UC-NRLF 5> m m ^ jn jra p;i } 1 i ) 1; j;: 1 i: ! 1' '•'■ Ti If |. |\ 1, }: , i; It i, i !;•!.«;• < V 1; «; nJi't SOLUTIONS OF EXAMPLES CONIC SECTIONS, TREATED GEOMETRICALLY W. H. BESANT, D.Sc, F.R.S. FELLOW OF ST JOHN'S COLLEGE, CAMBBIDGE. THIRD EDITICN, REVISED. CAMBRIDGE : DEIGHTON, BELL AND CO. LONDON : GEORGE BELL AND SONS. 1890 ^, PRINTED BY C. J. CLAY, M.A. & SONS, AT THE UNIVERSITY PRESS. PREFACE TO SOLUTIONS. I HAVE frequently received requests for a book of Solutions of the Examples in my treatise on Conic Sections, but have never been able to find time to prepare them. Mr Archer Green, B.A., Scholar of Christ's College, volunteered to undertake the task, with the aid of my notes and his own, and, with the exception of a few at the end, wrote out the solutions entirely. Mr Green was however prevented by illness from completing the revision of the proofs, and I am much indebted to Mr J. Greaves, Fellow of Christ's College, who kindly undertook to examine the rest of the sheets. The book will, I hope, prove useful both to students and teachers, as a companion volume to the treatise on Conic Sections. W. H. BESANT. Sept. 1881. 35784V PREFACE TO THE THIRD EDITION. The solutions have been revised, and many ad- ditions have been made to them. They will now be found to be in complete accordance with the sixth edition of the Geometrical Conies. W. H. BESANT. Jan, 1890. CONIC SECTIONS. SOLUTIONS OF EXAMPLES. CHAPTER I. 1. XF the tangent at P meet the directrix in Z, and 8 JL be the focus, PSZ is a right angle ; .-. S lies on the circle of which PZ is diameter. 2. Let PN and QTIf be the ordinates at P and Q. Then PN : QM :: SP : aSQ :: XN : XJf ; .'. the triangles PXN and QXM are similar and PX, ^X equally inclined to XS. 3. Bv Art. 8, FS is the external bisector of the angle PSQ. 4. SP : PK :: ^^ : ^A^ :: SE : J5^A^; .-. EP bisects the angle SPK. 5. Since F, S, P and K lie on a circle, the angle XSF= the angle FPX= the angle PTaS^ 6. PiV^: P'N' :: /SP : ^K; .\XK : XiV:: XX' : XN' ; :, the angle XiV:A''=the angle XiV^X=the angle LN'N. B. C. S. 1 2 Coriir, Sections, 7. ,Let Q be the point ^vhere the tangent at R meets NP. Tlien NQ : NX :: JSR : aS'X :: SA : ^X :: /S^P : NX; :,SP = QN 8. Let SY be perpendicular to the tangent at P and G^Z perpendicular to SP, Then, since the triangles PSY^ GPL are similar, PG : PL :: SP : aST, or P(^ : SB :: /SP : SY. 9. If the tangent meet the directrix in Z, and SP be drawn such that ZSP is a right angle meeting the tangent inP, then P will be the point of contact of the tangent ZP. 10. If P, Q be the extremities of the chord, and PK, QL be perpendicular to the directrix, SP : PK :: SA : AX :: SQ : QL ; .-. SP + SQ :: PK-\-QL :: .S'^ : ^X Now the distance of the middle point of PQ from the directrix is equal to half PK+ QL, and is therefore least when SP + SQ is least, that is, when PQ goes through the focus. 11. If TP, TP' be the fixed tangents, and the tan- gent at Q meet them in E, E', the angle PSE=i\\Q angle ESQ, and the angle QSE'=^i\iQ axigleE'SP'; .-. the angle ^^^' = half the angle PSP\ }2. If perpendiculars from the given points PK, QL be drawn to the directrix and S be the focus, SP : SQ :: PK : QL, a constant ratio; .•. the locus of /S' is a circle. Conic Sections. 8 13. Let the normal at P meet the axis in G. Taking as the fixed point in the axis, it is obvious that the triangles OSR^ GSP are similar ; .-. SO : SR :: SG : SP :: SA : AX; .'. SR is constant, and R lies on a circle of which aS' is the centre. 14. AT : AX :: SR : SX :: SA : AX; .'.AT=AS. 15. aS'T' bisects the angle between SP and SQ, Art. 12, and SR bisects the angle between QS, and SP produced, Prop. II., Art. 5 ; /. RST is a right angle. 16. The triangles EA T, BRS are similar ; /. AT : SR :: EA : ER :: AX : SX; .-. AT : ^X :: SR : aSX :: SA : ^A-^; 17. If TL be perpendicular to the directrix, SR : 7X :: SA : AX :: SM : TL ; .'.SM=SR. 18. FS\s the external bisector of the angle QSP, and F'SofQSP'; .'. the angle i^^i^' = half the angle PSP\ 19. Since the triangles SPN^ SGL are similar, .-. GL : PN :: SG : aS'P :: ^^ : AX. 20. If the normals PG, P'G' meet in Q, and QK be drawn parallel to the axis to meet the chord in F, Vq-, VPv.SG'.SPv.SA\AXv.SG''.SP'v. VQ : VP' ; :. VP= VP\ or r bisects PP\ 1—2 4 Conic Sections. 21. DS is the external bisector of the angle PSQ, and ES ofpSQ; .'. DSE is a right angle. 22. The semi-latus rectum is an harmonic mean be- tween /S'P and aS'P' ; .-. 2SP . SP'=SR . PP\ 23. PE : PL :: PQ : P(y :: PV : P^ :: PP' : 2^P, see Ex. 20 ; :.PE : aS'/? :: SP' : .S'i?; .-.P^^aSP'. Similarly P'E=^SP, 24. The right-angled triangles DSQ, DSE have a common hypotenuse. Also SE^SR = SQ; .'. the angle e^^=the angle ESP. 25. Let S be the focus and P and Q the given points. Through P draw a straight line PK so that SP may bear to PK the given ratio of the eccentricity. Through Q draw a straight line QL so that SQ : QL in the same ratio. With centres P, Q and radii PA", QZ respectively describe circles. The perpendicular from aS' on a common tangent to these circles will be axis. 26. Let the tangents at P and Q intersect in T. Draw TN perpendicular to directrix and TM perpen- dicular to SP, Then SM : TN :: SA : AX. But aST bears a constant ratio to SM, since angle TSM= hnUPSQ; :. ST bears a constant ratio to TN, Conic Sections. 5 27. Let T be the intersection of the tangents at P and p. Draw TK perpendicular to Pp. Then TK : PL :: TP : PG^ and TK : jt?/ :: 7> : p^. Again, draw GM, gm perpendicular to SP, Sp respectivelv, and TN, Ta perpendicular to SP^ Sp respectively. Then TP : PG :: TN : MP :: Tn : mp :: Tp : pg; .-. TIsT : PL :: TIC : pi ; .'.PL=pL CHAPTER II. THE PARABOLA. 1. TiiE distance of the centre of the circle from the fixed point is equal to its distance from the fixed straight line, and therefore its locus is a parabola of which the fixed point is focus and the fixed straight line directrix. 2. Through the vertex draw a straight line making the given angle with the axis ; the tangent at the point where the diameter bisecting this chord meets the curve will be the tangent required. Or, draw a radius vector from the focus, making twice the given angle with the axis. 3. Since TA = AN, PN=2AY; .*. AY''=AS.AN. 4. Let jSY^ be drawn perpendicular to the line through G parallel to the tangent. Then in the right-angled triangles YST, Y'SG, ST=^SG, and the angles YST, Y'SG are equal ; .-. SY=:SY\ T), Draw S Y perpendicular to the tangent and YA perpendicular to the axis. Produce aS'^ to JT, making AX equal to SA. Then the straight line through X perpendicular to SX is the directrix. 6. Let the circle touch the fixed circle in Q, and the straight line in R\ let P be its centre, and S the centre of tiie fixed circle. Produce PR to M, making RM equal to SQ, then the The Parabola. 7 straight line MX drawn through M parallel to the given line is a fixed straight line. Then, since SP is equal to PM, the locus of P is a parabola of which S is focus and MX directrix. 7. Draw SY^ SY^ perpendiculars on the two tan- gents. Then, if SA be perpendicular to YY\ A is the vertex. Produce SA to X, making AX equal to AS ; X is the foot of the directrix. 8. If the tangent at the end of the latus rectum meet PN in Q, QN=XN=SP. 9. Since /S^FP and P^^^S^are right angles, P, N, S, Y lie on a circle ; . TY,TP=TS.TN. 10. aS'^ is half TP, and PT^=--PN^+TN^ = AAS.AN-\-AAN^', .'. SE'' = AN.XN=A]Sr.SP. 11. If aS'F be drawn perpendicular to the tangent and A be vertex, SA F is a right angle ; .*. A lies on the circle of which aS'F is diameter. 12. Draw -S'F perpendicular to the tangent, then if the circle described with centre S and radius equal to a qujirter of the latus rectum meet the circle described on /S' F as diameter in ^, ^ is the vertex. Produce SA to X, making AX equal to aS'^I, then X is the foot of the directrix. 13. SN : SP :: SN' : SP\ or AN-AS : AN+AS :: AS-AN' : AS+AN'; .'. AN : AS :: AS : AN'; .'. AN. AN = ASK 8 The Parabola, Again, AAS . AN : A.AS'' :: A.AS'^ : 4:AS,AN'', .'. PiV^ : SE" :: aS'T?^ : P'iV'^ or FN : /S^ :. SR : P'iV; .'. the latus rectum is a mean proportional between the double ordinates. 14. Let V be the middle point of the focal chord PSP', and let the diameter through V meet the curve in Q ; then, if QT, QM be the tangent and ordinate at Q, and VL be ordinate of F, VL = QM 2iii(}i TM=SL) .'. Vr- = QM^:=4AS .AM=2AS. TM=2AS . SL. Hence the locus of F is a parabola of which S is vertex and SL axis. 15. If P, P' be the given points, PK, P'K' perpendi- culars on the directrix, the focus is the point of intersection of a circle centre P, radius PK with a circle of which P' is centre and P'K' radius. In general two circles intersect in two points, there- fore two parabolas can be drawn satisfying the given conditions. 16. If PG be normal at P, the triangles PNG, pPR are similar ; .-. Pp : PN :: RP : NG ; .-. i^P = 2iV6^ = latus rectum ; .*. the locus of R is an equal parabola having its vertex A' on the opposite side of X, such that AA' i^ equal to the latus rectum. 17. Let P, P' be the given points, aS' the given focus. A common tangent to the circles described with centres P, P and radii PS, P'S respectively will be the directrix. 18. If SP be the focal distance and /S'F perpendicular to the tangent at P, Y lies on the circle of which SP is diameter. The Parabola. 9 Also the angle A Y"aS'= the angle SP F; .*. A Y touches the circle. 19. The tangents at the ends of the focal chord PSP' meet in F on tlie directrix at right angles : also the straight line through F at right angles to the directrix bisects PF' in V; :, FV^VP=VP'', .'. the directrix touches the circle described on PP' as diameter. 20. Draw the farther tangent to the circle parallel to the given diameter, then the locus of the point is a parabola of which the centre is focus, and the tangent thus drawn directrix. 21. Draw a straight line parallel to the given straight line, on the farther side of it, and at a distance from it equal to the radius of the circle, then the locus of the point is a parabola of which the centre of the circle is focus, and the straight line thus drawn directrix. 22. Let Q be the centre of the circle touching the sector in R and AG in M. Through C draw CB at right angles to AG, and on the same side of it as Q, and draw QN perpendicular to the tangent at B. Tlien NQ+QM^BG= GQ + QR; .-. GQ = QN; .*. Q lies on a parabola of which G is focus and BN direc- trix. 23. Y is the middle point of TP and Z of PG^ ; therefore YZ is parallel to the axis. 24. If SQ be perpendicular to the normal PG, PQ = QG, and if QM be the ordinate, NM=MG ; .'. SM=AN and PN=2QM; .-. QM' = AS. AN^AS. SM; .*. Q lies on a parabola of which S is vertex and SG axis. 10 The Parabola. 25. The triangle PSG is isosceles; therefore GL is equal to FN. 26. If the circle described with centre S and radius equal to the perpendicular from S on the tangent at P meet the circle of which SP is diameter in F, and the angle SYA be made equal to the angle SP F, then the foot of the perpendicular SA on YA will be the vertex. 27. Since SQ is double SA, ASQ (and likewise QSP) is equal to the angle of an equilateral triangle ; therefore SP and SQ are equally inclined to the latus rectum. 28. QX^ = SX^ + SQ^ + 2SX. SQ = 4AS'-+QG'-h2SQ.NG =-4AS^+QN^ + 2QN.NG + NG' + 2SQ,NG = 4AS^+ QN'' + NG^ + 2NG. SN = 4^^=^ + NQ'' 4- 2 AN. NG = 4AS^ + QN' -h PN' = 4AS^+ QP\ 29. The angle SPF= SPG - FPG = SGP - GPR= SHP ; therefore the triangles SPF, SHP are similar ; /. SF.SH^SP^=^SGl SO. A, B, CyS lie on a circle ; therefore, if D be the end of the diameter drawn through S, DA, DB, DC are per- pendicular to SA, SB, SC respectively. 31. Since PQ and PR are equally inclined to the axis, the circle through P, Q, R touches the parabola at P; therefore PQ is a diameter of this circle. Therefore PRQ^ the angle in a semicircle, is a right angle. 32. Let MR and AQ meet in V. Draw the ordinates VW, RZ. The Parabola. 11 Then MW : MZ :: WV . RZ :: AW : AN-, :, MW : AW :: MZ : AN; .'. AN : AW:: AZ : AN, Again, VW' : QN^ :: AW' : AN^; /. VW^ : BZ^ :: AW : AZ; ,\ V lies on the curve. 33. Let P, Q be the given points. Bisect PQ in F, and draw FT parallel to the axis meeting the given tan- gent P in T. Draw PS, QS such that TP, TQ may be equally inclined to the axis and to SP, SQ respectively. Pas', QS meet in the focus. Through P draw a straight line PA" parallel to the axis, making PK equal to SP, then the straight line through K at right angles to PK will be the directrix. 34. Let P be the vertex and Q VQ' be corresponding ordinate. Take M in VP produced such that QV'^4MP.PV. Make angle TPS equal to the angle MPT, PT being parallel to QQ, and make PS equal to PM. Then S is the focus, and the straight line drawn through M at right angles to PM is the directrix. 35. Pil/2 : QN' :: AM : AN, QN being the ordinate of Q; .-. A3f=4AN3ind3AM=4NM; .-. 3AT=4QN=2PM. 36. Draw PiV perpendicular to AB. Then AN : NP :: CQ : AC :: NP : AC ; .\ PN^=AC,AN Therefore the locus of P is a parabola of which A is vertex and AB axis. 12 The Parabola. 37. The triangles LKP, PSK, KSA and TKA are similar ; .-. KL^ : /SP^ :: KP^ : KS'^ -^KA^ : AS^ :: TA : AS ::SP-AS : AS. 38. With centre S and radius one-fourth of the chord describe a circle meeting the parabola in P. The chord through S parallel to the tangent at P will be the chord required. 39. PN-^ = 4:AS.AN=4AS.AN' + 4AS^ = P'N'^-^N'G'^ = P'G'-\ 40. If Pp, P'p' be two parallel chords, and V^ V their middle points, FF' is a diameter. Let FK^meet the curve in Q, Draw QT parallel to Pp^ then QT 'w> the tangent at Q. Produce FPto a point J/ such that Pr^ = 4Qil/. QV, then the straight line drawn through M at right angles to MV\9> the directrix. Make the angle TQS equal to the angle TQM. Then if QS be made equal to Qi¥, S is the focus and the straight line through aS' perpendicular to the directrix is the axis. 41. Let the tangents at P and P^ intersect in T. Then 4.SP .PV : 4SP' . P' V ::P'V' : PV"' :: TP^ : TP'^ :: SP : SP' ; .-. PV^P'V. 42. If in the preceding Example P' 7" meets PV'm Z and the sides of the triangle ABC are parallel to ZP, PT and rZ respectively, AB : AG :: TZ : TP :: Tr : TP. 43. If Uf rbe the vertices of the diameters bisecting Pp, Qq. PS.Sp : QS.Sq :: ^^ : SV :: Pp : Q^. The Parabola. 13 44. Draw R W, LZ parallel to QQ\ Then PU : PR^ :: LZ' : RW :: QV^ : RW^ :: PV : PW :: PiNT : PR-, .-. PL'^=PR,PN. 45. This question is solved in Conies, Art. 212, p. 217. 46. PN^ : AN^ :: ^il/^ . qj^s . .-. 4.AS : ^iV :: AM : 4AS. 47. Let AP, Ap meet the latus rectum in L and I respectively. Then PN"" : SL^ :: ^iV^^ . ^^2 .. ^^ . ^^^^ (Example 13) :: PN^ : pri"; .: SL=pn. In like manner Sl = PN. 48. If PK^ QL be perpendicular to the directrix, and QU to P^ produced, the angle SPQ=^i\iQ angle QPL' ; .-. PL'^SP = PK', :, SQ = QL^KL' = 2PK=2SP. 49. Is equivalent to Example 32. 50. Let Q be the point of intersection, and let QK be the ordinate of Q. Then AK '. QK v. PN : NT ; .-. 2AK,AN=QK.PN=PN'' = ^AS.AN; .-. ^^=2^.S', or C lies on a fixed straight line parallel to the directrix. 51. Let 75?, Tq be the fixed tangents, and let PQ touch the curve in R. Then SP^=:^Sp . SR = Sq . SR = SQ'' ; .-. SP--^^Q. 14 The Parabola. 52. Let TMhQ the ordinate of T, and T^F perpendi- cular to SP. .'. the locus of 2" is a parabola of which X/S'is axis. If Tir=2AS, TM'' = 4AS. XM, or X is the vertex. 53. Let the chord FQ meet the axis in 0, and the tangent at A in V, Then by Art. 48, VO^^ VP ,VQ\ :, F is a fixed point, and the locus of A is the circle of which F is diameter. 54. Let the diameter Trmeet the curve in R. Then the tangent at i?, being parallel to PQ, meets TP at right angles in Z on the directrix. Also TZ : ZP :: TR : RV \ .-. TZ^ZP, Therefore yand P are equidistant from the directrix. 55. Let PTmeet the axis in U Then PQ : PT :: 1PV : PT :: "IPG : Pt, :: 2PN : Nt :: PN : AN. 56. If the tangents 7LP, TQ are equal, T lies on the axis. Let the tangent at R meet them in p and q. Then, since T, jt?, ^ and S lie on a circle, the triangles Sq r, SpP are similar ; .