid is a circle. 17. A, B, C are three circles and a, h, c their inverses with respect to any other circle. Shew that if A and B are inverses with respect to C, then a and b are inverses with respect to c. 18. A circle *S' is inverted into a line, prove that this line is the radical axis of S and the circle of inversion. 19. Shew that the angle between a circle and its inverse is bisected by the circle of inversion. 20. The perpendiculars, AL, BM, CN to the sides of a triangle ABC meet in the orthocentre K. Prove that each of the four circles which can be described to touch the three circles about KM AN, KNBL, KLGM touches the circumcircle of the triangle ABC. [Invert the three circles into the sides of the triangle by means of centre K, and the circumcircle into the nine-points circle.] 21. Invert two spheres, one of which lies wholly within the other, into concentric spheres. 22. Examine the particular case of tlie proposition of § 151, where O the centre of inversion lies on *S'. 23. If A, P, Q be three collinear points, and if P', Q' be the inverses of P, Q with respect to 0, and if P'Q' meet OA in ^,,. then AP. AQ OA'- A.P'.A^Q' " UI} ■ 24. A circle is drawn to touch the sides AB, AG of a triangle ABC and to touch the circumcircle internally at U. Shew that AE and the line joining A to the point of contact with BC of the ecircle opposite to A are equally inclined to the bisectors of the angles between AB and AC. [Invert with A as centre so that C inverts into itself.] 267 CHAPTER XX SIMILARITY OF FIGURES 270. Homothetic Figures. If i^ be a plane figure, wliicli we may regard as an assemblage of points typified by P, and if be a fixed point in the plane, and if on each radius vector OP, produced if necessary, a point P' be taken on the same side of as P such that OP : OP' is constant (= k), then P' will determine another figure F' which is said to be similar and similarly situated to F. Two such figures are conveniently called, in one word, honiotlietic, and the point is called their homothetic centre. We see that two homothetic figures are in perspective, the centre of perspective being the homothetic centre. 271. Prop. The line joining two jyoints in the figure F is parallel to the line joining the corresponding points in the figure F' ivhich is homothetic with it, and these lines are in a constant ratio. For if P and Q be two points in F, and P', Q' the corre- sponding points in F' , since OP : OP' = OQ : OQ' it follows that PQ and P'Q' are parallel, and that PQ : P'Q' = OP : OP' the constant ratio. In the case where Q is in the line OP it is still true that PQ : P'Q' = the constant ratio, for since OP : OQ = OP' : OQ' .-. OP -.OQ - OP =0P' : OQ' - OP'. 268 SIMILARITY OF FIGURES .-. OP:PQ=OP':P'Q'. .'. PQ.P'Q'=OP:OP'. Cor. If the figures F and F' be curves *S^ and *S" the tangents to them at corresponding points P and P' will be parallel. For the tangent at P is the limiting position of the line through P and a near point Q on S, and the tangent at P' the limiting position of the line through the corresponding points P' and Q'. 272. Prop. The homothetic centre of two homothetic figures is determined by two pairs of corresponding points. For if two pairs of corresponding points P, P'; Q, Q' be given is the intersection of PP' and QQ'. Or in the case where Q is in the line PP', is determined in this line by the equation OP : OP' = PQ : P'Q'. The point is thus uniquely determined, for OP and OP' have to have the same sign, that is, have to be in the same direction. 273. Figures directly similar. If now two figures F and F' be homothetic, centre 0, and the figure F' be turned in its plane round through any angle, we shall have a new figure Fi which is similar to F but not now similarly situated. Two such figures F and F^ are said to be directly similar and is called their centre of similitude. Two directly similar figures possess the property that the SIMILARITY OF FIGURES 269 Z POPi between the lines joining to two corresponding points P and Pj is constant. Also OP : OP^ is constant, and PQ:PiQi = the same constant, and the triangles OPQ, OPjQi are similar. 