-. Tq : it?P :; TS : SP ; .-. Tq=pP. So TiP = ^e. The Parabola, 15 57. AN. NL = PN^r=4AS. AN; .\ NL = 4AS, But NG = 2AS; .*. LG = half the latus rectum. 58. By Art. 5, P'S, Q'S are the external bisectors of the angles PSA, QSA ; therefore P'SQ' is a right angle. 59. The angles TCS, DRS are equal, being supple- ments of equal angles SCP, SRC, Art. 35. And the angle CTS=^ TQS=RDS; .'. the triangles TOS, DRS ore similar; .-. BR : TC :: RS : SO :: RC : GP ; :. PC : CT :: OR : RB. Similarly TD . DQ :: GR : RD, 60. Let A D and XP intersect in Q, and let QM be the ordinate. Then QM : DS :: AM : ^.S' and QM : PN :: XM : XN; .-. ^iJf : XM :: ^/S' : XN ; :. AM : AS :: A>S : ^iV^; .-. QM^ : PiV^2 .. ^j/2 . ^^2 .. jij^ . ^jv, or Q is on the parabola. 61. By Example 18, YY\ the tangent at the vertex, is a common tangent. SY^ = AS.SP, SY'^ = AS.Sp; .'. YY'^=SY^ + SY'^=AS.Pp. 62. If P r be the diameter bisecting A Q, AM=iAN. 16 The Parahola. Also AM,MR = QM'^ = AAS,AM', .-. MR = 4 AS. Now focal chord parallel to AQ = 4:SP = 4:XN=-4AS + A3f=AR. 63. Let AR, CP meet in p. Draw pN^pD perpendicular to CA^ CR, and let Dp meet the tangent at A in M, Cp : CP :: CD : Ci2 :: iVi? : CR :: ^i\^ : AC-, :. Cp = AN=^pM. Therefore the locus of p is a parabola of which C is focus and AM directrix. 64. If QMQ' be the common chord, ^AS'' = 4.AQ^ = AAM'' + 4.QM-=4AM'' + IQAM . AS ] .-. ^iV/ishalf^AS'. 65. Let the fixed straight line ^^meet the tangent at P in K. Draw KY^ at right angles, and S Y^ parallel to KP. Draw Y'A' perpendicular to the axis, and KL parallel to BA. Then, since A^F=^F; SA' = KL=^BA therefore -4 ' is a fixed point. Therefore KY^ touches the parabola of which S is focus and A' vertex. QQ. QD . DR = QM^-DM'^ = Q3P-PN'^ = 4 AS, AM- 4 AS. AN=4:AS. PD. 67. Draw the double ordinate QMq; then, if the diameter through Q' meet Qq in D', QD\D'q = 4:AS.Q'D\ Tlie Parabola. 17 Now NT : PN :: QU : QD' :-. Uq : 4^^; .-. D'q : \AS :: ^AN : PN :: PN : 2AS ; .-. 2PN=D'q-=D'M+Mq = QM+QM\ Therefore the line through P bisecting QQ' is parallel to the axis. Hence the locus of the middle points of a series of parallel chords is a straight line parallel to the axis. 68 \ Take CP, CQ two tangents such that PCQ is two- thirds of a right angle ; join jSC cutting the curve in li\ and draw the tangent ARE. Then, Art. 38, CSP^-CSQ = 120'', and CAR = \CSQ^m''', .'. CAB is equilateral. 69. Draw AZ, AN perpendicular to the tangent and SY respectively, and draw S3I perpendicular to ZA. Then SM^ = AN^ = YN. NS= ZA . AM, Therefore the locus of /S' is a parabola of which A is vertex and ZM axis. 70. If GZ be drawn parallel to PF and SZ to PG, then SY, SZ are equal. Therefore, ifZB be perpendicular to the axis, BS=AS. Hence GZ touches an equal parabola of which B is vertex and S focus. 71. If pqr be the triangle formed by the given straight lines, describe a parabola passing through p, q and r having its axis parallel to AS. (Ex. 45.) If 5 be the focus of this parabola, draw SP parallel to sp, PQ to pq, and PR to pr. Then PQ : pq :: SA : sa :: PR : pr, and the angles QPR, qpr are equal. ^ If a parabola touch the sides of an equilateral triangle, the focal distance of any vertex of the triangle passes through the point of contact of the opposite side. B. c. s. 2 18 The Parabola. 72. Let RWhQ. the ordinate of R. Then AN^ : ATV'^:: PN^ : RW^ :: Pm : QJ/2 .. AN : AM -, .\AN : AFT :: ATV : AM, or WN : AN :: MTV '.AW; .'. RL : QR :: AN '. AW v. PN -. NL. 73. Let PVhe the ordinate to the diameter RQM. Then PiJf : i^iif :: PN : 7W :: 2PN. AS : 4^aS' . ^iV^ :: 2^^ : PN ; .-. PM.PN=2AS.RM. But PM'^ = 4AS.QV=4AS. RQ ; .-. i^Jf : i^Q :: 2PN : Pil/ :: PP' : P3f ; .'. QM : QR :: P'M : Pi^. 74. Let PP be the chord, TWV its diameter, RQM the line parallel to the axis. Then PM : RM :: PF : TF :: PF : 2Wr :: 2VP.SW : 4aS'^. ^F :: 2^^^ : FV ; .: PM.PV=2SW.RM. But PM.MP' = 4SW.QM, .'. RM : QJf :: 2PF : i!/P, or i?Q : QM :: PJf : MP\ 75. >S^i2, xS'r are the exterior bisectors of the angles PSQ, pSQ respectively. Therefore RSr is a right angle. Therefore SD, which is half the latus rectum, is a mean proportional between DR and Dr. 76. Let P VP^ be parallel to the given straight line, Q VQ' the chord joining the two other points of intersection of the parabola and circle. Let the diameters through F and V meet the curve in p and p\ The Parabola. 19 Then pp' is a double ordinate ; draw V'H parallel to pp' to meet p V. VV is perpendicular to QQ\ and therefore parallel to the normal at p' ; .-. VW : p'g :: V'H : p'n ; .-. Vr = ^p'g. 77. The arcs Q U and R V are equal, since Q Fand UR are parallel. Therefore QR and ^F are equally inclined to Q V, that is to the axis. But QR and the tangent at P are equally inclined io the axis ; therefore UV\9> parallel to the tangent at P. 78. VR : VR' :: VR : TQ' :: PV : PP .'. VR. VR = QVK 79. If FR, QE meet the tangent at P in V and T, TE : EQ :: Ti? : i^P :: PF : FQ. (Ex. 74.) Therefore PPis parallel to TP. 80. If Q be the vertex of the diameter bisecting the chord Rr which meets the diameter PJVin W^ RW. Wr=r.4SQ.PiV. Therefore the rectangle under the segments varies as the distance of the point of intersection IV from P. 81. QSj Q'S are equally inclined to SP, and therefore to the axis. Therefore Q'S meets the curve at the end of the double ordinate QMq, and, since AM. AM =- AS^, the semi-latus rectum is a mean proportional between QM and Q'M'. Also, since the diameter through P bisects QQ', PS is an arithmetic mean between QMsmd QM'. 2—2 20 The Parabola, 82. BB' will bisect CA' in V. Let V be the middle point of B'B", VV is parallel to the axis. And BB" is parallel to VV, and therefore to the axis. ►Similarly AA" and CG'^ are parallel to the axis. 83. Let (7 be the centre of the circle. The angle between tangents to circle = PCP' = 2PSP' = 4 times angle between the tangents to the parabola. 84. The tangents at the ends of the focal chord PSP' will meet in T on the directrix. If the normals at P and P' meet in Q, TQ will be parallel to the axis. Let TQ meet the curve in ^? and PP' in V. Let QM be the ordinate of Q. Then XM= TQ = 2TV= 4Sp = 4Xn. Therefore, if we take B in XM such that XB = 4.AS, BM=4An, QM^=pn^ = 4AS,An = AS.BM. Hence the locus of § is a parabola of which B is vertex and BM axis. 85. Produce PA to P, making AP' equal to A P. On AP^ as diameter describe a circle meeting the tangent at P in T. Join TA and produce to iV, making AN equal to A T. In AN take a point S such that PN^ = 4:AS.AN, then S is focus. 86. If G be the intersection of the normals and Q vertex of the diameter bisecting PSp, ps.sp=AS. pp=As, tg: 87. If pq be a tangent parallel to P$, Tp=pP, and T, ;?, qy S and lie on a circle. Therefore the angles TSO, TpO are equal, and TpO is a right angle. The Parabola. 21 88. SM^ : AN' :: QM^ : PN^ :: AM : AN', .-. SM'' = AM.AN So ^^'2^^M'..4iV; .-. MM' . AN^MM' . (^il/-/SJ/') ; .-. MM' : SM-SM' :: ^P : ^iV^; 89. If P, Q, P', Q' be the points of intersection, PQ, P'Q! are equally inclined to the axis. Hence the middle points of PQ and P'Q' are equidistant from the axis. Therefore, if P, Q be on one side of the axis and PQ' on the other, the sum of the ordinates of P and Q is equal to the sum of the ordinates of P' and Ql. If P' be on the same side of the axis as P and (?, the ordinate of Q! is equal to the sum of the ordinates of P, ^, and P'. 90. Let the diameter through T meet the curve in W^ PQ in F, and PN in t. Let WZ be the ordinate of W \ draw Qq parallel to the axis to meet PN. QM.PN=PN.qN=tm-Pt''-^WZ'^-^AS, WV = 4AS.AZ-4:AS.LZ=4AS,AL. 91. pX : XA :: PN : ^iV^ :: 4AS : PiV^ :: 4^.^. QM : 4^.5'. ^Z. (Ex. 90.) So qX : XA :: PN : ^X ; .-. pX-rqX : XA :: PN+QM : AL ; .-. ^X+gX : PN+QM :: X^ : AL :: iX : 7X. But NP+QM = 2TL', :. pX+qX=1tX, or pt = ^^. 22 The Parabola. 92. Let TF, TD be drawn parallel to PE, QE normals at P and Q. The angle TFQ = PEQ = supplement oiPTQ=. TSQ ; .'. Q, S, F, T lie on a circle. Therefore TSF is a right angle. So TSD is a right angle, and DF goes through aSI 93. \ipq be a tangent parallel to PQ, Tq^qQ. Also, 7", j9, 2' and /S lie on a circle ; therefore the angles Tpq, TSq are equal. Therefore 2'Sq is a right angle. 94. Let i^O be the diameter through the given point 0. Take T in OR produced such that TR \ RO in the given ratio. If TP be a tangent, the chord POQ will be divided as required. (Ex. 74.) 95. If QN be the ordinate, BP + PQ = QN+BX-NX=BX+QN-SQ, which is greatest when QN=SQ, that is when Q is on the latus rectum. 96. If SZ and PG meet in Q siud QT be ordinate, TA : ^^ :: QZ : ZaS' :: QP : PG :: 7W : NG; .-. TN=2TA, 97. If Q F be the ordinate of the point of contact, TP^PV, Therefore the distance of V from TQ is twice the distance of P, or the locus of F is a straight line parallel to TQ. 98. If TPSQ be the parallelogram, the angles TSP, TSQ are equal; therefore TPSQ is a rhombus and jTlies on the axis. Therefore TSP is the angle of an equilateral triangle. The Parabola. 23 99. If /S'Z, SZ' be the perpendiculars on the second tangents, TQ, TQ' and PB be the common tangent, SY perpendicular to it, then angle A'S Y= AS Y= YSP ; .-. A' lies in SP^ and A in SP' ; .-. SP^SP". Now SQ.sp= sr-=SQ' . sr ; .-. SQ = SQ'; .-. sz=sz\ 100. If the tangent meet ^ F in F and the other parabola in Q, QM^ = ^AS,AM, AY^ = AS.AT, QM : AY=MT : AT; :. 2TM^=AT,A3L This can be constructed by tsikmg AM = 3fT, or by taking AM=2AT, the two solutions corresponding to the two points in which the parabola is cut by the tangent. CHAPTER III. THE ELLIPSE. 1. SD' = BC' = CS.SX, Therefore CDX is a right angle. 2. ST, SP are equally inclined to PT, since pST is parallel to S'P. Therefore ST^SP. 3. PN : PG' :: SY : SP :: ^C : CZ> :: PF : AG :: AG : PG\ Therefore PN=AG. 4. T lies on a circle of which QQ^ is a diameter and F centre ; therefore VT= VQ. Now QF2 : GP^-GV^ :: Ci)^ : CP^, or Kr^ . CV.GT-GV^ :: (7i>^ ; CPl Therefore TV : VG :: GD^ : GP\ Therefore GT : GF :: GJ)'+GP^ : GP\ or (7^2 : GF, GT :: AG' + BG^ : C/P-^. Therefore GT^=AG^ + BG'\ 5. Through T draw a straight line at right angles to AA' meeting AP, A'P in JS, E'. Then ET , PN v, AT \ AN :: GT-GA : GA-GN The Ellipse. 25 Now CT : CA :: CA : AN-, /. CT+CA : CT-GA :: CA+AN : CA-CN] .-. ^r : PiV^ :: ^'T : A'N :: ^T : PN, Hence PT bisects any straight line parallel to ET termi- nated by A'P, AP. 6. Draw CD parallel to the given line, and CP parallel to the tangent at £>. The tangent at P will be parallel to CD and the given line. 7. SB : XE :: SA : AX ;: aS'T? : aS'X Therefore XE=:SX, and AT=AS. 8. Draw 6^Z perpendicular to aS'P. Then PL = SB, and aST : SP :: PZ : PG^ :: SR : PG. The angle aS'PaS" is greatest when SP Y is least, that is when SY : iSP or BO : CD is least. Hence SPS' is greatest when CD is greatest, that is when CD = Aa Hence SPS' is greatest when P is on the minor axis. 9. CE^ = CP' + PE^ + 2PF. PE = CD^ + (7P2 ■¥2CD . PF=AC^-¥ BC^ + 2AC . BC ) :. CE=AC+BC So GE=AC-BC. {CP + GDf=AC^ + CB'' + 2CP . CD, which is greater than {AC+BC)% since (7P . CD is greater than PP. Ci> or ^(7.^(7. Similarly CP-CD is less than AC-BC, 10. Let /S"Q drawn parallel to SP meet the normal in A^and/S^Fin^. 26 The Ellipse. Then S'K^ S'P and KQ = SP ; therefore S'Q = AA\ 11. SYiS'Y' :: YP :PY' :: TP-TY: TY'-TP ::PG-SY:S'Y'-PG, 12. PS'Q is the supplement of QPS' + PQS\ and is therefore equal to the excess of twice QPT+PQT over two right angles, that is, is the supplement of twice PTQ. 13. Since CZ and SP are parallel, the angle CZP = SPY=SNY] therefore Y", Z, (7, N lie on a circle. 14. Let ^ Q and /S'P meet in i^. Then SA : /Si? :: SG : /S'P :: /S'^ : AX. Therefore R lies on a circle of which S is centre. 15. Since KPt is a right angle, t lies on a circle which passes through S, P, S', K, therefore GK : SK :: .S"^ : /S'P :: SA : ^X, and aS^^ : tK :: /S'F : SP :: ^(7 : CD. 16. If /S'P meet S'Y' in Z, then since S'Y'= YZ, SY' will bisect PG. 17. Let the circle whose centre is P touch the circles whose centres are /S', H in Q, i2. Then SP + PH= SQ-^QP + PH= SQ + HR. Hence the locus of P is an ellipse of which S and H are foci. 18. TN : TO :: PN : Ct. Therefore TN.NG : VT.NG :: PiV^ : ^.PA^. But PN^= TN.NG. Therefore CT.NG^Ct. PN= CB\ The Ellipse. 27 19. TP : TQ :: CD : AC :: BC : PF :: PG : BC, 20. PN^ : AF.A'r :: 7W^ : TA . T^' :: TN^ : CT'-CA^ :: 7W : CT :: CT-CN : CT :: CA^-CN^ : C^^ . .'. AF.A'F' = BCl 21. The perpendiculars from Ton /S'P, /S'Q, //P, HQ are all equal. Hence a circle can be described with centre T to touch SPy SQ, HP, HQ. 22. If P, Q be two points of intersection, PC bisects the angle ACa and QC bisects A'Ca. Therefore PCQ is a right angle. 23. If aS'P, i7Q meet in i2, PSQ + PHQ = 2PRQ - AS'()Zr- /S^Pi^, and SQH+ SPH+ 2RQ T+ 2RP T= 4 right angles, .-. PSQ + PHQ = twice the supplement of QTP. 24. Since t, P, aS', g lie on a circle, the angle PSt = P^^ = STP. 25. Q'ilf : PM :: ^(7 : ^(7 :: PN : QN. Therefore Q'M : CN :: CM: QN. Therefore QQ' passes through (7. 26. SY : SP :: BC : CD. Therefore SY. CD = SP. BC. 27. If T be intersection of tangents at A and P, then, since TC bisects AB, it is a diameter of the conic. Therefore the tangent at C is parallel to AB. 28. The angles SPT, HPt are equal. Also TP.Pt^ CD' = SP . PH, or TP : SP :: HP : Pt. Therefore SPT, HPt are similar. 28 The Ellipse. 29. PE=PE'=Aa Therefore SE= HE\ and the angles SCE^ HCE' are equal. Therefore the circles circumscribing SGE^ HCE' are equal. 30. The angles KPG, GPL are equal ; therefore KL is a double ordinate of the circle of which PG is diameter. 31. If e be the centre, QN the ordinate, and T, T the points where the tangent at P meets the tangents at the vertices, QN^ : SN.NH :: AT. AT : AH. AS :: BCT- : A'Sl (Ex. 20.) 32. Since the tangents are equally inclined to SP, S'P respectively, the bisector of the angle between them bisects SPS'j and therefore passes through the point where the axis minor meets the circle. 33. If PQRS be the quadrilateral, p, q, r, s points of contact, H the focus, the angle pHP - PHs, pHQ = QHq, SHr = SHs, rHR = RHq. Therefore PHQ + SHE = PHS + QHR = two right angles. 34. SG : SO :: SP :: SY{see Ex. 15) :: CE : EC :: PV ; VA. 35. The normals at P and Q will meet on the minor axis in K. Then angle between the tsmgexits = PKQ^PSQ. 36. The a\ixiliary circle lies entirely without the ellipse except at A and A' ; therefore AA' is the greatest diameter. The circle described on BB' as diameter lies wholly within the ellipse ; therefore BE' is the least diameter. The Ellipse, * 29 37. Let any circle passing through N and T meet the auxiliary circle in Q. Then CN,CT=CA^ = CQ\ Hence CQ touches the circle at Q, and the circle cuts the auxiUary circle orthogonally. 38. The angle PNY=PSY=PS'Y' = PNY', Therefore PY : PY' :: NY : NY, 39. PQ'' : TQ2 :: SY.S'Y' : TY. TY. But TQ'=TY.TY\ Therefore PQ" = SY.S'Y' = BC\ Therefore PQ^BC. 40. If QN and PM be the perpendiculars on the given lines passing through (7, R their point of intersection, RN : QN :: CP '. GQ\ therefore the locus of R is an ellipse of which the outer circle is the auxiliary circle. 41. SP : S'P :: SY : S'Y' :: SY'' : BC\ and S'Q : SQ :: S'Z' : SZ :: BC^ : SZ\ Therefore SP.S'Q : S'P.SQ :: SY'' : SZl 42. Let Cay Ch be the conjugate diameters, and Pm, Pn ordinates of P. Then Cm,CM=Ga\ and Cn^CN^Ch''', .-. CM. Pm : Co" :: CW : Pn.CN-, .'. the triangle CPM varies inversely as the triangle CPN. 43. CA V : CPT :: CA^ : CT^ :: CN : CT :: C/W : CPT; therefore the triangles CA F, CPN are equal. 30 The Ellipse. 44. Let TPQ, Tpq be the tangents intersecting the auxiliary circle in P, Q, p, q. Let E, e be their middle points. PE''-¥p6^ = ET^+ Te"- TP . TQ^Tp . Tq 45. Let OQ' be a diameter equally inclined to the axis with the conjugate to PP. Then the circles described through P, P\ Q and P, P\Q' will touch the ellipse at Q and Q\ Hence Q, Q' are the points at which PP' subtends the greatest and least angles respectively. 