274. Prop. If P, P^; Q, Q^ be two pairs of corresponding points of two figures directly similar, and if PQ, P^Q^ intersect in R, is the other intersection of the circles P-RP^, QRQj. For since Z.OPQ=zOP,Q^ 270 SIMILARITY OF FIGURES .•. Z OPR and Z OPiR are supplementary. .-. POP.R is cyclic. Similarly QiOQR is cyclic. Thus the proposition is proved. Cor. The centre of similitude of two directly similar figures is determined by two pairs of corresponding points. It has been assumed thus far that P does not coincide with Px nor with Qj. If P coincide with Pi, then this point is itself the centre of similitude. If P coincide with Qi we can draw QT and Qi?\ through Q and Qi such that z P,Q,T, = Z PQT and Q,T, : QT= P,Q, : PQ ; then T and 1\ are corresponding points in the two figures. 275. When two figures are directly similar, and the two members of each pair of corresponding points are on opposite sides of 0, and collinear with it, the figures may be called antihomothetic, and the centre of similitude is called the anti- homothetic centre. When two figures are antihomothetic the line joining any two points P and Q, of the one is parallel to the line joining the , corresponding points P' and Q' of the other ; hut PQ and P'Q' are in opposite directions. SIMILARITY OF FIGURES 271 276. Case of t^vo coplanar circles. If we divide the line joining the centres of two given circles externally at 0, and internally at 0' in the ratio of the radii, it is clear from § 25 that is the homothetic centre and 0' the antihomothetic centre for the two circles. We spoke of these points as ' centres of similitude ' before, but we now see that they are only particular centres of similitude, and it is clear that there are other centres of similitude not lying in the line of these. For taking the centre A of one circle to correspond with the centre A^ of the other, we may then take any point P of the one to correspond with any point Pi of the other. Let S be the centre of similitude for this correspondence. The triangles PSA, P^SA^ are similar, and SA : SA, = AP : A,P, = ratio of the radii. 272 SIMILARITY OF FIGURES Thus S lies on the circle on 00' as diameter (§ 27). Thus the locus of centres of similitude for two coplanar circles is the circle on the line joining the homothetic and anti- homothetic centres. This circle we have already called the circle of similitude and the student now understands the reason of the name. 277. Figures inversely similar. If ^ be a figure in a plane, a fixed point in the plane, and if another figure F' be obtained by taking points P' in the plane to correspond with the points P of ^ in such a way that OP : OP' is constant, and all the angles POP' have the same bisecting line OX, the two figures F and F' are said to be inversely similar ; is then called the centre and OX the axis of inverse similitude. Draw P'L perpendicular to the axis OX and let it meet OP in Pi. Then plainly, since OX bisects Z POP' AOLP'=AOLP„ and OP, = OP'. ,'. OPi : OP is constant. SIMILARITY OF FIGURES 273 Thus the figure formed by the points Pj will be homothetic with F. Indeed the figure F' may be regarded as formed from a figure F^ homothetic with F by turning F^ round the axis OX through two right angles. The student will have no difficulty in proving for himself that if any line OY be taken through in the plane of F and F', and if P'K be drawn perpendicular to F and produced to P.2 so that P'K = KP.2 then the figure formed with the points typified by P^ will be similar to F; but the two will not be similarly situated except in the case where OY coincides with OX. 278. If P and Q be two points in the figure F, and P', Q' the corresponding points in the figure F', inversely similar to it, we easily obtain that P'Q' : PQ = the constant ratio of OP : OP', and we see that the angle POQ = angle Q'OP' (not P'OQ'). In regard to this last point we see the distinction between figures directly similar and figures inversely similar. 279. Given two pans of corresponding points in tivo inversely similar figures, to find the centre and axis of similitude. To solve this problem we observe that if PP' cxrt the axis A. G. 18 274 SIMILARITY OF FIGURES OX in F, then PF : FP' = OP : OP' since the axis bisects the angle POP'. .