46. Draw the tangent Qr. Then, since the angles PSQ, QSr are equal, Q always lies on the tangent at the end of the focal chord RSr, 47. The triangle YCY' will be the greatest possible when YCY' is a right angle : P will then lie on the circle of which SS' is diameter. This intersects the ellipse in four points, provided SS' is greater than BB\ 48. The points where the lines joining the foci of the two ellipses meet the common auxiliary circle are points through which the common tangents pass. 49. The circle passing through the feet of the perpen- diculars is the auxiliary circle of the ellipse. 60. Draw QiV perpendicular to ^^. Then QN : NA :: BP : AP :: CA : AT, and QN : BN :: AT : AB. Therefore QN'^ : AN. NB :: CA : AB. Therefore the locus of Q is an ellipse of which ^^ is major The Ellipse. 31 51. PG, GN, NP are at right angles to CD, BR, RC respectively. Therefore the triangles CDR, PGN are similar. Therefore PG : CD :: PN : CR :: BC : AC. 52. Let Pas', QS meet the elUpse and circle again in^, ^. And let P'Cjp' be the diameter parallel to SP, Then, since pq is an ordinate, SQ : SP :: Sq : Sp :: Qq : pP :: AA' : P'p\ Again, P.S'./S'/; : ^.S'.aS'^' :: CP'' : C^l Therefore SR.Pp : 2i^(7*^ :: Py2 . j^/2^ or Pit? : A A' :: Py^ ^ ^^^2^ Therefore SQ \ SP w A A' : Qq, and Qq = P'p', 5;^. If /S'P meet S' Y' in Z', SL' = AA'; therefore SR = AC. 54. Since the directions before and after impact are equally inclined to the tangent at the point of impact, tlie lines in which the ball moves will touch a confocal ellipse or hyperbola. 55. Let the tangent at P meet the tangents at A and ^'inPandP^ Then, since the angles P.S'P, FSA and PaSP', F'SA' are respectively equal, S (and similarly S') lies on the circle of which FF' is diameter. 56. P', D\ the two angular points, will lie in PN, DM respectively. Therefore P'N : NC :: DM : NC :: DC : AC. Therefore P' lies on a fixed straight line through C. Similarly Q' lies on the other equi-conjugate diameter. 67. The angles aS'PaS", STS' are equal by Ex. 15. A SPS' : STS' :: SP , S'P : ST.S'T :: CD"" : ST\ 32 The Ellipse, 58. If T be the centre, then, since the angles TSP, TSA are equal, T lies on the tangent at A, 59. Let QL be the ordinate of Q, Then QL : LS :: CN : NP, and QL : LS' :: CM : MD. QL' : SL.SL' :: CM.CN : PN,DM :: ^(7^ : BC\ or Q lies on an ellipse of which SS' is minor axis. 60. If P is the corresponding point on the ellipse, and SZ the perpendicular on the tangent to the circle, SZ : AC :: CT-CS : CT :: AC^-CS. CN : AC^ :: aSP : ^(7; .-. SZ=SP. 61. The tangent at Q is parallel to the normal at P ; .*. the tangent at P is parallel to the normal at Q. 62. If PQ P'Q' be the parallelogram, the angle DHE is the supplement oi HPQ' + HQP'', that is, of SPQ + SQP) that is, oi DSE. Hence aS', //, i>, E lie on a circle. 63. Let the line through C parallel to the tangent meet the directrices in Z, Z'. Since the auxiliary circle is fixed, S Y, S' Y' are fixed straight lines meeting ZZ ' in fixed points y y\ And Cy . CZ= CX.CS= CA\ Therefore Z and Z' are fixed points. 64. The angle S'TZ=STY=SZY=com^\QmQ\\i of YZT, Therefore FZand /S"Tare at right angles. Qb. If G be the centre of the circle, GL bisects SP at right angles. Therefore SP is equal to the latus rectum. The Ellipse. 33 ^Q. If CZ be perpendicular to YY', the perimeter of the quadrilateral is equal to SS' together with twice CZ+ZY, which is greatest when CZ=ZYy that is when SPS' is a right angle. 67. Draw SZ perpendicular to S'Z the straight line on which aS" lies. I^et PS'r be the chord parallel to SZ. Produce PP' both ways to M and M\ so that S'M=S'M' = AA\ Then the lines drawn through M and M' perpendicular to SZ are fixed, and SP = PM, SP' = P'M\ Hence the ellipse will touch two parabolas having S for focus. 68. Let TQ, TQ! be tangents, Tthe middle point of Then QV. FQ'=CP'-CV' = CF. FT. Hence Q, Q', C, T lie on a circle. 69. Draw QM perpendicular to the minor axis. Then QG^ : AC^-CN^ :: BC^ : AC'', or QG-' : BG' :: AG'-GN'- : AG\ Therefore BG^-GN'-QN'' : GN^ :: BG' : AG^, or BG'^-MG'' : QM"" :: AG^+BG^ : AG\ Therefore Q lies on an ellipse of which BB' is minor axis. 70. If QM be the ordinate of Q, AM^ : GN^ :: QM^ : PN^ :: A3f,3fA' . AG^-GN\ Therefore AM . AA' : AM^ :: AG"" : GN\ or ^GN'^ = AG,AM. But ^Q.^0 : GP^ :: ^J/.^C : C7iV« ; therefore AQ,A0 = 2GP\ B. C. S. 3 34 The Ellipse, 71. SP : SN :: SO : CQ :: SC : AC-QR. Therefore SP .AG=SP .QE + SN, SO, But /S'P : XS+SN :: /S^ : CA or /S'P.^(7=XAS'.AS'(7+/S'iV^.>S'a Therefore SP.QE^XS. SC=BG\ 72. If the tangent meet the tangent at A in T, and aS^'F, /S"Z be perpendicular to TS, and the tangent, T,Aj F, /S'', Z lie on the circle of which S'T is diameter. The angles YTZ, ATS' are equal since ATS.S'TZ are equal. Art. 68. Therefore the chords FZ, AS' on which these angles stand are equal. 73. If P, Q, P', Q' be the parallelogram, p, q, p\ q' the points of contact, pq^ p'q' are parallel focal chords bisected by PCP\ But QGQ' bisects pp\ qq' and is therefore conjugate to PCP' and parallel to pq^ p'q'. Therefore CQ = CQ' = CA 74. If T be the point from which the tangents are drawn, ST, S'T are perpendicular to TP\ TP respectively. Therefore SP, S'P' are both parallel to CT, 75. CS'^ : CA'' :: CG : GN :: GG XT : GN , GT Therefore GG,GT=GS\ 76. If PG be the normal at the point of contact, GG,GT=GS\ Therefore 6r is a fixed point and P lies on the circle of which GTi^ diameter. 77. Let the given straight line pq meet the axis t. Let the tangents at p and q meet in Q. Let GQ meet pq in V and the curve in P Through P, Q draw P6^, Q(^' perpendicular to pq, meeting the transverse axis in G and G\ The Ellipse. 35 Then CG' : CG :: CQ : GP :: GP : GV :: GT:Gt', :. GG\Gt=^GG.GT=GS\ or G' is a fixed point. 78. Draw SY perpendicular to the tangent ; produce SY io L making L F- YS. The point of intersection of the circles described with centres L and P', and radii A A' and AA'- SP respectively will be the second focus. 79. Draw SY, SY' perpendicular to the given tan- gents. The point of intersection of circles described with centres Y and Y\ and radius equal to GA will be the centre. 80. If GS be drawn perpendicular to PQ, S will be one focus. If aS'P, PS' be equally inclined to GP and SQ, QS' to GQ, S' will be the other. Bisect SS' in G, and take GA in SS' such that 2GA = SP + PS'. If Jl be the foot of the directrix, GX,GS=GA-^. 81. Qq : Aq :: PN : AN, and Er : rA' :: PN : NA'. Therefore Qq.Rr :Aq. A'r :: PN"" : AN. NA' v.BG'^'.AG'v.SL-.AG. Now Aq : qA' :: Aq^ : Qcp, and A'r : rA :: A'r^ : Rr^. Therefore Aq.A'r : Ar.A'q :: AG'^ : SLl 82. By Ex. 75. GT : GS :: GS : GG; therefore TS : GS :: SG : GG. But TY : PY '.: TS : SG -, 3—2 36 * The Ellipse, 83. If be the intersection of the lines 84. TP : TQ :: Cr : CQ\ and the angles PTQ, P'CQ! are equal. Therefore PQ is parallel to P'Q', 85. /Si P, if, /S' lie on a circle, and the triangles SCt^ P YS are similar. Therefore St : Ct :: SP : SY :: CD : BG, or St.PN : Ct.PN :: CD . BG : BC\ Therefore St . PiV^= (7i) . BG 86. If the tangent at Q meet the minor axis hi f , the angle SQS^ = SfjS\ or ^^ is on the circle. Now QM.Cf = BG' = PN. Ct. Therefore QM : PN :: Ct : Clf :: Ct : Ct + St :: BG : BG+ CD by Ex. 85. 87. If S'Z be perpendicular to TY, the angle >S'2'F= complement of TZY\ (Ex. 64) - half supplement of YCY^ the angle at the centre, = GYY' ; and STY=S'TY\ 88. The tangents at L and L^ are perpendicular to the tangent at P, and therefore D and Z>' where they meet the tangent at P are on the director circle. Now DL : DP :: D'D : DP; therefore PQ bisects the angle LPL\ Therefore LP + PL' - diameter of director circle. 89. ABfAE are equally inclined to BCy and AB' = AD.AE, Therefore AB is 2i tangent. 90. If the tangent at P meet the tangent at A in T, TS, TS' bisect the angles PSA, PSA. The Ellipse. 37 91. If the chords of intersection NO, PQ 3neet in T and CD, CE, C'D\ G'E' are parallel radii, CD^ : (7^2 .. TN, TO : TP . TQ :: CD"" : C'E\ 92. IfPiVmeet (7i)in^, PK : PQ :: SG : ^P :: .S'^ : AX, and PiV^. P^= PG . PP= ^(72. Therefore PQ varies inversely as PiV. 93. Draw perpendiculars S Y, CE, S' Y' , SZ, CF, S'Z' on tangents TP, TQ, then CT'' = CF^+ TF^=^CZ^+TZ. TZ = GA'^^SY. S' Y' = CA'^ + CB\ 94. If the circle meet the minor axis in K and L, the tangents at P and Q meet either in jSTor L, see Ex. 15. 95. This problem is equivalent to Ex. 45. 96. Let CF bisecting the chord QSQ' meet the curve in P and directrix in T Let DCB' be the parallel diameter. Then SR : SG :: GS- GR : GS :: GT- GV : GT :: GS^-GG"" : SG^ :: SG. GS' : GS^ :: SP.PS' : GA'- :: CD' : (7^^ .. ^^2 . 2)/) 2 by Art. 76. 97. Let the tangent at Q meet PN in P' and the axis in U. Then GT . GN= GA' = GM. GU ; therefore GT : GM :: GU : 6W, or TM : 6'Jf :: iV^Z7 : GN. But PiV^. Q'Jf : Q'M^ :: 2W : TM, or PN.Q'M : GM.MU :: GN.NT : GM.NU. Hence PiV^ . Q'ilf : 6W. iV^r :: (7Jf . ^iff : (7Jf . PW :: QM^:P'N,QM, Now PiV"2 . ON, NT :: BC'' : AG' :: ^^2 . ^^^^^2, Hence PN.Q'M : PiV^2 .. q/j/2 . p']S[,qM. 38 The Ellipse. Therefore P'N : PN w AC \ BC, or P' is on the auxiliary circle. 98. The diameter bisecting PQ is fixed, hence V the centre of the circle, is a fixed point. VM bisecting RS at right angles, is a fixed straight line ; PQ and BS are equally inclined to the axis ; .'. CM and CV are equally inclined to the axis. Therefore Jf is a fixed point, and RS a fixed straight line. 99. The angle ^/S'a = BAG+ SB A + SCA =BAG+ HBG+ HCB = BAC+ supplement of BHG. Hence if BHG is constant, BSG will be constant. 100. The angles SPT, HPt are each equal to SQH j also STP^tQS-PtH = HPt-PtH=PHt. Therefore TP : SP :: HP : Pt, or TP.Pt=SP.RP = GD\ Therefore GT^ Gt are conjugate. 101. GT bisects PQ and is parallel to SP ; .'. T is the foot of the perpendicular from S' on PT. See Cor. (3), Art. (66). 102. SG : GYis a given ratio and SY is fixed. 103. Take p a point near and let the focal chord p'Sq' meet pq in O; PQ : p'q' .'. pO.Oq : p'O . Oq :: ST. Oq : Sp' . Oq' :: ST.pq : /SJ^ . j^V ultimately. The Ellipse. 39 104. Dropping perpendiculars from the focus on the sides, their feet are the middle points, and, as they lie on a circle, form a rectangle ; the diagonals, intersecting in H, are therefore at right angles, and SAD can be proved equal to HAB, 105. If CD, CP meet the directrix in E and G, ES is perpendicular to the chord of contact of tangents from E, which is parallel to CP. 106. If CP, DC meet the tangent at ^ in Q and R^ prove that AQ.AR = BC^=AS.AS\ Then QSA =ABS', and QBA = A QS\ CHAPTER IV. THE HYPERBOLA. 1. If the circle whose centre is P touch the circles whose centres are S and H m Q and R, SP'-HP=SQ'-HR. Therefore P lies on an hyperbola of which S and // are foci. 2. SD'^=BC'= CS^- CA'^CS^-CX, CS= CS, SX. Therefore the triangles SCD, SDX are similar. ,3. If the straight line meet the curve in P and the directrix in F, SF : SX :: CS : CA :: SA : AX :: SB : SX. Therefore SF=SE. Draw PX perpendicular to the directrix. Then PF : PX :: SO : CA :: SP : PK, Therefore SP^PF. 4. Draw SD^ SD' perpendicular to the asymptotes. Then DD' is the directrix. 5. If the asymptote meet the directrix in Z>, then DS drawn at right angles to CD meets the axis in the focus. The Hyperbola, 41 6. If PK, QL be the perpendiculars from the given points on the directrix PS—SQ-PK-QL wliich is con- stant. Therefore S lies on an hyperbola of which P and Q are foci. 7. If the circle inscribed in the triangle ABC touch the sides in D, E, F-, D, (7, D being given, BA-CA=BF-EG=BD-DC. Hence A lies on an hyperbola of which B and C are foci and D a vertex. 8. FN : Pg :: SY : SP :: BC : CD :: Pi^ : ^(7 :: AC : P^. Therefore PN=AC, 9. Draw Ci) parallel to the given line and CP parallel to the tangent at Z>. Then the tangent at P is parallel to CD and the given line. 10. Let A'P and P'A meet in Q, and draw the ordinate QM. Then QM : A'M :: FN : NA\ and QM : ^3f :: PW : NA. Therefore QM' : AM, MA' :: PiV^ . aN,NA' :: ^(72 : ^C^, or Q lies on an hyperbola having the same axes. 11. Let the tangent at P, AF and A'P meet the minor axis in t, E and E\ Then (7^ : FN :: C^ : ^iV^ :: CA.A'N : AN,NA\ and Ci^' : PiV^ :: C^.^JV :: AN.NA'. Hence CE-CE' : FN :: 2(7yP :: y4i\^. iV:^'. Now P-.V2 : AN.NA' :: PiV. (7if : ^(72; therefore C^- CE' = 2Ct, Therefore Ft bisects every line perpendicular to AA' terminated by A'F^ A P. 42 The Hyperbola. 12. SPTh an isosceles triangle since pST is parallel to S'F. Therefore SP = ST. 13. Draw JSD perpendicular to the asymptote and /SK parallel to it. If TS bisect the angle PSK, T being on the asymptote, TP is the tangent at P. Draw aS'F perpendicular to it. Then CM which bisects i> F at right angles will meet the asymptote in the centre G. DX drawn perpendicular to CS will be directrix. 14. If the tangent at P meet the tangents at A and A^ in Z and Z', ZS and Z'S are the internal, and external bisectors of the angle ASP. Hence the foci lie on a circle of which ZZ^ is diameter. 15. Draw P^ perpendicular to the directrix and DS at right angles to the asymptote. Draw cxs at right angles to the directrix meeting it in iv. With centre P and radius PS such that SP : PKwcs : cZ), describe a circle meeting Ds in S. Then S is the focus. 16. Draw Qq', Pp and Rr, Qq parallel to the asymp- totes. Then GP : GQ :: Gp : Gq' :: Gq : Gr :: GQ : GB. 17. QG^-GB^ : GN^-GA^ :: BG^ : AG^ :: PN^ : GN'^-GAK Therefore PN^ = QB . Q^'. 18. DN : NA :: QM : MA and ^iV^ : NA' :: QJ/ : MA'. Therefore ND.NE : AN.NA' :: ^3/2 . ^ jf . 3f^' :: PiV2 : AN.NA', or PN' = ND.NE. 19. ^, ^', F, Z lie on the auxiliary circle. Therefore AT. TA'= YT . TZ. The Hyperbola, 43 20. If the tangent at P meet the asymptotes in L and L\ CD-PL = PL' \ therefore Q divides LL' in a constant ratio. Draw QH, QK parallel to the asymptotes. Then QH. QK varies as CL . CL' and is therefore constant. Therefore Q lies on an hyperbola having the same asymp- totes. 21. This is equivalent to the preceding. 22. Since SK= S'K, JTlies on the circle passing through S, S' and P, and since KPt is a right angle, t lies on the same circle. Therefore GK : S'K :: SG : SP :: SA : AX, and aS'^ : tK :: SY : SP :: BC : CD. 23. Let P, P' be the points of trisection of the arc SS' and let XM bisect aS'aS'' at right angles, then SP = 2PM ^ndS'P' = 2P'M. Hence P and P' lie on hyperbolas of which S smd S' are foci and XM directrix. If C\ a be the centres CS=4.CX and CS' = 4.CX. Therefore C and C are the points of trisection of the chord SS'. 24. Draw SZ parallel to the asymptote : the angle STQ=TSZ=TSP. Therefore SQ = QT, 25. Since the hyperbolas have the same asymptotes the ratios CS : BG : CA are constant. Let NP be the fixed line parallel to an asymptote, and PQ proportional to an axis. Then PQ'^ varies as CS^, that is, as CN .NP, that is, as NP. Hence Q lies on a parabola having NP for a diameter. 44? The Hyperbola. 26. PY.PY' = AC^-CP''=CD'^-BG'^=CS^ fSP--S'P\^ ^y=( 27. Let TP meet the other asymptote in T'^ then PT=PT\ Therefore PQ = R'P = QR. 28 Draw OrR parallel to PQ, meeting the ellipse and hyperbola in r and R. Let Oa, Oh be the axes, then since OP, Or are conjugate in the ellipse, and OQ, OR in the hyperbola, if PN, QM, rly RL be the or din at es, 0N.0b = rL0a', PN .0a = 0l.0h', QM.Oa ^OL,Ob', OM. Oh ^RL . Oa. Therefore PN : ON :: QM : OM since rl : 01 v. RL : OL, or OP and OQ are equally inclined to the axes. 29. Through S draw SO parallel to the bisector of the angle between the asymptotes meeting the asymptote which is given in position in G. Draw SD perpendicular to that asymptote, and DX to GS. Then if A be taken in GS such that GA'^^GX.GS, A is vertex. 30. If the tangents at P and Q meet in T, then since the perpendiculars from T on SP, SQ, HP, HQ are all equal, a circle can be described with centre T to touch SP, SQ, HP and HQ, 3L If GL, GU be the asymptotes, S will lie in the bisector of the angle LGL\ Draw PL, PH parallel to the asymptotes to meet them in L and L' ; and take S in GS such that GS^'^^GL . GL\ then iS' i3 a focus. The Hyperbola. 45 32. If the conjugate diameters PCP^, DOB' be given, complete the parallelogram LML'M' formed by the tangents at Z>, P, D' and P*. The diagonals LL'^ MM are the asymptotes and the axes bisect the angles LCM^ LCM'. 33. Let QT'and RQ meet the asymptotes in L and M, Then QL : PR :: RH : 7Z :: CR : CT :: RM : T^ :: PA" : QM -, therefore QL . QM^ PH . PA', or Q is on the curve. 34. Let CL bisecting the angle ACE' meet PN in Z, draw Q J/ parallel io LC. Then CX is proportional to CN ; therefore (7A . CM is proportional to CiV". NT+ CN"", that is to CA\ Hence Q lies on an hyperbola of which CL and CB are asymptotes. 35. Draw S Y perpendicular to the tangent and produce it to Z making YZ-^SY. Then if Q be the point of contact, and P the fixed point, HP-PS=HQ'-QS=HZ; therefore HP - HZ = PS, or the locus of H is an hyperbola of which P and Z are foci. 36. If P T, Pt the tangents to the ellipse and hyperbola meet BC in T and t, then since the curves have the same conjugate axis, for CA''= CS"" + GB^ Ct.PN=BC^=CT.PN, or CT^CL 37. This problem is the converse of Ex. 3. 46 The Hyperbola, 38. If (r be the point of intersection CG=i GP, or G lies on an hyperbola having the same asymptotes. 39. The angle GYT = S'P Y' = S'NY' since S', P, N and Y^ lie on a circle. Therefore Y, Y\ C and N lie on a circle. 40. U SY meet .S^'P in Z, jSY=YZ; therefore S'Y bisects PG. Similarly jSY' bisects PG. 41. BC^ : ^6^2 .. jsTG : ON :: CT.iVG^ : CN. CT; therefore CT.NG = BCi 42. The angle STP = TS'P + S'PT= TS'P + SPT = PS'S+ jSS't = supplement of PSt. 43. Pt.PT= CB" = SP . S'P, or SP : PT :: Pt : S'P; and the angles SPT and tPS' being equal, the triangles SPT, i^P/S'' are similar. 44. The circles SC^, B'GE stand upon equal chords SG, S'G and contain equal angles SEG, jS'E'G, since CB is parallel to the bisector of jSPjS\ 45. If the tangent at P meet the tangent at Aj the vertex of the branch on which P lies, in T, Tis the centre of the circle inscribed in the triangle jSPS', since T/S, TS' bisect the angles ASP, AS' P. 46. GT : GA :: GA : (7iV^ :: GP : (7Q, or ^$ is parallel to PT, 47. GE : (7^ :: ^/S^ : (7^ ; therefore GE^ GS; but GD = GA. Therefore AD and /S'^ are parallel. 48. If E be the centre of the circle and EK its radius, EK : GE :: ^(7 : SG, The Hype^'hola. 47 or EK : CA :: BG : SG+BG :: {SG-BG)BG : C^* And SR' : CaS' :: BG : ^C :: aST? : BG; therefore ER' : jSR :: GS-BG : ^(7, or RR' : GS-BG :: aS72 : ^(7 :: J5(7 : GA, Therefore EK=RR', 49. PM=^PL; therefore GL = (ralf. 50. If Ga, Gb be the conjugate diameters and one hyperbola touch the ellipse in P, the tangent at P will meet Ga, Gb in T, t, such that TP = Pt = GD, Hence PD is bisected by Gt, and tD touches the other hyperbola and is parallel to GP. 51. If LL and MM' be the tangents, GL : GM :: GM' : GL\ or Zilf' and Z'iltf are parallel. 52. If the tangent at P meet the tangents at A and A' in F and P' and QM be the ordinate of the centre of the circle, QM : MS :: .S'^ : AF, and QJf : MS' :: /S"^' : ^'P'. Hence QM'^ : SM.MS' :: /S'^^ . aF.A'F' :: aS'^^ . ^(72 (Art. 126). Hence the locus of Q is an hyperbola of which S and S' are Tertices. 53. If PM be perpendicular to the directrix, PK : PJIf :: GS : (7^ :: SA : -4X :: /S'P : PM, or PK=SP. 48 The Hyperhola. 54. Let PD meet an asymptote in riy draw PI, Dm parallel to Cm Then Dm . Dn = Pn . PI ', therefore Dn = Pn. Therefore if LPL' is tangent at P, LD is tangent at Z>, and (7P, CD are conjugate. 55. If QP meet the asymptotes in q and r, qQ = rR ; therefore if PPe be the tangent at P, CL : CN :: qQ : Pe :: P^ : gi2 :: CN : CJf. 56. If the circle intersect the axis in h, B, CB,Cb=CS^ or CB^-\- CB,Bb = CA^-hOB^; therefore CB.Bh = CA\ 57. Let the straight line q'Q'APQq meet the asymp- totes in Q\ Q. Draw POP' parallel to ^P terminated by A'q, A'q, Then PQ' = AQ^CP^ Qq, and Q'q'=CP'=AQ' = PQ; therefore Pq ' = Pq. 58. T is the centre of the circle inscribed in the triangle PS'Q, therefore the difference between PTQ and half PS'Q is a right angle. 59. Draw CD, CE parallel to OA, OB and PH, PK parallel to and terminated by CE, CD. Then PH : OD :: CK : CE :: CP : CA :: CB : CP :: CD : CH :: OE : P/r; therefore Pi7. PK= OD . OE, or P lies on an hyperbola having CD, CE for asymptotes. 60. Draw PH, PK, QH\ QK' parallel to the asymp- totes. Then PL : QM :: PH : QH :: QK' : PK :: QiV : PR, or PL.PR = QM.QN. The Hyperbola. 49 61. If TK, TN be perpendicular to the directrix and SP, TK=SN. Therefore ST : TK :: ST : /SiV a constant ratio, and the angle between the asymptotes is double PST, that is, double the external angle between the tangents. 62. Q'V^-RV^ = GD'^ = RV^''QV\ or QV^ + Q'r' = 2RV\ Again CT, CV= CP^ = CV. CT ; hence CT= CT' 63. If r be the middle point of PQ, then since B, V are the middle points of LRU and LPQl^ RF is parallel to the asymptote GUI. Hence P3f+QN=2RB, 64. If TP, TQ be the tangents, PTQ, STS' have the same bisector which passes through the point where the circle meets BCB', 65. The tangents at P and Q intersect in t on the circle and BCB'. Hence the angle PtQ = PSQ. 66. The angle PNY= PSY= PS' Y' = supplement of PNY\ 67. The triangle YCY' is greatest when YGY' or SPS' is a right angle. In that case P 7" meets BC in t such that Ct = CS; therefore CS. PN= Ct . PN= BGK 68. The triangle SPS' : triangle StS' :: SP . PS' : St^ 69. S'P is parallel to CFand S'Q to CZ. Therefore S'T is parallel to bisector of YCZ and is perpendicular to YZ. 70. If G be the centre of the circle, GL bisects SP ; therefore SP = 2PL = 2SR, B. c. s. . 50 TJ^e Hyperhola. 71. The tangent at Q is parallel to the normal at P, therefore the tangent at P is parallel to the normal at Q, or CP is conjugate to normal at Q. 72. If F be the point from which the tangents are drawn, SP and S'P' are both parallel to GY. 73. SC"" : AC"" :: CG : CN :: CG . CT : CN . CT, or CG.CT^SCl 74. By Ex. 73, G the foot of the normal is a fixed point ; therefore P lies on the circle of which TG is diameter. 75. If TP, TQ be the tangents, CT will bisect PQ in and CT.CV=CT\ or PC is a tangent at F. 76. Let GQ meet the conjugate in G\ Then C(^' : QG :: CiV^ : NG :: ^C^ : 56^2, Therefore, by Art. Ill, QG' is normal at Q. 77. If PM be drawn perpendicular to the directrix of the parabola the angle PTQ = SPT-SQT=h8iU SPM- 78. If abed be the quadrilateral and S lie on the circle the angle Hcd = Scb = Sab = Had, or H is on the circle. 79. If PP' be the chord of contact and CV bisect PP' then CF, PP' are parallel to a pair of conjugate diameters in both conies. Hence if from a common point Q, a double ordinate QVQ' be drawn parallel to PP\ Q' must lie on both curves. Similarly RR' the line joining the other two common points is parallel to PP\ The Hyperlola. 51 80. If SD, SD' are perpendiculars from the common focus. on the asymptotes, DD' is the tangent at the vertex of P and a directrix of H, If P be a common point, and PM perpendicular t'O SP : PM :: SG : CA, but SP = PM+SX. Therefore SP : SX :: CS : CS-CA :: CS : AS. Hence ^/S'. SP = SX. CS=BC^ = AS. SA \ or SP=A'S. Therefore A'P touches the parabola at P, 81. With centre P, the given point and radius of the given length describe a circle meeting the other asymptote iwp. Then pPQq is the line required. 82. Let CB, GA be semiaxes of the ellipse, Ga, Gb of the hyperbola. Let PN meet the asymptote in Q, then QN"" : GN^ :: Gh^ : Ga% or QN^+GN^ : Ga^ + Gb^ :: CW^ . ^^2. but SP + S'P = 2GA, and SP-S'P = 2Ga. Rence 4:GA.Ga=SP^-S'P^ = SN^-S'N^ = 4.GN.GS; therefore (7^^ . CS^ ... (7jv^2 . (7^2 .. qn^ + cN'- : (7a^ + C&2 :: GQ^ : (7>S^. Therefore Qlies on the auxiliary circle of the ellipse. 83. Let Q be a common point. Then SQ - QH=AA' and SQ-QP=SP- 2PH =^AA'-PH, Therefore QP = QH+ PH, or Q must be the other extremity of the focal chord PH. 4—2 52 The Hyperbola, 84. li A'K meet the directrix in F^ then, since SA' = 2A'X, FA'S is an isosceles triangle and FS is parallel to KD, Also A'F : FP :: AX : XN :: ^'/S' : SP or /^aS' bisects the angle ASP ; tlierefore if SP and i>^ meet in Q, QSD is an isosceles triangle. Therefore Q lies on the circle of which A'D is diameter. 85. This problem is a particular case of Ex. 61. 86. PL.PL' = PL'=CD'^^PG.Pg', therefore G, g, L, L' lie on a circle of which Gg is dia- meter. C is on this circle since GCg is a right angle. The radius of this circle varies as Gg and therefore as CD and therefore inversely as the perpendicular from C on LL\ 87. If PCP\ DGD' be conjugate diameters and Q any point on the curve, QP' + QP'' = 2CP^+2CQ^; QD'' + QD"'- = 2CD'^-¥2CQ\ Therefore eP- + QP"" - Qlf^ - QU^ = 2CP'^- 2GD'^ = 2AC^'- 2BCI 88. If S^U, S'M' be drawn parallel to the asymptotes LS\ MS' bisect the angles PS'L, PS'M\ Hence ZAS^'i^^half the angle between the asymptotes. 89. If Pr meets the tangent at A in F, T.S' bisects the angle ASP\ therefore SP : ST :: PV : VT :: AN : AT. 90. If P is a point of intersection, let the tangent and normal of the ellipse at P meet the transverse axis in T and G, and the conjugate axis in t and g. The Hyperbola, 63 Then, PT being the normal of the hyperbola, the semi- axes of which are A'C and B'C, CT : CN :: SC^ : A'C\ Art. Ill, .-. AC^ : CN^ :: SC ; A'C; .',CN,SG^AC.A'a Again, Pg being the normal of the ellipse, Cg : PN :: SC^ : BG\ Art. 72, and Cg,PN=EC\ .'. B'C^ : PN' :: SC^ : BC^ and 'PN,SC=BC.BV. Hence, if PN meet the asymptote in Q, QN : CN :: B'C : AV, and it is easily deduced that QN : PiV :: AC : ^01 91. Let ABCD be the quadrilateral. A, B, and (7 being fixed points. Then AB+CD=BC+AD, or CD-DA = CB-AB, Hence D lies on an hyperbola of which A and C are foci. 92. Since Q, S, C, t lie on a circle, the angle tQC=tSS' = SPt, hence CQ is parallel io SY and CF, aS'Q are equally inclined to SY-, therefore SQ = CY=CA. 93. Draw SY perpendicular to the tangent and pro- duce to Z making SY= YZ. Then if P be the point of contact HZ= HP -SP = AA\ Hence the locus of ^ is a circle of which Z is centre. 94. RS and VS' bisect the angles PSQ and PS'Q; let QS, S'P meet in Z. 54* The Hyperbola, Then RSP + FS'Q = half PSQ + half PS'Q = half QSP + half SZS'-hsiU SQS'=QSP + hBlf SPS'-holf ^QS' = QTP+ TQS-SQT=PTQ. 95. CgP is an isosceles triangle, and the angle CGt = CPT^TCP; therefore PG = Ct and CU' =-PG . Pg=Cg .Ct= CS\ 96. Since the asymptote CD bisects BA, CD is parallel to the axis of the parabola and BA is parallel to the other asymptote. If QPVP'Q' parallel to BA meet CD in F, er=rC' and PF=FP'; therefore QP = Q'P\ 97. Let EL be the ordinate of E, and draw EF perpendicular to PN. Then, CD being conjugate to CP, the triangles CZ>J/ and PFE are similar and equal. /. CL = CN+ EF^ CN-\- DM, .\CN'.CL','.AC:AC+BC, and similarly EL : PN :: BC : BC- AC; /. EL^ : {BG-ACf :: PiV^^ : BC^ ::CN^-AC^:AC^ : : CL' - {A C+ BCf : {A C+BCf. 98. If PM be drawn from the centre perpendicular to BC AP : PM :. PC \ PM, a constant ratio; therefore P lies on an hyperbola of which A is focus and BC directrix. If S be the other focus and SP meet the circle in Q SQ = SP-PA = constant, or, the envelope is a circle of which S is centre. 99. The conies will be confocal having their foci H and H' on PG, such that PH^=PT.Pt = CD\ For their locus see Ex. 9 on the ellipse. The Hyperbola, 55 100. If SY, SZ, S' Y\ S'Z' be perpendiculars on tan- gents at right angles GT'^-GA''=-TY, TY' = SZ.SZ' = CL'K If SYZ, S'Y'Z' are perpendicular to parallel tangents and CWW be the perpendicular through the centre 2CW=SY^-S'Y'; ^CW' = SZ^S'Z\ and S Y- SZ=^ S' Y' + S'Z' ; Hence CW^-CW'^=CB^^CB'\ 101. The chord QR is inclined to the axis at the same angle as the tangent at P and is therefore always parallel to a fixed line. 102. TP and the asymptote subtend equal angles at S' ; :. PS'T=S'TC=STP. 103. SF'- = FX"" + /S'X' = CF^ - (7X2 ^ ^X* = CF^ + CS^-2GS.GX = GF'^ + GS^-2GA'^ = GF^-GA'^ + GB^ = square of tangent from F = FA.FB. 104. If S be the focus of the ellipse and S' of the hyperbola, GS : GA :: GA : GS'; .'. aS' and S' coincide with the feet of the directrices. The relation, GN. GT=GA\ proves that S and S' are the feet of the ordinates, and the relation, Gt,PN-BC\ proves that t and t' are on the auxiliary circle. Also Gif : GS=AG : GS=GS' : Gt; .*. the tangent intersects at right angles. 105. For PG.Pg=GD^, Art. 123, = PL'^- PL .PL\ and the diameter of the circle is Gg, which varies as CJJ, Art. 123, and therefore inversely as GY, CHAPTER V. THE RECTANGULAR HYPERBOLA. 1. The angle PCL = (7ZP= complement of ZCF. 2. QV^=VP. Vp, hence VQ touches at Q the circle QPp^ 3. LP = PM=GD = PG = Pg. Hence LGM is a right angle. 4. If X if be the straight line, and G be the corner of the square, GL . GM is constant, hence LM touches an hyperbola of which GL, GM are asymptotes. 5. Let AP^ A'P' meet in Q, and draw the ordinate QM, Then QM : MA :: PiV : NA, and Cif : MA' :: PW : iVL4'. Hence QM^ : ^if/. JO' :: PN"^ : AN.NA' ; therefore QM^ = AM. MA'. Hence Q lies on a rectangular hyperbola having AA' for transverse axis. 6. If P, P' be joined to Q meeting an asymptote in R and B', the angle QRL = (7XP - QPL = LGP'-PP'Q= GR'P' = QEL, The Rectangular Hyperbola. 57 7. Produce LP to M making PM=PL, then if MG be drawn perpendicular to the given asymptote CL^ C is the centre. In CS the bisector of the angle LCM take aS* such that CS is a mean proportional between CL and CM. Then aS' is focus, and X the middle point of CS is the foot of the directrix. 8. The angle DCL^PCL, and D'CL = P'CL'y hence DCD' = PCP'. 9. A diameter is a mean proportional between the parallel focal ciiord and AA', therefore focal chords parallel to conjugate diameters are equal. 10. As in the preceding, focal chords at right angles are equal, since diameters at right angles are equal. 11. If CD, Cd be conjugate to CP in the ellipse and hyperbola, (7i)2 ^ SP . PS' = (7^2 = CP\ 12. PN=CM) DM=CN', and CD = CP. 13. SP.PS' = CD'' = CP^. 14. QV^=CV''-CP^=CV^-CT.CV=CV.VT. Hence Q V touches the circle CTQ. 15. If i) be the intersection of tangents at A and B, CD'^^AC'^BC^SCK Hence D lies on the circle of which SS' is diameter. 16. If LPM^ G'PG be the tangent and normal to one, LP = PM=CD = CP = PG = PG'; therefore GG' is the tangent to the second hyperbola. 17. The angle CRT=CQT+RTQ = 2CLQ + LTL' = CLM+ CLT= CLE! + L'Cq = TEQ. Hence C, T, R and R' lie on a circle. 58 The Rectangular Hyperbola, 18. QR^ = CN^=CA'' + RN^=CA^ + CQ^=AQ\ 19. Let AQjAEhe the fixed straight lines, and P the middle point of QOR. Through G the middle point oi AG draw GH, (7A^ parallel to AQ,AR, and through P draw PHM, P^iV^ parallel to AB, AQ. Then the complements AG, HK about the diagonal MN are equal. Therefore PH . PK'i^ constant, and P lies on a rectangu- lar hyperbola, having GH and GK for asymptotes. 20. Draw QB perpendicular to AB, and make AB equal to GD, then ^ is a fixed point. Then AD : AB :: BG : GD :: QB : DP, or PD.DA^AB.BQ. Hence P lies on a rectangular hyperbola of which AB is one asymptote. 21. If Z>, ^, P, be the centres of the escribed and inscribed circles, OG,GF=DG.GE, since the triangles OGE, DGF are similar. Hence the hyperbola is rectangular since diameters at right angles are equal. 22. If the diameters of the parallelogram LML'M' meet in C, the angle SLS'^SL'S'. Hence S and S' lie on a rectangular hyperbola circum- scribing LML'M'. (Art. 137.) 23. If PSq, SQq be the chords and Z> be a point on the directrix, such that DS bisects the angle QSp, then D will lie in pq. But DS is perpendicular to the asymptote since Pp, Qq are equally inclined to it, therefore D lies on the asymp- tote. Similarly Pq, Qp meet in D', the foot of the perpendicular from S on the other asymptote. The Rectangular Hyperbola, 59 24. If V be the middle point of PQ and POP' be a diameter, the angle VNP = NP V= QP'P = PO V, Hence O, P, F, N lie on a circle. If OQ meet the given tangent in T, produce OQ to V, making OV b. third proportional to OT and OQ ; with centre F and radius, a mean proportional between KO and VT, describe a circle meeting the given tangent in P its point of contact. In the tangent measure off PL = PM=OP, then OL and OM are the asymptotes. 25. Draw PD perpendicular to the base QR, then, since PD'- --= DQ . DB, DP is the tangent at P. 26. Let a circle on DE meet the hyperbola in P and Q, draw the diameters PCP', QCQ\ Then, since the angle DPE is either equal or supplemen- tary to DPE and DQE to DQ'E, the similar circle on the other side of DE, will meet the curve in P' and Q'. 