-. PF:FP'=PQ:P'Q'. Hence if P, P' ; Q, Q' be given, join PP' and QQ' and divide these lines at F and G in the ratio PQ : P'Q', then the line FG is the axis. Take the point Pj symmetrical with P on the other side of the axis, then is determined by the intersection of P'Pj with the axis. Note. The student who wishes for a fuller discussion on the subject of similar figures than seems necessary or desirable here, should consult Lachlan's Modern Pure Geometry, Chapter IX. EXERCISES 1. Prove that homotlietic figures will, if orthogonally projected, be projected into homothetic figures. 2. If P, P' ;, Q, Q' ; R, R' be three corresponding pairs of points in two figures either directly or inversely similar, the triangles PQR, F'Q'R' are sin)ilar in the Euclidean sense. 3. If S and *S" be two curves directly similar, prove that if S be turned in the plane about any point, the locus of the centre of similitude of S and .*>" in the different positions of S will be a circle. 4. If two triangles, directly similar, be inscribed in the same circle, shew that the centre of the circle is their centre of similitude. Shew also that the pairs of corresponding sides of the triangles intersect in points forming a triangle directly similar to them. 5. If two triangles be inscribed in the same circle so as to be inversely similar, shew that they are in perspective, and that the axis of perspective passes through the centre of the circle. G. If on the sides BC, CA,ABoi-e. triangle ABC points X, Y, Z be taken such that the triangle X YZ is of constant shape, construct the centre of similitude of the syltem of triangles so formed ; and prove that the locus of the orthocentre of the triangle XYZ is a straight line. SIMILARITY OF FIGURES 275 7. If three points X^ T, Z be taken on the sides of a triangle ABC opposite to A, B, C respectively, and if three similar and similarly situated ellipses be described round A YZ, BZX and CXY, they will have a common point. 8. The circle of similitude of two given circles belongs to the coaxal system whose limiting points are the centres of the two given circles. 9. If two coplanar circles be regarded as inversely similar, the locus of the centre of similitude is still the 'circle of similitude,' and the axis of similitude passes through a fixed point. 10. P and P' are corresponding points on two coplanar circles regarded as inversely similar and S is the centre of similitude in this case. Q is the other extremity of the diameter through P, and when Q and P' are corresponding points in the two circles for inverse similarity, »S" is the centre of similitude. Prove that SS' is a diameter of the circle of similitude. 11. ABCD is a cyclic quadrilateral ; AC and BD intersect in E, AD and BC in F ; prove that EF is a diameter of the circle of similitude for the circles on AB, CD as diameters. 12. Generalise by projection the theorem that the circle of similitude of two circles is coaxal with them. 18—2 276 MISCELLANEOUS EXAMPLES 1. Prove that when four points A, B, C, D lie on a circle, the orthocentres of the triangles BCD, GDA, DAB, ABC lie on an equal circle, and that the line which joins the centres of these circles is divided in the ratio of three to one by the centre of mean position of the points A, B, C, D. 2. ABC is a triangle, the centre of its inscribed circle, and Jj, 5i, G■^ the centres of the circles escribed to the sides BC% CA, AB respectively ; Z, M, JV the points where these sides are cut by the bisectors of the angles A, B, C. Shew that the orthocentres of the three triangles LB^C^, MC^A^, JVA^B^ form a triangle similar and similarly situated to Aj^B^C^, and having its orthocentre at 0. 3. ABC is a triangle, Z, , Mj, N-^ are the points of contact of the incircle with the sides opposite to A, B, C respectively ; L.^ is taken as the harmonic conjugate of L^ with respect to B and C ; M.2 and N^ are similarly taken ; P, Q, R are the middle points of ZjZg, MjM^, iVi#2- Again AA-^ is the bisector of the angle A cutting 3C in A^, and A^ is the harmonic conjugate of A^ with respect to B and C ; B^ and Cg are similarly taken. Prove that the line A.^BJJ^ is parallel to the line PQR. 4. ABC is a triangle the centres of whose inscribed and circum- scribed circles are 0, 0' ; Oi, 0^, 0^ are the centres of its escribed circles, and OjOo, Oa^s meet AB, BC respectively in L and M ; shew that 00' is perpendicular to LM. 5. If circles be described on the sides Qf a given triangle as diameters, and quadrilaterals be inscribed in them having the inter- sections of their diagonals at the orthocentre, and one side of each passing through the middle point of the upper segment of the corresponding perpendicular, prove that the sides of the quadri- laterals opposite to these form a triangle equiangular with the given one. MISCELLANEOUS EXAMPLES 277 6. Two circles are such that a quadrilateral can be inscribed in one so that its sides touch the other. Shew that if the points of contact of the sides be P, Q, B, S, then the diagonals of PQRS are at right angles ; and prove that PQ, ES and QR, SP have their points of intersection on the same fixed line. 7. A straight line drawn through the vertex A of the triangle ABC meets the lines DF, DF which join the middle point of the base to the middle points E and F of the sides CA, AB in JT, Y ; shew that BY \s, parallel to CX. 8. Four intersecting straight lines are drawn in a plane. Re- ciprocate with regard to any point in this plane the theorem that the circumcircles of the triangle formed by the four lines are con- current at a point which is coucyclic with their four centres. 9. E and F are two fixed points, P a moving point, on a hyper- bola, and PE meets an asymptote in Q. Prove that the line through E parallel to the other asymptote meets in a fixed point the line through Q parallel to PF. 10. Any parabola is described to touch two fixed straight lines and with its directrix passing through a fixed point P. Prove that the envelope of the polar of P with respect to the pai'abola is a conic. 11. 8hew how to construct a triangle of given shape whose sides sliall pass through three given points. 12. Construct a hyperbola having two sides of a given triangle as asymptotes and having the base of the triangle as a normal. 13. A tangent is drawn to an ellipse so that the portion inter- cepted by the equiconjugate diameters is a minimum ; shew that it is bisected at the point of contact. 14. A parallelogram, a point and a straight line in the same plane being given, obtain a construction depending on the ruler only for a straight line through the point parallel to the given line. 15. Prove that the problem of constructing a triangle whose sides each pass through one of three fixed points and whose vertices lie one on each of three fixed straight lines is poristic, when the three given points are collinear and the three given lines are con- current. 16. A, B, C, D are four points in a plane no three of which are collinear and a projective transformation interchanges A and B, and 278 MISCELLANEOUS EXAMPLES also C and D. Give a pencil and ruler construction for the point into which any arbitrary point P is changed ; and shew that any conic through A, B, C, D is transformed into itself. 17. Three hyperbolas are described with B, C ; C, A; and A, B for foci passing respectively through A, B, C. Shew that they have two common points P and Q ; and that there is a conic circumscribing ABC with P and Q for foci. 18. Three triangles have their bases on one given line and their vertices on another given line. Six lines are formed hy joining the point of intersection of two sides, one from each of a pair of the triangles, with a point of intersection of the other two sides of those triangles, choosing the pairs of triangles and the pairs of sides in every possible way. Prove that the six lines form a complete quadrangle. 19. Shew that in general there are four distinct solutions of the problem : To draw two conies which have a given point as focus and intersect at right angles at two other given points. Determine in each case the tangents at the two given points. 