27. Let 0^2), 0^(7 be the fixed straight lines, PM, PN, PL perpendiculars from the centre of the circle on J5(7, AD and the bisector of the angle AOB. Let PM, PN meet OL in m, n. Draw X/, LI', Pr, Pr' perpendicular to BG, AD, LI, LV respectively. Then, BW^ 4- MP'^ = Am-¥ NP\ or Pm-PH'^^BM^'-AN'' which is constant, P]Sr' = {Ll' + Lr'f = {Ll-Lrf + ALl'.Lr' = PM'' + 4.Ll.Lr, Hence the rectangle LI . Lr is constant ; but LI : OL in a constant ratio and Lr : PL is constant. Therefore PL . LO is constant, or P liies on a rectangular hyperbola having OZ for an asymptote. 60 The Rectangular Hyperbola. 28. If P be the point of contact CL = 2PN, CL' = 2CN; therefore CL .CL'^2 Ca^ - 4.PN. CN. Hence CN.CL : Oa^ :: Ca^ : PN.GL, or AC^ : Ca^ :: Ca^ : C£\ 29. Draw CF conjugate to PQ, Then, the angle • TPQ = PP'Q = PCV= GPQ' ; therefore the angles CPQ, TPQ' are equal. 30. Let DB, DC meet the asymptotes in h and c : draw AH, AK parallel to the asymptotes. Then OBb, OAK are similar triangles, also OCc, OAK. Hence OB'.OA :: Oh : OK :: OH : Oc :: AK : Oc :: OA : OC ; therefore A lies on the circle of which BC is diameter as does D, 31. Let the tangents at P and Q meet the asymptote in L and M. The angle PCQ = PCL - QGM^ PLC- CMT^ L TM = supplement of PTQ. 32. Each hyperbola passes through the orthocentre of the triangle ABC, Hence D is that orthocentre. Now the line joining the middle point of ^^ to the middle point of CD is a diameter of the nine point circle. And ^ 52 +(72)2= square q^ diameter of circumscribed circle. Hence the circles intersecfc at right angles. 33. PN''=CN^-CA^=CN^-CN.CT=CN.NT. Hence the triangle GPN is similar to PTN, and therefore iotTG The Rectangular Hyperbola. 61 34. This problem is a particular case of Ex. 61 on the hyperbola. 35. CM, CN which bisect PP\ PQ' are conjugate being equally inclined to the asymptotes; therefore PQ' is a diameter. 36. If AB he a diameter of the hyperbola, CD subtends, at A and B, angles which are both equal and supplementary, and are therefore right angles. 37. If AQ\ BQ meet in R, the angle Q^Q' = supple- ment of QBQ' = RBP', therefore R hes on the circle. 38. Let CV, CV bisect PQ, PQ. The angle PRQ^PQL= VCQ = CQ V, so PR'Q=.CQV. Draw VM perpendicular to CQ, then PQ : QR :: VM : CV, and P'Q : RQ :: VM : QF. Now PQ = 2CV, and PQ = 2QV', therefore QR = QR\ 39. Let PN meet CF in IT, then CF varies as CG, and PF varies as PK, or Ct, or CT. Hence PF, FC is proportional to CG, CT or CSl Hence P lies on a rectangular hyperbola having CF for asymptote. 40. If the tangent at Q meet in V the line joining the fixed points A and B, VQ^=VA,VB. 41. If the chord QR meet the tangent at P in B, RPL = QPE--^PRQ. 62 The Rectangular Hyperbola. 42. If D be the point, CD.CT=2AC'^^ndLCD = Aa 43. CP = CD and are equally inclined to the asymptote. 44. Let BAD be the given difference, and draw CL parallel to^Z>, meeting BA in X. CAL, CBL are similar triangles ; CL^=AL.BL, 45. If tQT, t'QT' be the tangents, and SM perpen- dicular to the axis, SQM=SQT-MQT=S'Qf-CtT = QS'M+ QT'C- CtT= QS'M; /. QM^=SM.S'M. 46. If V is the middle point of OP, OTV= VOT, v DTP is a right angle, = OCT, Art. 136; ' /. O, F, (7, 7" are concylic, and .-. VCT=VOT=OCV. Similarly, if U is the middle point of OQ, OTU= UTQ. CHAPTER VI. THE CYLINDER AND THE CONE. 1. Take two points E and A on the generating line, and draw EX at right angles to the axis, making ^X equal to^^. Then the plane containing AX, and perpendicular to the plane EAX will cut the cylinder in an ellipse of the required eccentricity. 2. Take two points E and A on the generating line and with centre A, and radius twice EA, describe a circle meeting EF in X. Then the plane through AX perpendicular to the plane EX A will intersect the cone in an ellipse of the required eccentricity. 3. Take two points EA on the generating line, the least angle ot the cone will be, when EA is double the perpendicular from A on EF, that is when the semi-vertical angle is equal to the angle of an equilateral triangle. 4. A tangent plane to a cone touches it along a generating line OF, hence OF is parallel to all sections parallel to the tangent plane which are therefore para- bolas. If C be the centre of the sphere FES, the ratios CS : CA and CA : CO are constant, and the angle OCS is constant, therefore COS is constant, and S lies on a cone of which is vertex and OC axis. 64 The Cylinder and the Cone, 5. Through the flames of the candles which are treated as points, draw planes intersecting in the ceiling, in a straight line, since these fixed planes must always be tangent planes to the ball, the locus of the centre of the ball ia a horizontal straight line. 6. The triangles A EX, A^E'X' have all their sides parallel; therefore aS'^ : AX :: EA : AX :: E'A' : A'X' :: S'A' : A'X\ 7. If C be the centre of the sphere FES, the angle OCAS' is constant. And the ratios CE : CA, CE : CV, and CS : CA are all constant ; therefore CS : CV \^ constant and the angle CVS is con- stant, or VS is a fixed straight line. 8. Take two points E and A on the generating line and with centre A and radius AX, such that EA : AX in the ratio of the eccentricity, describe a circle intersecting FE in JT, the section of which ^X is axis will have the required eccentricity. 9. XS, XS' are tangents to the same sphere FES of which C is centre. Let SS\ EF meet the axis in V\ L, and let CX meet SS' in M. Then CL,CV'^ CM. CX= CE\ Hence Fand F' coincide. 10. Draw CiV perpendicular to the axis, then 2CN=A'D''-AD, and 20N=0D+0D\ In CN take a point Q, such that QN : CN :: DO'' : DA^; Tlie Cylinder and the Cone. 65 then since OD'-OD : 2CN :: OD : DA, 2QN : OD'-OD :: DO : DA ; also AD + A'D' : 20N :: ^i> : DO, and ^^'2 = Z>Z>'24-(^Z> + ^'i)7. Hence QO^ : CA' :: 0Z)2 : DAI Hence Q lies on a sphere of which is centre, and there- fore C lies on a spheroid, having OD for its axis, which is oblate or prolate according as DO is greater or less than DA, that is according as the vertical angle of the cone is greater or less than a right angle. 11. Draw a plane CT through C, the centre of the sphere perpendicular to the axis intersecting the tangent plane at S in T. Then since the angles COB and CTS are equal, and CB= CS, it follows that CT= CO. Hence S lies on the surface generated by the revolution of the circle of which CT is diameter. 12. Only two circles can be described passing through Sj and touching the generating lines in which the plane through S and the axis intersects the cone. If ST, ST' be the tangents, and OS meet the circles in D and D'j the angle DST^^ half SCD = half SC'D' = D'ST. Hence the planes of the corresponding sections make equal angles with OS. 13. If the plane of section meet the plane through O perpendicular to the axis in the hne ZZ', and PK be per- pendicular to ZZ' OP ov OQ bears to P^a constant ratio. Hence if P' in the projection corresponds to P OP : P'K in a constant ratio. But OP' is equal to the perpendicular from P on the axis which bears to OP a constant ratio. Hence the ratio OP : P'K is constant. B. C. S. n 66 The Cylinder and the Cone, 14. A'Q — QA = SS' and a right cone can be constructed of which Q is vertex, such that the generating lines inter- sect the ellipse. Hence PQ + AS=AS'¥QR + RP = AE+EQ + SP=AQ + SP. 15. Through the vertical straight line which is the locus of the luminous point draw vertical tangent planes to the ball intersecting the inclined plane in O Y, OZ. The locus of G the centre of the shadow will be the straight line bisecting YZ, at right angles, SY, SZ being perpendi- culars from the point of contact of the ball on OY, OZ which are tangents to the elhptic shadow. 16. The given plane to which the sections are perpen- dicular must be supposed to contain the axis of the cone. Take OK on the generating line equal to AS. Draw KG perpendicular to OK^ meeting a line through O at right angles to the axis in G, then (r is a fixed point. Draw GM perpendicular to AS, then CE : EO :: OK : KG, or KG : KA :: AE : CE. Hence the angle KAG^ACE^h^MKAM. Hence AS touches a circle centre G and radius GK 17. VP=VQ=VA+AQ = 2AE+AN=2AS+AN. 18. EX\ X'A' will be parallel to EX, XA. Hence if AF be drawn parallel to A'E\ the ratio of the eccentricities is AE : AF which is constant. 19. The volume varies as the area A VA' and BC; and the area AVA' varies as ^F. VA', or AD . A'D', that is, BG\ Hence BG is constant. The Cylinder and the Cone, 67 20. A'O -0A= A'E' -EA= A'S- SA = SS'. Hence the locus of is an hyperbola of which A and A' are foci. 21. B, R, P, P' lie on a circle ; hence PR, DP' are equally inclined to the axis. So DQ, PD' are equally inclined to the axis ; hence PR, DQ are parallel, since DP', PD' are parallel. 10. PG=CD^CP. Hence the radius of curvature varies as CP^, 11. LP, PG^ PT,Pt = CD\ Hence PL is equal to half the chord of curvature in direction PC. CP,CL = CP,PL + OP''=CD'^+CP^=AG' + BC\ 12. The circle on PE as diameter touches the curve at P and goes through Q ; when Q coincides with P the circle becomes the circle of curvature. 13. If the tangents at two near points P and Q meet m T, TP : TQ :: CD : CE, Curvature. 71 Hence the difference of TP and TQ is very small com- pared with either, and if a circle be described, of which the intersection of normals at P and Q is centre, to touch TP and TQ at P and Q, when P and Q coincide this becomes the circle of curvature at P. 14. If (7 be the centre of curvature at the vertex, AG = 2AS. If PR be the tangent, PR^ = CP^-CE' = PN^ + CN^-4AS^ = 2AC.AN+CN'--'AC' = AN'i 15. PGP' and the tangent at P are equally inclined to the axis. 16. If be the centre of curvature, PO''--^AG .G^PF.GD. But PO,PF=GD\ hence PO = GD=PF, Hence, if with centre G and radius GP, such that GP^ = AG'' + BG^-AG. BG, a circle be described, it will meet the curve in points, at which the radius of curvature has the required value. 17. If PF be the chord and GQ be drawn parallel to Pt, 2GD^ = PQ.PV=^Gt.PV, But Gt,PM=^BG\ hence PV : PM :: 2GD^ : BG\ 18. If PQ be the common chord of the ellipse and circle of curvature, TPt must make equal angles with both axes, or GT= Gt, Make the angle AGP such that PN : NG :: BG'^ : AG^) then GP will meet the ellipse in the point required. 19. SP is one-fourth of the chord of curvature through Sy hence PQ is half the radius of curvature. 72 Curvature, 20. If Pr be the chord, PV.PD = 2CD\ But PN=ND, hence PV : CD :: (7i> : PiV. 21. P(9 : PP :: PK : P(7 :: PP : Pg, or PP. PP= PG.Pg = CD' ; hence P is the centre of curvature. 22. Draw SQ at right angles to SP to meet the normal in Q ; let the tangent at P meet the directrix in Z. Then PL : PS :: ZP : ZS .: PQ : PS. Therefore PL = PQ = half the radius of curvature. 23. If CB bisect PQ, PCE is a right angle and PF.PE=CP'' = CD''] hence PQ is equal to the diameter of curvature at P. 24. OP : (7P :: PQ : PF, or OP.PF=CP''=CD^', hence is the centre of curvature at P. 25. If the normal at P meet BG in ^; ^, P, H, K lie on a circle ; hence, vrhen P coincides with B, K becomes the centre of curvature at P. 26. If HT be produced to H' making iPr= TH, TQ, PH' will be parallel ; hence PR : RS :: iP^ : T^S' :: TH : TOl Therefore PR.PS wTH .ITG :: ZTP : 2(7F:: HP : 2CA, or 2PR .AC^PS .HP=CD\ Hence PR is one-fourth of the chord of curvature through S. Curvature. 73 27. If be the centre of curvature at A, SO : SA :: SA : AX. Hence curvature of ellipse is greater than that of parabola, and curvature of parabola is greater than that of hyper- bola. 28. By Ex. 6, SP--^ACy and SP : CY' :: FE : PC. Hence PB : EP' :: 3 : 1. 29. Let CQ' be conjugate to CF and CP' parallel to PQ. Then OJ^^ : OF^ :: OQ.OP : OF' :: CP'' : CQ'"" :: CZ>2 : CQ'^ :: TP^ : TF"" :: T^2 . 77^2, Hence TFOF is cut harmonically. 30. Project the angle between the common diameters into a right angle ; the ellipses obtained will be inscribed, symmetrically, in a square, and will therefore be equal. 31. SB^ = Py^ + {EP - Syf = SP^ + EP'-2EP.Sy ^EP^-S.SP^ Art. 160. 32. Let F be the middle point of PE, the radius of curvature at P. Then PF.SY= SP^, Art. 160. .-. SY : SP :: SP : PF, ,*. SYP, SPF are similar triangles, and the angle PSFis a right angle, so that the locus of S is the circle, diameter PF. CHAPTER VIII. PROJECTIONS. 1. The theorems are obtained by projection from the following properties of the circle. Art. (65) Every diameter of a circle is bisected at the centre, and the tangents at its extremities are parallel. (70) If the chord of contact of tangents from T to a circle meet CT in N, GT . CN= CA\ (71) If M,t correspond to M\ V on the circle, CM : CM' :: BC : AC :: Ct : Ct' \ therefore CMXt=BC\ (73) If the chord of contact of tangents from T to a circle meet CT in V and CT meet the curve in P, CT,CV=CP\ (74) A diameter of a circle bisects all chords parallel to the tangents at its extremities. (75) If a diameter of a circle bisects chords parallel to a second, the second diameter bisects all chords parallel to the first. (76) Art. 77 is meant. If PCP\ BCD' be diameters of a circle at right angles, QF2 : PV, VP' :: CD'^ : CP\ Projections, 75 (78) If CA, CB be radii at right angles, CM : PN ;: AG : BG :: GN : DM, (81) The area of a square circumscribing a circle is constant ^nd equal to the rectangle contained by diameters at right angles. (82) If (7Z>, GP be radii at right angles and the tangent at P meet a pair of radii at right angles in T and if, PT,Pt=GL\ (83) Cor. 1. The two tangents TP, TQ from any point are equal, and the parallel diameters AGA\ BGB' are equal, .-. TP : AGA' :: TQ : BGB; and these ratios are unaltered by projection. Cor. 2. In the circle TGT' is a right angle, and .-. PT,Pr=GP^ = GD'', GD being parallel to TPT. (84) Taking ABA' as a semicircle, EGF is a right angle; project on any plane parallel to the line AGA\ 2. If a parallelogram be inscribed in a circle its sides are at right angles. The greatest rectangle than can be inscribed in a circle is a square having its area equal to 2AG^ ; hence the greatest parallelogram that can be inscribed in an ellipse has its area equal to 2 AG. BG. 3. The theorem is true in the case of a circle, and follows by projection. 4. The greatest triangle which can be inscribed in a circle is an equilateral triangle of which G is the centre of gravity. Produce PC to F, making 2GF=PG; then, if Q VQ' be the ordinate, PQQ! is the greatest tri- angle which can be inscribed in the ellipse having its vertex at P, 76 Projections, 5. If a straight line meet two concentric circles, the portions intercepted between the curves are equal. 6. The locus of the point of intersection of tangents s^t the extremities of diameters of a circle at right angles is a concentric circle. 7. The locus of the middle points of lines joining the extremities of diameters of a circle at right angles is a concentric circle. 8. If CP, CD be radii of a circle at right angles and CA bisect the angle PCD, the tangent at A meets CP in T such that PD^ = 2 A T\ 9. If a chord ^Q of a circle be produced to meet the diameter at right angles to CA in and CP be parallel to^e, Aq.A0^2CP\ 10. If OQ, OQ! are tangents to a circle and R be a diagonal of the parallelogram of which OQ^ OQ' are adjacent sides, then if R be on the circle the locus of is a concentric circle. 11. If a parallelogram be inscribed in a circle and from any point on the circle straight lines are drawn parallel to the sides of the parallelogram, the rectangles under the segments of these lines made by the sides are equal to one another. 12. If a square circumscribe a circle and a second square be formed by joining the points where its diagonals meet the circle, the area of the inner square is half that of the outer. And if four circles be inscribed in the spaces between the outer square and the circle, their centres will lie on a concentric circle. 13. If a rectangle be inscribed in a circle so that the diameter bisecting one pair of sides is divided in a constant ratio, the area is constant. 14. If a parallelogram circumscribe a circle and one of its diagonals bear a constant ratio to the diameter it contains, the area is constant. Projections, 77 15. PtQ/2 is a triangle inscribed in a circle, the centre being the intersection of lines joining the angular points to the middle points of opposite sides. If PC, QC, EG meet the circle again in P\ Q, R\ the tangents at P\ Q\ R\ will form a triangle similar to Pi^Ry its area being four times as great. 16. The locus of the middle points of chords of a circle passing through a fixed point is a circle of which the line joining that point to the centre is diameter. 17. The ellipse which touches the middle points of the sides of a square (i.e. a circle) is greater than any other in- scribed ellipse. 18. If a polygon circumscribe a circle, its area is a minimum when any side is parallel to the line joining the points of contact of adjacent sides. 19. The greatest triangle which can be inscribed in a circle has one side bisected by a diameter and the others cut in points of trisection by the diameter at right angles. 20. ^^ is a given chord of a circle, C any point of the circle, the locus of the intersection of the straight lines joining A, B, C to the middle points of BC, CA, AB is a circle. 21. If CP, CD are radii of a circle at right angles, the circle on PjD as diameter will go through (7. 22. The theorem is true in the case of a circle inter- secting a concentric rectangular hyperbola, and follows generally by projection. 23. If V is the middle point of Qq, project CVQ into a right angle. 24. If PT, pt are tangents at the extremities of a diameter Pp of a circle, then if any diameter meet P 2" in T and the diameter at right angles meet pt in t, and any tan- gent meet PT in T and jt?^ in t\ PT : PT wplf \ pt. 78 Projections. 25. If (7P, CD be radii of a circle at right angles and P/>, Dd be drawn parallel to any tangent, and any line through G meet Pp^ Dd and the tangent in p, d and ty Cp^ + Cd^=Ct\ 26. If AC A', BC and CD, CP be pairs of radii of a circle at right angles and if BP, BD be joined, also AD, A'Pj the latter intersecting in 0, BDOP is a parallelo- gram. 27. If TM be perpendicular to SP, TM : TP :: SY : SP :: BG : CD; hence TP : (72) is constant. If a point be taken on a tangent to a circle such that its distance from the point of contact is constant and therefore proportional to the parallel radius, its locus is a concentric circle. CHAPTER IX. CONICS lA^ GENERAL. 1. TN : XN :: SR : SX :: SP : XN; hence SP=TN. Also TP . TP' = 7W^- PN^ = aS'P^ _ PiV^s ^ ^;v^2^ 2. Draw Pm, Qn perpendicular to the directrix. Then PR :QN::KP:KQ::Pm:Qn::SP:SQ::PM:QN, or PR = PM. 3. PS.SQ : AS.SA' :: (7p2 : C^^^ and PQ.SR^ISP.SQ. Hence PQ varies as Cp\ 4. Let Pj9, Qq intersect in 0, Then QO.Oq : PO.Op :: rp2 : TQ^ :: QO^ : POl Hence TO bisects pq as well as PQ, and is a diameter and goes through t 5. Since RS is the exterior bisector of the angle P'SQ\ SP" : SQ' :: RP' : /2Q'. 80 Conies in General, 6. Let S'T, ST meet PQ in EE\ and let TF, TG be the common interior and exterior bisectors of the angles ETE\ PTQ. Bisect FP in 0. Then OE . OW = OF^ =OP.OQ. Kow RP : EQ :: PE : EQ, and RP : RQ :; P^' : E'Q. Again, OP : OE :: OE' : OQ; hence P^ : 0^ :: E'Q : OQ. Also OP : OE' :: 0^ : OQ ; hence P^' : OE' :: ^Q : OQ. Therefore RP,PR : RQ.QR :: PE.PE' : EQ, QE :: 0^. 0^' : OQ^ :: OP^ : OQK Hence TF bisects the angle RTR\ and the angles RTP, RTQ are equal. 7. TS and ^aS' are the interior and exterior bisectors of the angle PSR, Hence if ST meet PP' in P, RK : TP :: ^P : EP :: ^P : EP' :: ^P' : PT, or RK^KR, 8. If PP, D'E' are perpendicular to aS'P, SP', then SE=SE'. Hence PP, D'E intersect on the bisector of the angle PSP\ which is ST. 9. If PP' meet the directrix in K, PP' is har- monically divided at S and K, Hence any chord through S is harmonically divided by the directrix and the tangents at P and P'. 10. Draw /SF perpendicular to the tangent, then since SO : CY :: SA : AX, the locus of (7 is a circle. Conies in General. 81 11. Let Fp meet the curve in P', and let QP' meet the directrix in q\ Then since pS, q'S are the exterior bisectors of the angles PSP\ QSP\ the angle pSq' = \i2i\i PSQ=pSq; hence «$ and gf coincide. 12. SL : SP :: FT : FP :: TN : P^. Therefore aSX : TN :: aS'P : P^ :: SA : ^X 13. FN' : AC^-CN^ :: <7i?2 . q^-i .. (752 . q^'^ : : /??i2 : Cn^ - (7a^. Hence, if FN=pn, AG'^ CN^= Cn^- Ca\ or CN^ + Cn^^CA^-^Ca\ 14. Qi? : LG :: PQ : PG^ :: PM : PiV. Hence QR : Pi^ :: Z(5^ : FN :: /SG^ : SP :: aS'^ : ^X 15. Draw KV parallel to the axis. Then VK : VP :: GS : /S^P ;: SG' : aS'Q :: VK : TQ, or PV=VQ. Also PZ : Pil/ :: PK : PG :: PT : PS; hence 2PZ . PS= PM. PQ==SR.PQ = 2SP . SQ, or PL = SQ. Hence /ST= FZ, and the diagonal of the parallelogram SL goes through V. 16. If PQi2 be the triangle and S the focus, make the angles QSr, QSp each equal to the supplement of PSR, Then, if PSq = PSr, p, q and r are the points of contact. 17. By Ex. 20, Chap, i., if KV be drawn parallel to the axis, PF=Fe. Hence FN : PL :: PK : PG :: PV : PS, Hence 2PiV^. PS= SR,PQ = 2SP . SQ, or PN = SQ. Hence SN=2SV, and the locus of iV is a similar conic. B. c. s. 6 82 Conies in General. 18. Let CT, CT meet the curve in p, d. Then CT.PR = 2Cp\ and Cr,QR=2Cd\ Hence the triangle CTT : 2 triangle Cpd :: 2Cpd : PRQ, or the triangle CTT : AC. EG :: ^(7. ^(7 : the triangle PRQ. 19. Let S be the centre of the circumscribed circle, H the ortho- centre, then the feet of the perpendiculars from /S'and H on AB, BC, CA lie on the nine-point circle, and the angle /S'^4^ = complement of C—HAC. Therefore with S and H as foci a conic can be inscribed in ABC. 20. Let SV meet the directrix in Q and PK'vn Z, let QP meet the axis in G. Then PZ : SP :: SG : SP :: SA : AX; hence PZ : PK :: .S'^' : AX^ Now /S'(7 : CX :: PZ : PK :: SA" : ^X^, or (7 is the centre. 21. DE.DF : DG^ :: ^^2 . ^(72 .. 2)^3^2 . i>(72, or DG^= DE.DF. 22. By Ex. 74 on the parabola, if PGQ be the chord of contact, DF : i^(? :: PG : G^Q :: G^P : FE, or FG''=FD.FE, 23. If ^^^ be drawn parallel to DTP, DP'^ : .^jc» . ^g :: DF : j&P. Conies in General. 83 For DF is parallel to a generating line VM of the cone of which the hyperbola is a section. Draw Dim, LEM in the plane VFEM to meet the cone. Then the sections of the cone by the parallel planes, IPniy pLQ are similar, and Dm = EM. Hence DP': Ep .Eq :: Dl.DmiEL, EM :: Dl : EL :: DF : EF. Again, Ep.Eq : EQ' :: PT" : TQ^ :: EK^ : Eq^\ hence ^JT^ ^ Ep.Eq if ^jt?^ meet PQ in iT. And DG" : G^2 .. j)^2 . ^^2 .. 2)p2 :Ep.Eq::DF: EF. Therefore FG^ = FD . FE. 24. Let the tangents at P, Q and R meet JS^J5 in p, q and r. Then Ep.Eq^ EF\ if PQ meet EB in F; also EB^=Er . Ep, Ea^ = Er . Eq. Hence EB^ : EC^ :: Ep : Eq, a constant ratio. By Ex. 23. the same proposition is true in the case of an hyperbola if EB be parallel to an asymptote. 25. GK being perpendicular to SP, Pk : PK :: Pg : PG :: AC* : BC ; .'. Pk is constant. Also kL : Pk :: SG : SP ; .'. kL is constant. 26. Let the fixed line meet the curve in P and Q, and let the tangent at P meet SL in D, and the directrix in F; then, Art. 11, aS'Z) : SF :: /SZ : SX. The angle PS'P is a right angle, so that SF is a fixed line, and, SX being a fixed line, the ratio of iSF to SX is constant ; .*. SD is constant and D is fixed. The envelope of PG is therefore the parabola, of which D is the focus and PQ the tangent at the vertex. 6—2 CHAPTER X. HARMONICS, POLES AND POLARS. 1. If a OB be the common chord and PQOpq any transversal, F0,0p = A0,0B = Q0.0q. 2. OA is perpendicular to B'C and meets it in D, OA . OD = square on radius of concentric circle = OB . OB = OC,OF, Therefore OD = 0E= OF. 3. Draw ABC to be bisected by OB in B, BEG to AO produced to be bisected by (9(7 in E, and BFK to CO produced to be bisected by OA in F. Then any one of the straight lines drawn through O parallel to AC, BG^ BK will form a harmonic pencil with OA, OB, OC. Draw BL, J5i»f parallel to OA, OC to meet OC, OA in L and M, Then since OE=EL, OF=FM, EF is parallel to LM and is therefore bisected by OB and is also parallel to ABC. Hence the pencil BC, BE, BO, BF is harmonic. 4. Let the circles meet in P, bisect AG in E. Then EB.ED = EC = EP% hence the circles cut at right angles. Harmonics f Poles and Polar s. 85 6. By Art 182, PQ, AE, BD intersect in A, and A {B, Ey Q, F} is harmonic. 6. AP, BQ and PB^ AQ meet each pair on the polar of on which G lies ; And the pencil formed by OB, CO, CA and the polar of O is harmonic. 7. If ^, ^ be points of contact and the third conic meet AB in C and />, A and B are the foci of the involu- tion, P, Q are conjugate points. Hence ACBD is a harmonic range. 8. Let the common chords meet in E, and let EPRQ be a tangent at R ; then since the common chords are one conic of the system, E and R are foci of the involution EPRQ and EPRQ is a harmonic range. 9. Let rp, TQ be the tangents, TE any line, then F the pole of TE lies on PQ and PEQF is a harmonic range. 10. If the tangent at P meet the asymptotes in L and L\PL = PL\ Hence the pencil CD, CL, CP, CL' is harmonic. 11. A diameter is bisected at the centre: and the polars of the extremities of a diameter intersect at infinity. 12. If jTbe the pole of QR and H the second focus of the conic which touches the ellipse at Q, PQ + QH=SQ-¥QS''y or HS'=SP, Therefore HS' + S'P = S'P + SP ; or /S" is on the conic of which ^is focus. Again the angles TS'R, TS'Q are equal, and PS\ S'H are equally inclined to TS' ; hence TS' is a tangent at/S'. Therefore T lies on the directrix of the conic of which P, H are foci. 86 Harmonics, Poles and Polars. 13. Let ST, S'T meet FQ in E, E' \ then the angles JSrrPjj&'T'Q are equal. Now the ranges RPEQ, R'QEP are harmonic and QTP is common to the two pencils; hence the angles i2'rQ,i2rP are equal. 14. The middle points of all chords of the cone parallel to the given line, lie in a plane through the vertex, let this plane meet the given line in P and any section through it in A and A ', Then Q the pole of the given line lies in AA'P and AQA'P is a harmonic range. Since VA, VA' are fixed generating lines, VQ is a fixed straight line. 15. If Tpq be the chord, P its pole, then PN the ordi- nate of P is the polar of T. Let GP meet pq in v and the curve in Q, Let QM, QG' be the ordinate and normal at Q, If PG be drawn perpendicular to pq, it is parallel to QG'; hence GG : GG' :: GP : GQ :: GN : GM; Therefore GG : GN :: GG' : GM :: SG"" : -^ICl Hence G is a fixed point. 16. Let the polar of Q meet the conjugate GFD in B. Draw QQ' parallel to GP. Then P^ : PQ :: EF : PP; and PG : PQ :: (7P : PF; hence ^G^ : PQ :: GE : PF; or EG.PF=PQ,GB. Now Q is on the polar of ^, since 22 is on the polar of Q, Hence PQ,GB=GQ\ GB = GD\ Hence EG is equal to the radius of curvature at P. Harmonics, Poles and Polars. 87 17. If be the orthocentre of ABC and A'B'C the reciprocal triangle, B'C\ C'A', A'B are perpendicular to OA, OB, OC respectively, and BG, CA, AB are perpen- dicular to OA', OB', OC respectively. Hence ABC and A'BC have their sides parallel and O is the orthocentre of each. 18. If pPSQq be the focal chord and the tangents at P and Q meet in T, TS is perpendicular to PQ, hence the tangents at p and q meet in T. 19. This theorem is the reciprocal of the following : if two circles intersect they have two common tangents : if one circle lie entirely within the other, they have no common tangents. Reciprocate with respect to a point on one circle and within the other. 20. If /, C be the centres of the inscribed and circum- scribed circles, and CI meet them in r, r and B, B ' re- spectively, then if AA' be the major axis of the ellipse into which the circumscribed circle is reciprocated, IA.IB = Ir\ IA'.IB'=Ir\ Cr'=CB^-^CB,Ir. Hence lA : Ir :: Ir : CB-CI :: CB + CI \ ICB, and lA' : Ir :: Ir : CB^CI :: CB-CI ; 2CB. Hence A A' : Ir :: 2CB ; 2CB ; or AA' = Ir. 21. The four circles which circumscribe the triangles of a complete quadrilateral meet in a point. 22. See Ex. 30 on the parabola, or by reciprocation. 23. See Ex. 46 on the ellipse, or by reciprocation. 24. Let CPP\ COO' be perpendicular to the polars of P and O. Draw OX, OA perpendicular to CP and the polar of P ; P Y, PB perpendicular to CO and the polar of O. Then CO' : CP' :: CP : CO :: CY : CX. Hence CO'-CY : CP'-CX :: CY : CX :: CP : CO or FB : OA :: CP : CO, 88 Harmonics, Poles and Polars, 25. See Ex. 18 on Chapter I. 26. (1) If a quadrilateral circumscribe a conic a pair of opposite sides subtend at the focus angles which are together equal to two right angles. (2) If we reciprocate with respect to the focus S the new theorem is, if Q be taken on a circle and QL be drawn such that the angle SQL is constant, QL envelopes a conic of which S is focus. 27. The envelope of chords of a circle which subtend a constant angle at a fixed point on the circle is a smaller concentric circle. 28. Two circles, such that a point can lie within both cannot have more than two common tangents. But if the circles be such that all points lie without both, or within one and without the other they may have four common tangents. 29. If a straight line meet the sides of the triangle A'B'C in X, M, N the circles circumscribing the triangles A'BC, A'NM, B'NL, G'LM meet in a point. 30. If points P', Q' be taken on a circle of which C is the centre, P'G will meet the line drawn through Q^ at right angles to P'Q' and Q'G will meet the line drawn through F' at right angles to P'Q' on the circle. 31. If S be the orthocentre of the triangle ABC and circles be described with centres A and B passing through C, S will lie on the radical axis of the two circles. If we reciprocate with respect to S we see that if with the orthocentre of a triangle as focus we describe two conies each touching a side of the triangle and having the other two sides as directrices, the conies will have a parallel pair of common tangents and therefore their minor axes equal. 32. If a system of circles have two points in common the locus of their centres is a fixed straight line, and the polar of a fixed point meets the radical axis in a fixed point. HarmonicSy Poles and Polars. 89 33. If the tangent and normal at P meet QR in T and 6?, the range TRGQ is harmonic, since TF, PG bisect the angle QPR. Hence PG is the polar of T, Hence the pole of QR lies on PG since the pole of PG lies omQR. 34. If the tangents at P and Q meet in T and TA meet P^ in Z, the range DPLQ is harmonic ; hence the pencil TD, TP, TL, TQ and the range DBAO are har- monic. Therefore ABBC is a harmonic range. 35. If the pencil joining BPA Q to any point on the curve is harmonic, the pencil formed by joining them to any other point on the conic is harmonic. For if BK, PK^ AK^ QK meet the directrix in hpaq, hpaq is a harmonic range, provided KEBPAQ be a har- monic pencil. And the angles hSp, pSa, aSq, qSb are half the angles BSP, PSA, ASQ, QSB ; Hence the pencil Sb, Sp^ Sa, Sq is the same wherever A" be taken on the curve. Now PQ goes through the pole of AB : let PQ meet AB in R. Then if The the pole of PQ, TARB is a harmonic range. Therefore the pencil joining Q to BPAQ is harmonic ; hence the pencil joining q to BPAQ is harmonic. Hence Pq bisects AB since AB, qQ are parallel 36. Four circles can be described so as to touch the sides of a triangle, and the reciprocal of the radius of the inscribed circle is equal to the sum of the reciprocals of the radii of the other three. If the triangle be equilateral the inscribed circle touches the three escribed circles. 90 Harmonics, Poles and Polars. 37. If the tangents at P and Q meet the axes in T and Vf the angle PSQ = SQ V- SPT= SVQ- STP ^ VST. li SW ho perpendicular to PQ\ the tangents at the vertices intersect in W. Draw aS'FZ perpendicular to the tangents at P and Q. Then WSP\ WSQ' are supplementary to WYP\ WZQ, Hence P'SQ', PSQ are supplementary. If two circles intersect in P, Q the angle between the tangent at P, Q is equal to the angles which the centres subtend at S and supplementary to the angle which PQ subtends at the other point of intersection. 