20. An equilateral triangle A BO is inscribed in a circle of which is the centre : two hyperbolas are drawn, the first has C as a focus, OA as directrix and passes through B ; the second has C as a focus, OB as directrix and passes through A. Shew that these hyperbolas meet the circle in eight points, which with C form the angular points of a regular polygon of nine sides. 21. An ellipse, centre 0, touches the sides of a triangle ABC, and the diameters conjugate to OA, OB, OC meet any tangent in D, E, F respectively ; prove that AD, BE, CF meet in a point. 22. A parabola touches a fixed straight line at a given point, and its axis passes through a second given point. Shew that the envelope of the tangent at the vertex is a parabola and determine its focus and directrix. 23. Three parabolas have a given common tangent and touch one another at P, Q, R. Shew that the points P, Q, R ai'e collinear. Prove also that the parabola which touches the given line and the tangents at P, Q, R lias its axis parallel to PQR. 24. Prove that the locus of the middle point of the common chord of a parabola and its circle of curvature is another parabola whose latus rectum is one-fifth that of the given parabola. MISCELLANEOUS EXAMPLES 279 25. Three circles pass through a given point and their other intersections are A, B, C. A point £> is taken on the circle OBC, E on the circle OCA, F on the circle OAB. Prove that 0, D, E, F are concyclic if AF.BD. CE = - FB . DC . EA, where AF stands for the chord AF, and so on. Also explain the convention of signs which must be taken. 26. Shew that a common tangent to two confocal parabolas subtends an angle at the focus equal to the angle between the axes of the parabolas. 27. The vertices A,B oi& triangle ABC are fixed, and the foot of the bisector of the angle A lies on a fixed straight line ; determine the locus of 6'. 28. A straight line ABCD cuts two fixed circles X and Y, so that the chord AB oi X is equal to the chord CD of Y. The tangents to X at A and B meet the tangents to F at C and D in four points P, Q, R, S. Shew that P, Q, P, S lie on a fixed circle. 29. On a fixed straight line AB, two points P and Q are taken such that PQ is of constant length. X and Y are two fixed points and XP, YQ meet in a point P. Shew that as P moves along tlie line AB, the locus of i2 is a hyperbola of which AB is an asymptote. 30. A parabola touches the sides BC, CA, AB of a triangle ABC in D, E, F respectively. Prove that the straight lines AD, BE, CF meet in a point which lies on the polar of the centre of gravity of the triangle ABC. 31. if two conies be inscribed in the same quadrilateral, the two tangents at any of their points of intersection cut any diagonal of the quadrilateral harmonically. 32. A circle, centre 0, is inscribed in a triangle ABC. The tangent at any point P on the circle meets BC in D. The line thi'ough perpendicular to OD meets PD in D' . The corresponding points A", F' are constructed. Shew that AU , BE' , CF' are parallel. 33. Two points are taken on a circle in such a manner that the sum of the squares of their distances from a fixed point is constant. Shew that the envelope of the chord joining them is a parabola. 34. A variable line PQ intersects two fixed lines in points P and Q such that the orthogonal piojection of PQ on a third fixed line is of constant length. Shew that the envelope of PQ is a para- bola, and find the direction of its axis. 280 MISCELLANEOUS EXAMPLES 35. With a focus of a given ellipse (A ) as focus, and the tangent at any point P as directrix, a second ellipse (B) is described similar to (A). Show that (B) touches the minor axis of (A) at the point where the normal at P meets it. 36. A parabola touches two fixed lines whifh intersect in T, and its axis passes through a fixed point D. Prove that, if S l)ethe focus, tlie bisector of the angle TSD is fixed in direction. Shew further that the locus of *S' is a rectangular hyperbola of which D and T are ends of a diameter. What are the directions of its asymptotes ? 37. If an ellipse has a given focus and touches two fixed straight lines, then the director circle passes through two fixed points. 38. is any point in the plane of a triangle ABC, and X, Y, Z ai'e points \xi the sides BG, CA, AB respectively, such that AOX, BOY, COZ are right angles. If the points of intersection of CZ and AX, AX and BY be respectively Q and R, shew that OQ and OR are equally inclined to OA. 39. The line of collinearity of the middle points of the diagonals of a quadrilateral is drawn, and the middle point of the intercept on it between any two sides is joined to the point in which they intersect. Shew that the six lines so constructed together with the line of collinearity and the three diagonals themselves touch a parabola. 40. The triangles A-^B^C^, A.>B./J^ are reciprocal with respect to a given circle; B-^C^, C-^A^ intersect in P^, and B^C^, C^Ao in P.,. Shew that the radical axis of the circles which circumscribe the txnangles P^A^B.-,, P^A.^B^ passes through the centre of the given circle. 41. A transversal cuts the three sides BC, CA, AB oi a. triangle in P, Q, R ; and also cuts thi'ee concurrent lines through A, B and C respectively in P\ Q', R'. Prove that PQ' . QR' . RP' =-P'Q . Q'R . R'P. 42. Through any point in the plane of a triangle ABC is drawn a transversal cutting the sides in P, Q, R. The lines OA, OB, OC are bisected in A', B' , C \ and the segments QR, RP, PQ of the transversal are bisected in P', Q' , R'. Shew that the three lines A'P', B'Q', C'R' are concurrent. 43. From any point P on a given circle tangents PQ, PQ' are drawn to a given circle whose centre is on the circumference of the MISCELLANEOUS EXAMPLES 281 first : shew that the chord joining the points where these tangents cut the first circle is fixed in direction and intersects QQ' on the line of centres. 44. If any parabola be described touching the sides of a fixed triangle, the chords of contact will pass each through a fixed point. 45. From D, the middle point of AB, a tangent DP is drawn to a conic. Shew that if CQ, CR are the semidiameters parallel to AB and DP, ' AB:CQ = 2DP:CR. 46. The side BC of a triangle ABC is trisected at M, JV. Circles are described within the triangle, one to touch BC at M and AB at H, the other to touch BC at iV and AC at A'. If the circles touch one another at L, prove that CH, BK pass through L. 47. ABC is a triangle and the perpendiculars from ^, ^, C on the opposite sides meet them in L, M, xi respectively. Three conies are described ; one touching BM, CN at M, N and passing through A ; a second touching CX, AL at N, L and passing througli B ; a third touching AL, BM at Z, J/and passing through C. Prove that at A, B, C tliey all touch the same conic. 48. A parabola touches two fixed lines meeting in T and the chord of contact passes through a fixed point A ; shew that the dii-ectrix passes through a fixed point 0, and that the ratio I'D to OA is the same for all positions of A. Also that if A move on a circle whose centre is T, then AO is always normal to an ellipse the sum of whose semi-axes is the radius of this circle. 49. Triangles which have a given centroid are inscribed in a given circle, and conies are inscribed in the triangles so as to have the common centroid for centre, prove that they all have the same fixed director circle. 50. A circle is inscribed in a right-angled triangle and another is escribed to one of 'the sides containing the right angle; prove that the lines joining the points of contact of each circle with the hypothenuse and that side intersect one another at i-ight angles, and being produced pass each through the point of contact of the other circle with the remaining side. Also shew that the polars of any point on either of these lines with respect to the two circles meet on the other, and deduce that the four tangents drawn from any point on either of these lines to the circles form a harmonic pencil. 51. If a triangle PQR circumscribe a conic, centre C, and ordinates be drawn from Q, R to the diameters CR, CQ respectively, 282 MISCELLANEOUS EXAMPLES the line joining the feet of the ordinates will pass through the points of contact of PQ, PR. 52. Prove that tlie common chord of a conic and its circle of curvature at any point and their common tangent at this point divide their own common tangent harmonically. 