38 and 39. If from any point P in the radical axis tangents be drawn to the circles, and a circle be described, with centre P and radius equal to the tangent, this circle will intersect the line of centres in two points JE and F which are the limiting points of the system. Take A at centre of one of the circles, and M at the point where the radical axis intersects the line of centres. Then, P U and P U' being the tangents from P to the circle PM^ + ME^ = PE'^ = PU^ = PA^-AU^', .-. ME^=AM''-AU'^MF'; ,\ AE.AF=AM^-EM^ = AU\ .'. the polar of P passes through E, Reciprocating with regard to F, the pole of uU\ i.e. of the fixed line through E, is the centre, which is therefore fixed, and the conies are confocal. Therefore, if we reciprocate with regard to either limiting point we obtain confocal conies. 40. If perpendiculars be drawn from A, B, C to BC, CAf AB these lines will meet in a point 0, and the circles circumscribing ABC, OBG, OCA, GAB are equal. 41. If the tangents at P and Q, points on a circle intersect at a constant angle, and lines be drawn through Harmonics, Poles and Polars, 91 P aiid Q making constant angles with the tangents at P and Q respectively, this pair of straight lines will intersect on a concentric circle. 42. If two circles intersect in A and B and PQ be a common tangent and QB, PA meet the circles in C and 2>, then PC, QD are parallel. 43. If from any point on a circle circumscribing a triangle perpendiculars be drawn to the sides of the tri- angle, the feet of these perpendiculars lie on a straight line. 44. Since the orthocentre is on the hyperbola, DEF is a self-conjugate triangle and the pole of EF lies on BC, Hence the pole of BC lies on EF, 45. If AB, CD meet in the fixed point E, CA and BD in F, and BC and AD in G, then FG is the polar of E, Hence the centre of the circle lies in a straight line through E perpendicular to FG the polar of E with respect to both curves. 46. The radius of an escribed circle of an equilateral triangle is | the radius of the circumscribed circle, and if SE be the tangent from the centre of the circumscribed circle to the escribed circle whose centre is D ; SD = DE+iDE= ^DE. The proposition in the question is obtained by recipro- cating with respect to the circumscribed circle. 47. If AD be drawn parallel to the axis to meet BC, AD is bisected at D^ where it meets the curve. Hence the tangent at Z>' is parallel to BC and bisects ' AB and AC Since a straight line intersects a conic in two points and two tangents can be drawn from a point, the reciprocal polar of a conic with respect to another conic is a third conic. 92 HarmonicSy Poles and Polars, Now by Ex. 44, if a rectangular hyperbola circumscribe a triangle DEF it will go through the ortho-centre and ABC the triangle formed by joining the feet of the perpen- diculars is a self- conjugate triangle, and O is the centre of the circle inscribed inABG. If we reciprocate with respect to O the reciprocal conic is a parabola, since it has one tangent at an infinite distance and ABC is a self-conjugate triangle. The tangents from are at right angles, since the hyperbola was rectangular, hence is on the directrix. The locus of the poles of the lines at an infinite distance, that is, of the centres of the hyperbolas, was the circle cir- cumscribing ABC. Hence the envelope of the polars of O with respect to the parabolas is an ellipse inscribed in ABG having O for a focus. Since is now the centre of the circle circum- scribing ABCy the auxiliary circle of the eUipse is the nine point circle. MISCELLANEOUS PROBLEMS. 1. If S and H be the rifle and target, and P the hearer, the difference of the times in which sound travels from iS' and ^ to P is equal to the time of the bullet's transit from S to II. Hence HP-SP is constant, and the locus is a hyper- bola of which S and // are the foci. 2. Let tp^ tq be the tangents parallel to PQ and P'Q and let qt meet in r the diameter through p) then qt^tr, and Pi22 . PQ2 .. tr^ . fp2 :: tq^ : tp^ :; SP'.SQ' : SP.SQ :: P'Q' : PQ ; Art. 17. .\PR^=PQ.P'Q\ 3. QN : CM :: BC : AG :: DM : CN, hence QN+DM : NM :: BG : AG, 4. If GVBD be conjugate to PQ, PQ,Qp=PV^-QV\ Hence PQ.Qp : CD^-CE^ :: Ci?^ : (72)2, GR being parallel to PQ. 5. CiV^ : iVX :: .S'^ : AX :: /S'i? : SX. Hence Q lies on the tangent at the extremity of the latus rectum. 6. PN^ : AN. NA' :: BG"- : -4(72 and QN.PN=AN.NA\ Therefore QN'' : AN.NA' :: AG^ : BG\ 94 Miscellaneous Problems. 7. Let AP, QB meet in i?, and draw RV parallel to POQ. Then RV : VA :: PO : AO, and RV : VB=QO : OB. Hence RV'^^VA.VB, since AO.OB^PO.OQ. Therefore QR lies on a concentric rectangular hyperbola. 8. The line joining T to the intersection of the normals at P and P' bisects PP' and therefore passes through the centre. 9. If the tangent at P meet the tangents at A and A' in Tand T and TS, T'S' meet in Q, the angle SS'Q = TS'A'= T'S'P ; QSS'^AST=TSP', and SPT=jS'Pr. Hence S, P, S' are the feet of the perpendiculars of the triangle TQT. Therefore QP is perpendicular to TP. 10. Make the angle PSF a right angle, then the tangent at P meets the directrix in i^: if a circle be described with centre P and radius PK such that the ratio SP : PK is equal to the eccentricity the directrix is a tangent from F to this circle. Two tangents can in general be drawn. If the angle SPF be such that SP : PF :: SA : AX^ only one conic can be constructed ; there are two positions of PF equally inclined to SP corresponding to this case. If the eccentricity be unity, one conic is a line parabola through S. 11. If PK be drawn perpendicular to the directrix of the parabola SP = PK, hence HM=^HP + PJr= AA\ Miscellaneous Problems. 95 Tlierefore the directrix touches a circle of which H is centre. 12. Draw UN perpendicular to the minor axis. Then CN.AC=SR.AG=BC^=AC^-S(P^AC^-RN^; Hence BN'=AC{AC-CN}, or R lies on a parabola. 13. If ST, FQ meet in H, HTRS is a harmonic range. But OH, OT,OV, OS is a harmonic pencil. Hence V passes through R. 14. Let AC A' be the diameter bisecting the parallel chords QNy etc. in iV, etc. Then PN^ varies as QN\ that is as AN. NA\ Hence the locus of P is an ellipse. The locus will be a circle if PN=QN, that is if the vertical angle is a right angle. 15. If PP\ QQ' be the double ordinates of the given points, P, P\ Q, Q' are fixed points, and since the ellipses are similar, the corresponding points of the auxiliary circle, at which the major axis subtends a right angle, are likewise fixed points. 16. The ordinates of the point and of the end of one of the radii are in the ratio of the radii. 17. If P be the centre of the circle and P^ perpen- dicular to the fixed straight line, the ratio SP : PK is constant. 18. Let pqr be a triangle touching the parabola in Parabolic area PQR = | triangle PqR. .-. Triangle PQR^%{PqR- PrQ-QpR), 3PQR-^2{pqr^-PQR); .*. PQR = 2pqr. 96 Miscellaneous Problems. 19. If a rectangular hyperbola circumscribe a tri- angle the orthocentre is on the curve, if we reciprocate with respect to the orthocentre we have the case of a parabola inscribed in a triangle the tangents from the orthocentre being at right angles. See Ex. 47, Ch. 10. 20. Produce HA to K making AK equal to AH, then PK and AL are parallel. Hence SQ : aSP :: SA : SK :: SA : SA + AJI, Therefore the locus of Q is a similar ellipse of which S is focus. 21. If KLMN be the quadrilateral and k, /, m, n the points of contact, KM will bisect nky Im : and LN will bisect kl, mn. Hence klmn, and therefore KLMN, is a parallelogram. 22. The locus of the second focus is a circle of which the radius = AA- SP. The locus of the centre which bisects SH is similar, that is, a circle. 23. Since LC, LL' are tangents, the angles HLC and SLL are equal. Again CL . CL' = GH\ hence the angle CHL = GL'H^ SL'L. Therefore CL : HL :: SL : LL\ the triangles CLH^ SLL being similar. 24. If the theorem be true in the case of a circle, it will follow by orthogonal projection for any ellipse. If PQ be the chord of contact of tangents drawn to a circle from a point on a concentric circle, the angles PAQ, PA'Q will be constant, A, A' being extremities of a fixed diameter. Let APy A'Q meet in R, and A'P and AQ'mR. The angle AEA' = APA'-PA'Q, and the angle ABA'= APA' + PA Q. Miscellaneous Problems, 97 Hence the loci of R and R' are circles passing through A and A'. 25. Circumscribe circles to two of the triangles formed by the intersections of the tangents, these circles intersect in the focus JS: the pedal line of S is the tangent at the vertex. A parabola can be drawn to touch five straight lines, if the circles circumscribing the triangles formed as above all meet in the same point S. 26. PF is the same for both curves, and therefore CD is also the same. 27. Prove that SY. S'Y' is constant. 28. By reciprocation. 29. P, Qj R, R' lie on a circle of which PQ is diameter and PQ, LL' are equally inclined to the axis. If jt?, p' are the vertices of the diameters bisecting PQ, RR' in V and V\ pp' is a double ordinate. Let VV which is parallel to the normal at /?' meet the axis in 0. Let VM, V'M' be the ordinates of Fand V\ 'l]ie\iLL'=L'M' + M'0 + MO-LM=20M-=^2ng = 4AS, 30. The bisectors are tangent and normal to a con- focal conic. Hence CG,CT=CS\ 31. Reciprocate the following theorem : li S, A, B, C be points on a circle and with centres A, B, Cand radii AS, BS, CS circles are described, they will intersect two by two in points which lie in a straight line. 32. OE : EG = SP : SG = Sp : Sg = OE : Eg. 33. If an ellipse be reciprocated with respect to its centre, the reciprocal is a similar ellipse having its major axis in the minor axis of the original ellipse. B. c. s. 7 98 Miscellaneous Problems. If we reciprocate an ellipse circumscribing a triangle and having its centre at the orthocentre with respect to that orthocentre, the reciprocal is a similar eliipse, inscribed in a triangle having its sides parallel to those of the original triangle, the homologous axes being at right angles, and having its centre at the orthocentre. This reciprocal ellipse is similar and similarly situated to the ellipse inscribed in the original triangle having its centre at the orthocentre. 34. Q lies on the common circle of curvature, hence PQ = 4PT. 35. ABf BG are equally inclined to the axis, hence since the angles at A and G are equal, AD,DG are equally inclined to the axis. Hence the tangent at D and A G are equally inclined to the axis. Therefore the tangents at B and D are parallel. 36. The volume cut off varies as the area VAA' and BB' ; and the area VAA' varies as ^4 F. VA' or AD . A'D\ that is BG\ 37. SQ : Pg :: St : tg :: SY : SP :: BG : GD :: PF : AG. Hence SQ.AG=PF.Pg=AG^, or AG^SQ. Let QL be the ordinate of Q and let MQ meet the major axis in V. Then GV : PN :: GV : GM :: GL : GM-QL :: GL : PN-QL, and SP-AG : GN :: SG : AG, or AG, SP = AG''-rGN. SG=BG''+ GS. SN. Again GN-GL : SP-SQ :: SN : /S'/*, and SP-SQ : (7iV :: 6'(7 : AG] Miscellaneous Problems. 9^ hence CN-CL : CN :: SN.SG : SP.AC; therefore CL : CN :: BC^ : SP.AC, or CL : SP-SQ :: BC^ : aS^P.^S'^ Now SP-SQ : PN-QL :: ^P : PiV; hence (7Z : PN-QL :: i?(72 : PN.SC, Hence CV.SC=BC\ or F is a fixed point. 38. If /SX, aS'JI/, SN be drawn perpendicular to the given tangents, the circle circumscribing LMN is the auxiliary circle of the conic. 39. If S be the centre of the circumscribed circle, H the orthocentre, the centre of the nine-point circle bisects SH\ and if PQR be the triangle, the angles SPQ, HPR are equal : hence S and // are the foci. 40. If Pg, P'g' be the normals at P and P' and gL^ g'L' be drawn perpendicular to PP\ then PL = RL\ by Ex. 27, Chapter I. ; hence gG=Gg', Therefore 2SG : SP + SP' :: Sg + Sg' : SP + SP' :: SA : AX. 41. Reciprocate the following with respect to S: ASB is a diameter of circle meeting a concentric circle in >S', the opposite sides of the quadrilateral formed by tangents through A and B to the inner circle are parallel, and the tangents to the outer circle at the points where it meets the tangent at S are respectively parallel to them. 42. If P, Q be two points on a rod and PS, QS are at right angles to the directions of motion of P and Q, then if R be any point on the rod the direction of motion of R is at right angles to SR. Hence the directions of motion of all points on the rod envelope a parabola of which aS' is focus and the rod tangent at the vertex. 1—2 100 Miscellaneous Problems. 43. SP and HQ are parallel to CT. Hence PaS^ = complement of /S'Pp = complement of CTP. QHq = complement of iZQg = complement of CTQ. Hence PSp and QJTg are together equal to the supplement oiPTQ. 44. Let ST, S'T meet CP in ^? and p\ and (72) in d and c?'. Since the angle PTS=d'TD, and TPp = TDd\ the angles /S'j^C, (7J'/S" are equal, and c?, d', p and j9' lie on a circle. 45. If the asymptote and directrix meet in 2), SDG is a right angle and if DP be the tangent PSD is a right angle. Therefore SP is parallel to the asymptote. 46. If PTQ, ptq be two consecutive positions and TV, ty be drawn perpendicular to pt, TQ respectively, tV=Pp+pt-PT=PT+TQ + Qq-tq-PT = TQ + Qq^tq=Ty, Hence the tangent at Tis equally inclined to PT'and TQ. Hence T lies on a confocal ellipse. 47. CP and CQ are at right angles ; hence G, P, 2), Q lie on a circle. 48. If PF be the chord through the centre, and pp' the parallel focal chord, PV.CP = 2CD\ and pp\GA = 2GD\ Hence PV : pp' :: CA : GP. 49. SP and QH are both parallel to GT ; hence the angle pGq =pGT+ TGq = PHQ + QSP. 50. The angle SP F=half the supplement of SPS'= hsiUPSr. Miscellamous Problems. , , ; ; J.Ol 61. QSP, Q'SP' are right angles and the perpen- diculars from Q, Q' on SP, JSP^ are equal to the perpen- diculars on PP\ Hence QP, Q'P' are tangents at Q and Q,' to the parabola of which /Sis focus and PP' directrix. And the diameter parallel to PP' is tangent at the vertex, since it bisects SH. BB' is a tangent since SCB is a right angle. 52. 8E will evidently envelope a conic of which S is focus and the given circle auxiliary circle. 53. Join SP, the bisector of POS bisects SP in V\ since the locus of P is a circle, the locus of F is a circle. Hence VO envelopes a conic of which /S is a focus. 54. The angles RSP and QHV are the complements ofP/S'TandQlTT: Hence TJaS'P + C^r= supplement of half PSQ^PHQ = half the angle between the tangents at P and Q, by Ex. 23 on the ellipse. 55. TS, ZS are the interior and exterior bisectors of the angle QSR, and if TQ meet PZ in P, FS is the exterior bisector of the angle QST. Hence Q, T, R lie on a conic of which S is focus and PZ directrix and ZT is the tangent at T since ZST is a right angle. 56. If OE be the radius of the sphere, OE.AC=2irQ2iOAA' and varies as OA . OA' ov AD . A'D' that is as BC\ Hence the iatera recta of all the sections are the same. 57. Let PQ\ QP' meet the ellipse in ZZand F, then since TP^ = TQ . TQ' and TQ^=TP,TP\ TP : TP' :: TQ' ; TQ, or PQ', P'Q are parallel. 102 Miscellaneous problems. Hence if OF be parallel to P Q and PQ', P'V.P'Q : P'P'' :: CF^ : CD\ and Q'U.qP : Q^Q-^ :: (7i^^ : GEK JS^ow P'P^ : qq^" :: TP^ : T^Q'^ .. ^j^q . 7^^/^ and CE'^ : (7/>2 :: TQ^ : TP'^ :: TQ : TQ'. Hence P T. P'Q .qU.Q'P.: TQ^ : TQ 2 ;: p'Q2 . pqi^ Therefore PV : P'Q y. Q'U : QP. 58. If the tangents at P and Q meet in T, CT which bisects PQ is parallel to P'Q, P' being the other extremity of the diameter PCP\ Hence the angle PCT= PP'Q = TPQ. 59. /S', P, aS", g lie on a circle, hence Sg : P^ :: S'G : aS'P :: aS'^ : AX. 60. ^(7 is parallel to the polar of A, hence AD is parallel to the axis. Also the angles SAC, DAB are equal : and the angles ABDj A SO are likewise equal. Therefore AC : AS :: AD : AB. 61. RK : QN :: KC : NC and PN : RK :: iV^(? : KG. Hence ^C : AC :: KCNG : NC.KG, or ^(7 : ^(9 :: AC : ^(7. Therefore (7i2 : CQ :: (7^ : CN :: AC+BC: AC, or CR=^ AC+BC Hence if iVP meet /?// in iV^', PiV'^ QiV. Hence KL passes through P. Also PL = QC=AC, and KP = QR^BC 62. CT.CN^CA^ and CT',PN=BC'^) Miscellaneous Problems, 103 hence CT.CT : CA.CB :: CA.CB : CN.PN. Hence the triangle CTT' varies inversely as the tri- angle FCN, 63. By Ex. 6. Chap. VII. 2SP=3ACy aiid if CB be drawn parallel to the tangent at P, P£=AC. 64. Let CT which bisects PP' in V, meet the ellipse in Qy and let Cj& be conjugate to CQ. Then PV^' : (7F. F^ :: GE^ : CQ^ :: PF2 : CQ^-CV\ Hence ce2=(7r. rr+(7F2=6^r. cr, or 2LP, T'P' are tangents. 65. See Ex. 82 on the hyperbola. 66. If we reciprocate with respect to a focus the theorem that tangents to an ellipse at right angles intersect on a fixed circle, we find that if the sides of a quadrilateral A BCD subtend each a right angle at a fixed point S the sides envelope an ellipse of which /S'is a focus. If be the centre, the angle 0^5 = complement of half -4 05 = complement oiSGB = CBS=SAD. Hence is the other focus. 67. If A', B, C, D' be the points of contact and E\ F\ G' the points of intersection of A'G', B' D'\ A'D\B'G'\ A'B',C'D\ then E'F'G' is a self-conjugate triangle. If BA'A, G'DFmQQt in F the pole of A'G\ F will lie on F'G' the polar of E\ since E' lies on A'G' the polar of F. Similarly AD'D, BB'G meet in G on F'G' and AC, BD in E\ Let A'D\ CC'D meet in a, and B'G', ADD in /S. Then if be intersection of AC, GD\ a8 is the polar of 0. 104 Miscellaneous Problems, Now since GB. GD' are tangents and the tangent at C^ meets them, the range CC'DF is harmonic and therefore the pencil GB\ GC\ G^, GF\ Hence B'C ^F' is a harmonic range. So AD' aH is a harmonic range. Hence ajS passes through G'. Since G' hes on the polar 0, lies on BD, the polar of G. Similarly AB\ CA' intersect on BD, 68. By Art. 138 the four points in which two rect- angular hyperbolas intersect are such that any one of them is the orthocentre of the triangle formed by the other three : hence any conic through the four points is a rect- angular hyperbola. 69. If KVt be drawn parallel to the axis to meet, PQj ST in F and ^, and if KL be perpendicular to PQ, PL = SQ and Pr= VQ, hence tV= VK. Therefore ST SP, SK and the axis form a harmonic pencil. 70. Let DE meet PF in K) and let PD, P'E meet Then GH is the polar of K^ but K lies on DE the polar ofi^. Hence F lies on GH, and FG is parallel to the chords bisected by PP'. 71. Let RR the common tangent be bisected by PQ the common chord in 0. Then RO" = OR'^ =OQ,OP\ hence RR\ PQ are equally inclined to the axis. Hence PR is a diameter, and the diameter of curvature = 2PF=2CD, Therefore CD'- = CD . PF= A (7. BC. 72. Reciprocate with respect to S the following theorem : S is taken on the outer of two concentric circles ; S Y, SZ are drawn perpendicular to a pair of parallel tangents to the two circles ; YZ is constant. Miscellaneous Problems, 105 73. Let the tangents at P and Q meet in T. Then the angle P'/e(2' = PJ'Q- supplement of P(7Q, by Ex. 68. Hence P', C, Q' and li lie on a circle. 74. TN is half the difference of QM and Q'M' and i?i? is half their sum. Hence R'P = | Q'M' = ^J?. Now PN : Q'M' :: TK : 2A'^/2 :: Pil/ : QM :: QJf : 4SP, Hence PiV^ : Qi»f :: Q'M' : 4>S'P :: PM' : Q'M'. Therefore PN^ : PM'^ :: Qil/^ : Q'3f^ :: PJ/ : Pil/; or PN''=PM.PM', 75. Let i2i2' F be the diameter bisecting PQ. Then if PQ meet a common tangent /?p' in O, Oi?^ : OP. OQ :: Sp : .ST? :: S'p' : S'R' :: 0^^ : OP, OQ, or Op = Op', 76. Let 0, (7 be the centres of the hyperbola and ellipse to ibCh'f the tangent at (7, then since PN , Ct-=r.Clr and CN . CO = Crt^, the area of the ellipse will be a maximum, when CN,PN is maximum, that is, when CN,NO is a maximum, or ON=NC, Hence PP' is a tangent to a similar hyperbola. 77. RP .RP'=Rm-PN^=RN^-4.AS .AN. Now RN'^ : ^^aS'.^^' :: AN^ : AA'\ or i2iV^2 . 4^^. jijsf :: jijsf : aA\ Hence RN'^-PN^ : 4^AS'.^iV^ :: ^W : ^^'. Therefore RP.RP' : AN.A'N :: 4:AS : A A'. 78. Let the tangent at P meet the confocal conic in Q. Draw GEF parallel to PQ meeting the normal at P in F, Then OP . PP= (72>2 = ^s^p p^/ ^ ^^2^ (7^? being conjugate to CP. Hence is the pole of PQ with respect to the confocaL 106 Miscellaneous Problems. 79. Let the tangents naeet in U, SU meeting the curve in Q, and let the tangent at Q meet BR' in T, Then, V being point of contact of RB\ TSR=TSQ+ l/SP-RSP=TSV+ USP'-RSV = USP'-RST, .-. 2TSR=USP'=RSR', :. locus of T is tangent at Q. Or by reciprocation of the theorem, If ABC be a triangle inscribed in a circle, and DE the diameter perpendicular to AG, DB and EB bisect the angle B and its supplement 80. Reciprocate with respect to any point S the theorem that if two points on a circle be given, the pole of PQ with respect to that circle lies on the line bisecting PQ at right angles. 81. PQ.PR = PE.PF=AG,BG and QR = GE^AG-BG, Hence PQ=BG 2ind PR= AG. Also ER is parallel to GQG. Hence PG : GD :: PG : PE :: BG : AG and PG.PF=BGi Hence GQ, GR are the axes. 82. This is a particular case of Art. 195, since the second point where AE meets the curve is at an infinite distance, hence AE=EK. 83. The circle of curvature is greatest at the ex- tremity of the minor axis. Hence BO the direction of the minor axis is given. And BG.BO = AG'^=SB\ being the centre of curvature. Hence S lies on the circle of which BO is diameter. 84. Ga : Gh :: Ba.Ac : bA . cB and Oa : Gb' ;: Ba.Acf : h'A . c'B, by Todhunter's Euclid, Art. 59. Miscellaneous Problems. 107 And Ac . Ac' : Ab.Ah' in ratio of squares on parallel diameters. Hence BG is the tangent at a, 85. This depends on the fact that any chord is bisected by the diameter through tlie intersection of the tangents at the-ends of the chord. 86. Let P, Q be the points of contact of parallel tan- gents to the conic and circle. Then by Art. 132, the angles PGA, QGA are equal. 87. Let GL, GL be the fixed straight lines, /S'the fixed point. Then the angle LSL'^GSL' + GL'S, hence the angles GL'S, GSL are equal. Therefore GL : GS :: GS : GL\ or LU touches the hyperbola of which CX, GL' are asymp- totes and S focus. 88. Since the semi vertical angles are complementary they touch one another along their common generating line. Now EA : AX :: GE : EG :: OE' : G'E' :: EA : AS'. Hence S' coincides with X, and similarly JS with X'. 89. Draw GR F perpendicular to the tangent at P, GE bisecting PP' and PN parallel to GE. Then QO.OQ' : PO.OP' :: GL>^ : GR^ :: PO.PF: GN. GV V. PO \ PE V. "IPO : PP'. 90. A circle can bo described with centre T to touch SP, SQ, HP, BQ. Hence SN-NII=SM-MH, and TM^ TN bisect the angles at M, N. Hence TM, TN touch a confocal conic passing through M andiV; 91. If aS' and H are the given points, the locus of P is the conic in which the given plane through aS' intersects the 108 Miscellaneous Problems, surface generated by the revolution about SH of a conic of which S and H are foci. If ST be drawn perpendicular to the plane to meet the directrix plane corresponding to yS'in Ty the cone formed by joining H to all points of the locus of P is a right circular cone of which TJri is axis. 92. PG.Pq = BG^ = PQ^; ,'. PGQ and PgR are similar triangles. 93. If OL be the ordinate of 0, LG : GN :: OL : P'N :: CL : CN, or LG : LG :: CN : NG :: AG : BGK Hence CL : (76^ :: ^(7^ : AG' + BGK Therefore GO : GP' :: (7^ : CiV^ :: AG^^BG^ : AG^+BG\ 94. The polygons FaSP F and Z'HPZ are similar and the perpendicular from G on FZ bisects VZ ; hence if FLE^be taken on FF' such that VE=ZZ\ then GE=GV'=GZ\ Hence W ,ZZ= W . VB= VG^- rG^ = GA'-GA'\ 95. The centre is the middle point of GP, 96. TSQ and T'/S^'Q are right angles ; .*. the middle point of TQ is the centre of the circle TSQS' and is equidistant from S and S'. 97. 7^(2 : TP :: aS'Q : /S'r, and TP : TQ' :: ST : /SQ' ; .-. TQ,TP : TP,Tq :: aS^CaS^^ : SQ! , ST \\SQ,Pr '.sq:,pt. 98. Draw NE perpendicular to NM^ and prove that E is a fixed point in the axis. 99. P and Q are equidistant from the plane of the circular section of the cone, which contains the centre of the section. 100. Produce OG to E so that GE= OG ; then PE is parallel to GZ and EP Y= GZO = OPY; that is, the tangent to the curve bisects the angle OPE, Miscellaneous Problems. 109 101. If PEQ be one of the tangents, and ERV the chord, EP^ : ER.EF :: EQ^ : ER.EV, for each ratio is that of the parallel focal chords. 102. If i^^, F(? be the tangents, i^(ra:QGf) is harmonic, and EFG is a right angle. 103. Reciprocate with regard to C the theorem, that, if a circle centre C intersect another circle at right angles at the point E, and CPQ be any chord, CE^ = CP, CQ. 104. Reciprocate the conies into two intersecting circles. 105. Reciprocates into the theorem of the existence of the director circle. 106. If PEQ be the chord required, and P'EQ' a consecutive chord, the areas PEP\ QEQ' are ultimately- equal, and Ey which is the centre of curvature, is the middle point of PQ. PQ is therefore the diameter of curvature and is in- clined to the axis at the same angle as the tangent, i.e. half a right angle. 107. The pole F of the straight line is fixed, and P, the point of contact of a tangent, is the foot of the perpen- dicular from F on the normal. 108. The angle MNC=LCN=LCN, if I be the point where the tangent meets the other asymptote, .*. MN is parallel to Clj and passes through P the middle point of LI. 109. The diameter of curvature being the same for both, it follows that /SP : S'P is a constant ratio. 110. CK'^=CF^ + PK'^ + 2PK\PF=AC'^-hBC^ + 2AC.BC; 110 Miscellaneous Problems, 111. If P, V, R, Q be the points of contact of AB, BC, CD, DA, 2ASB = PSV-\-FSQ, and two right angles = P/S" V-h ASP + VSC= PSV+ ASQ + CSR ; :.PSV=QSE = 2QSD, and 2ASB = 2QSD + PSQ = 2ASD, .'. ASB = ASD = 2l right angle. Also ASP + DSE = ASD, .'. PSQ is a straight line and PA, ED intersect on the directrix. 112. If the tangent at P meet the director circle in E and T, perpendiculars to the tangent through E and T are tangents to the ellipse. Draw PB parallel to CE, meeting CT in V and take CQ^= CV. CT', similarly find the point D on CE ; then CQ and CD are conjugate diameters, and the con- struction is completed in Art. 216. 113. Q'E' meets the axis in T, the pole of NP ; .*. the tangents at Q', E', meet at a point E on NP, LetCPmeet Q'PMn F; then EN.PN=^EN^-EN.EP = EC''-CN'^-CY. CE, = EC.CY- CN^ =CA'- CN' = PNK /. EN : PN=PN : PN=AC : BC=Q'M : QM, and the tangent at Q passes through P'. 114. If S^ be the other focus of the fixed ellipse, aiKi H of the moving ellipse, S'P = HP and S'Q^-HQ. Join S'H meeting the chord in Z, and let fall SY the perpendicular on the chord ; then SY.S'Z^^SY. HZ^BC\ and in the chord touches a confocal conic. 115. Let be the centre of the circle, PQ a chord of intersection not perpendicular to the axis, meeting an asymptote in L, and the axis in K, Micsellaneous Prohlenis, 111 angle EOG^ 90« - LKG= 90" - LGA + CLK = 46<^ + CLK^ LGA + LGE= EGO, /. EGO is isosceles, and E lies on a fixed line perpendicular to the axis. 116. Project the ellipse into a circle, and prove that the angle DQP=QPR, observing that the tangent and common chord are equally inclined to the axis. 117. Reiprocates into the following : If /S' be a fixed point, and SK a tangent to a circle centre G, and if TE be any other tangent from a point T, and the angle GTE=GSK, the locus of T is a circle pass- ing through S, 118. SY, HZ being perpendiculars on the tangent, Pq : HZ :: PT : TZ :: TR : TG :: PR : PE; /. Pq : PR :: HZ : AG :: HP,£G : AG. GJJ :: HP.PG : GB' :: PG : S'P; .'. Rq is parallel to SG. 119. IfPTfQ T and pt, qt be two near positions and tMy Tm be drawn perpendicular to PT^ Qt, then tM : Tm in the ratio compounded of PT : QT or GD : GE and Pp.QO' : Qq,PO or Pi^ : QP' : Q0\ PO being the radii of curvature. Hence tM=Tm, or the normal at T to the locus of T bisects PTQ. Therefore T lies on a confocal ellipse. 120. Referring to the figure of Art. 148, and drawing the lines, a circle can be drawn through ADOG; .-. angle DOA = DGA = 90' -GVA =AEG, and the triangles AOD, AGE are similar. .-. AO : AD :: GE : AG, or AO,AC-=AD.A'D'-^BG\ 112 Miscellaneous Frohlems. 121. Referring to the preceding theorem, describe a sphere centre G and radius GA ; the tangent from any point of the ellipse to this sphere will be equal to the tangent from P to the circle of curvature. Describing a similar sphere with centre G\ the sum of the tangents = FA' = SS\ 122. In the second figure of Art. 144, take T smj point in the tangent at P and let C be the centre of the upper sphere. Then CRP and CSP are right angles, PR = PS, and TR — TS these lines being tangents. .'. T lies in the plane through CP perpendicular to the plane CRPS-, .-. angle SPT^RPT, and RTP = STP; similarly RTP = S'TP. Hence RTR' = STP + S'TP = STP + STQ^PTQ, TQ being the other tangent. 123. Produce ER to E' making RE' equal to RE. Then the polar of E' passes through E. Now C is the pole of PQ which passes through E. Hence CE' is the polar of E and is therefore parallel to ARB. Hence CE is bisected by AB. Again CE bisects in E the polar of E' which is parallel to^^. Therefore CE bisects AB, Therefore ACBE is a parallelogram. 124. Let be the centre of the conic which touches the sides AB, BC, CD, DA in E, F, G, H. Then since OA, OB, OG, OD bisect HE, EF, FG, GH the sum of the areas of the triangles AOB, COD is half the area of the quadrilateral Miscellaneous Problems. 113 Let AB, CD meet in K and let O, 0' be two positions of a Draw OM, OM' perpendicular to AB and ON, O'N' per- pendicular to CD. Also draw OK, O'L perpendicular to OM and ON: then OM.AB + ON, CD = 0'M\AB + 0'N\ CD, Hence OK. AB = 0L. CD. Therefore OL : OA^'in a constant ratio. Hence the locus is a straight line. 125. By Art. 241 the sum or difference of the tangents is proportional to the distance between the ordinates of the points where the circles touch the curve according as the point does or does not lie between those ordinates. 126. If SY, S^ Y^ be the perpendiculars from the focus on the tangent at P, CY' is parallel to SP ; and, if DK is the perpendicular on CY', DK : OD = SY : SP = BC :: CD; .'. DK=BC. 127. If Fand 7^ are contiguous corners of the parallelo- gram formed by the tangents, and if CV and CT meet in E and F the sides of the parallelogram formed by the points of contact, CE. CV= CP' and CF. CT=CD'; .'. (CE. CF) (CV. CT) = {CP . CDJ. 128. Taking the figure of Art. 12, let TL, TM, TN be drawn perpendicular to SF, SP, FF\ and let the circle intersect FF' in G and G'\ then TG : TN :: TL : TN :: SM : TN :: SA : AX, .'. TG is parallel to an asymptote. 129. For 7W bisects QSq, Art. 12, and FS bisects the outer angle, Art. 5. B. 0. s. 8 114 Miscellaneous Problems. 130. Let the chord Qq, normal at g, meet the directrix in F, and let T be the pole of Qq ; then S being the pole of the directrix, ST is the polar of F, and therefore T being a point in the directrix, TSF is a right angle. Taking V as the middle point of Qq^ let FS meet in L the polar of V, which is parallel to Qq. Then LSQ = TSQ - TSL = TSq - TSF =FSq=SqV-SFq = Pq V-STq, V T, S, q, V are concyclic, = PqV-QTV,KYi.m, =:PVq-QTV=^QVN-QTV ^TQV=LTQ, V 7Z,e Fare parallel. .•. L, T^ S, Q are concyclic, and TQL=TSL = 90^. 131. (1) If the plane through the axis and the given point P intersects the cone in VA, VB, describe a circle passing through P and touching VA, VB; then if APB is drawn touching the circle, AB is the axis of the sectien of which P is a focus. (2) Produce VP to Q making PQ = PF, and in the plane above mentioned, draw VK parallel to VB, meeting VA in ^, and VL parallel to VA, meeting VB in Z; Then APL is the axis of a conic of which P is the centre. 132. If aS', S' are the foci,, and X the foot of the directrix, VS'S is a straight line, and XSS' is an isosceles triangle. Taking A and A^ as the corresponding vertices, draw AL and A'L^ parallel to SS\ meeting AS' in L and AS inZ'. The latera recta are in the ratio of VS to VS', and VS : AS=A'L' : ^Z' and VS' : A'S'=AL : A'L. Now AX= LX, A'X= rX, and AL = AL'; but VS. AL'== AS. AT ^nd VS\A'L = A'S'.AL; .'. VS : VS' = AS.AX : A'S'AX. CAMBRIDGE: PRINTED BT C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. March 1802. 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