53. Shew that the point of intersection of the two comn)on tangents of a conic and an osculating circle lies on the confocal conic which passes through the point of osculation. 54. In a triangle ABC, AL, BM, CiV are the perpendiculars on the sides and MN, jVL, LM when produced meet BG, CA, AB in P, Q, P. Shew that P, Q, R lie on the radical axis of the nine-points circle and the circumcircle of ABC, and that the centres of the circumcircles of ALP, BMQ, CNR lie on one straight line. 55. A circle through the foci of a rectangular hyperbola is reciprocated with respect to the hyperbola ; shew that the reciprocal is an ellipse with a focus at the centre of the hyperbola ; and its minor axis is equal to the distance between the directrices of the hyperbola. 56. A circle can be drawn to cut three given circles orthogonally. If any point be taken on this circle its polars with regard to the three circles are concurrent. 57. From any point tangents OP, OP', OQ, OQ' are drawn to two confocal conies ; OP, OP' touch one conic, OQ, OQ' the other. Prove that the four lines PQ, P'Q', PQ', P'Q all touch a third con- focal. 58. P, F and Q, Q' are four collinear points on two conies JJ and V respectively. Prove that the corners of the quadrangle whose pairs of opposite sides are the tangents at P, P' and Q, Q' lie on a conic which passes through the four points of intersection of f/'and V. 59. If two parabolas have a real common self-conjugate triangle they cannot have a common focus. 60. The tangents to a conic at two points A and B meet in 7\ those at A', B' in T' ; prove that T(A'AB'B, = T'{A'AB'B). 61. A circle moving in a plane always touches a fixed circle, and the tangent to the moving circle from a fixed point is always of constant length. Pi'ove that the moving circle always touches another fixed circle. MISCELLANEOUS EXAMPLES 283 62. A system of triangles is formed by the radical axis and each pair of tangents from a fixed point P to a coaxal system of circles. Shew that if P lies on the polar of a limiting point with respect to the coaxal system, then the circumcircles of the triangles form another coaxal system. 63. .Two given circles S, S' inteisect in A, B ; through A any straight line is drawn cutting the circles again in P, P' respectively. Shew that the locus of the other point of intersection of the circles, one of which passes through B, P and cuts S orthogonally, and the other of which passes through B, P' and cuts S' orthogonally, is the straight line through B perpendicular to AB. 64. Four points lie on a circle ; the pedal line of each of these with respect to the triangle formed by the other three is drawn ; shew that the four lines so drawn meet in a point. 65. A, B, C, D are four points on a conic; EF cuts the lines BC, CA, AB in a, b, c respectively and the conic in E and F ; a, b' c are harmonically conjugate to a, b, c with respect to E, F. The lines Da, Db', Dc meet BC, CA, AB in a, /3, y respectively. Shew that a, /3, y are collinear. 66. Three circles intersect at so that their respective diameters DO, EO, FO pass through their other points of intersection A, B, C ; and the circle passing through D, E, F intersects the circles again in G, H, I respectively. Prove that the circles AOG, BOH, C 01 are coaxal. 67. A conic passes through four fixed points on a circle, prove that the polar of the centre of the circle with regard to the conic is parallel to a fixed straight line. 68. The triangles PQP, P'Q'L" are such that PQ, PR, FQ', P'R are tangents at Q, R, Q\ R' respectively to a conic. Prove that P{QR'Q'Ii) = P'{QJi'Q'Ji) and P, Q, R, F, Q', R' lie on a conic. 69. If A', B', C, D' be the points conjugate to A, B, G, D in an involution, and /-•, Q, R, S be the middle points of A A', BB', CO', DD', {PQRS) = {ABCD) . {AB'CD'). 70. ABC is a triangle. If BDCX, CEAY, AFBZ be three ranges such 'that (XBCD) . (A YCE) . {A BZF) = 1, and A D, BE, CF be concurrent, then A'', Y, Z will be collinear. 71. If ABC be a triangle and D any point on BC, then (i) the line joining the circumcentres of ABD, ACD touches a parabola: 284 MISCELLANEOUS EXAMPLES (ii) the line joining the incentives touches a conic touching the bisectors of the angles ABC, ACB. Find the envelope of the line joining the centres of the circles escribed to the sides BD, (7/) respectively. 72. Two variable circles S and a^S" touch two fixed circles, find the locus of the points which have the same polars with regard to >S^ and S'. 73. QP, QP' ai-e tangents to an ellipse, QMi?, the perpendicular on the chord of contact I'P and K is the pole of QM. If H is the orthocentre of the triangle PQP', prove that UK is perpendicular to QC. 74. Two circles touch one another at 0. Prove that the locus of the points inverse to with respect to circles which touch the two given circles is another circle touching the given circles in 0, and find its radius in terms of the radii of the given circles. 75. Prove that the tangents at A and C to a parabola and the chord AC meet the diameter through B, a third point on the para- bola in a, c, b, such that aB : Bh = Ah : bC ^ Bb : cB. Hence draw through a given point a chord of a parabola that shall be divided in a given ratio at that point. How many different solutions are there of this problem ? 76. If A, B, G be three points on a hyperbola and the directions of both asymptotes be given, then the tangent at B may be constructed by drawing through B a parallel to the line joining the intersection of BC and the parallel through A to one asymptote with the inter- section of AB and the parallel through C to the other. 77. A circle cuts three given circles at right angles; calling these circles A, B, C, CI, shew that the points where C cuts O are the points where circles coaxal with A and B touch O. 78. If ABC, DEF be two coplanar triangles, and »S' be a point such that SD, SE, SF cuts the sides BC, CA, AB respectively in three collinear points, then SA, SB, SC cut the sides FF, FD, DE in three collinear points. 79. ABC is a triangle, i) is a point of contact with BC of the circle escribed to BC \ E and F are found on CA, AB in the same way. Lines are drawn through the middle points of BC, CA, AB parallel to AD, BE, CF respectively ; shew that these lines meet at the incentre. INDEX The references are to pages. Antiparallel 37 Asymptotes 103, 171 • Auxiliary circle 147, 153, 165, 205 Axes 101, 104, 107, 166 Axis of perspective 61, 62 Brianchon's theorem 215 Carnofs theorem 116 Ceva's theorem 33 Circle of curvature 120, 138, 158, 187 Circular points 242 Circumcircle 1, 5, 133, 245 Coaxal circles 20, 226, 234 Collinearity 31, Di, 214 Concurrence 33 Coufocal conies 165, 235, 236, 248 Conjugate points and lines 15, 74 Conjugate diameters 151, 155, 174, 11(2 Conjugate hyperbola 170 Desargues' theorem 227 Diameters 106, 132 Director circle 150, 169 Double contact 246, 253 Ecircles 10, 262 Envelopes 130, 147, 212 Equiconjugates 156 Feuerbach's theorem 262 Focus and directrix 91, 94, 96, 108, 233, 246 Generalisation by projection 249 Harmonic properties 74, 75, 92 Homographic ranges and pencils 54, 59, 60, 62 Homothetic figures 267 Incircle 10, 262 Inverse points 13, 256, 261 Involuti(m criterion 81 Involution properties 84, 93, 216, 224, 227 Isogonal conjugates 36 Latus rectum 114, 126 Limiting points 21 Loci 24, 127, 137, 153, 179 Medians 8 Menelaus' theorem 31 Newton's theorem 117, 126, 136, 152, 177, 178, 179, 180 Nine points circle 3, 195, 262 Normals 113, 127, 145, 149, 164, 186 Ordiuates 107 Oithocentre 2, 133, 194, 232 Orthogonal circles 22, 73 Orthogonal involution 85, 94 Pair of tangents 111, 130, 147, 150, 168 Parallel chords 95 Parameter 136 Pascal's theorem 214- Pedal line 5, 133 Pole and polar 13, 92, 229 Projective propeities 45, 50, 53, 92 Quadrangle 76, 222, 224 Quadrilateral 75, 222, 224, 226, 248 Radical axis 17 Reciprocal figures 220, 237 Salmon's theorem 17 Self-conjugate triangles 16, 121, 236, 245 Signs 28, 119, 180 Similar figures 268 Simititude, centres of 24 Similitude, circle of 25, 271 Simson line 6 Subnormal 128 Symmedians 37 Tangents 108, 127, 145, 164 Triangles in perspective 64 CAMBRIDGE : PRINTED BY J. L. PEACE, M.A., AT THE UNIVERSITY PRESS •IHTORN TO mSK^l^^^^ BORKOW.O LOAN DEPT. _^!:!!!!l^ "' »«'>'•<« » imn.edi.,e recall. 5£P2 'DH^ .y- c ii6?a33 THE UNIVERSITY OF CALIFORNIA UBRARY