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ASSOCIATED CONICS. 
 
 eee ee eee 
 
 I. If an ellipse touches both the branches of an hyper- 
 bola and its conjugate, they are concentric, and the chords of 
 contact are conjugate diameters. 
 
 _ If the chord of contact PP’ (Fig. 1) does not pass through 
 the center, the tangents at P and P‘ meet at a point O, and 
 since the line joining O and the center of either curve bisects 
 PP’, the straight line joining the centers must pass through 
 QO; similarly if the chord of contact DD’ does not pass through 
 the center of either curve, the straight line joining the centers 
 passes through the point of intersection of the tangents at D 
 and D’; but PP’ and DD’ cannot be parallel, therefore the 
 tangents at P and P’ or at D and D’ must be parallel, and there 
 fore the curves are concentric. 
 
 _ If CD is not parallel to the tangent at P, let CEF be paral- 
 lel to it; then the tangents at E and F are parallel which is 
 absurd, because the tangent at F meets CD and the tangent at 
 E meets CD produced ; hence CP and CD are conjugate. 
 
 2. If an ellipse touches an hyperbola and its conjugate, 
 and a straight line is drawn from the center C to meet the el- 
 lipse and the hyperbola in P and Q, the triangles formed by 
 
- 
 
 e 
 es sneey ant” ; 
 
4 
 
 the tangents at P and Q and the asymptotes are in the dupli- 
 cate ratio of CQ to CP ; and if PP’ and QQ are the double or- 
 dinates of Pand @ to one of the diameters of contact, the 
 
 tangents at Pand Q are respectively parallel to the tangents 
 at WY and P’. 
 Because in Fig, 2 PN?: ON. NT: QM?2: CM. MT’, 
 and PN?; QM?2:: OCN2:CM2, 
 therefore NT; MT’:: CN: CM:; PN: QM, 
 and Q’M: MT’:: PN: NT; 
 hence PT is parallel to Q’T’. Similarly P’T is 
 parallel to QT’. 
 Because CN: CM:: CT’: CT:: CR’: CR, 
 therefore CP: CQ: CQ’: CR, 
 and the triangle ICI’: the triangle LCL’:: the triangle 1Cl’ 
 : the triangle FCE’ = the duplicate ratio of CK: CQ! = the 
 | duplicate ratio of CQ: CP. 
 
 Hence it follows that the diameters passing through the 
 extremities of a double ordinate to the diameter of contaet bi- 
 sect parallel chords of the two curves, 
 
 If the ellipse and the hyperbola have the same principal 
 axes, the triangles /C/’ and LCL’ are similar, and the. tangents 
 at P and Q are equally inclined to either axis. 
 
 8. If an ellipse touches an hyperbola and its conjugate, 
 and tangents are drawn from any point to both the curves, 
 the triangles formed by the chords of contact and the asymp- 
 totes are equal 1m area. | 
 
 Let O be the pole of the chords RR’ and FF’ in Fig. 3.with 
 respect to the ellipse and the hyperbola respectively. 
 
 Because CV: CP :: CP: CO, and CO: CQ:: CQ: CV’, 
 
 therefore CV : CV’ :: CP%:CQ?, 
 
 hence the triangle ICI’ = the triangle LCL’. 
 
5 
 
 If ON is drawn parallel to the common tangent at A, 
 CN.CT = CA2, and therefore l/ and LL’ meet the 
 diameter of contact in the same point T. 
 
 If the ellipse and the hyperbola have the same principal 
 axes the triangles ICV’ and LCL’ are equal in every respect, and 
 the chords RR’ and EE’ are respectively equal to KK’ and FF’. 
 Since the triangles ICI’ and LCL’ are equal in every respect, Il’ 
 and LL’ are equally inclined to either axis. 
 
 4. If an ellipse touches an hyperbola and its conjugate, 
 and a tangent is drawn from any point in either cwrve to the 
 other, the chord of contact towches the former curve. 
 
 Because in Fig. 2 the triangles FCF’ and LCL’ are equal, 
 therefore the triangles formed by the tangents at P and P’ and 
 the asypmtotes are equal (Art. 2), and because in Fig. 3 the tri- 
 angles ICI’ and LCL’ are equal,therefore LL’ touches the ellipse 
 if //’ touches it. Similarly it may be shown that if JU’ touches 
 the hyperbola, LL’ also tonches it. Or we may proceed thus. 
 Because ( Fig. 2) CP’: CQ’:: CQ’: CR, therefore P’ is. the 
 pole of the tangent PR with respect to the hyperbola. 
 
 Since the polar of P ( Fig. 2) with respect to the hyper- 
 bola is the tangent at P’, and the polar of Q with respect to the 
 ellipse is the tangent at Q’, the ellipse is its own reciprocal 
 with respect to the hyperbola, and the hyperbola is its own re- 
 _ ciprocal with respect to the ellipse. The ellipse and the hyper- 
 bola and its conjugate which have the same conjugate diame- 
 ters, are called associated conics. They may also be called self- 
 conjugate conics, since each is its own reciprocal with respect 
 to the other. . 
 
 5. If CP’Q’ is drawn bisecting KK’ (Fig. 3), the tangent 
 at P’ is parallel to the tangent at Q, because it is parallel to 
 KK’, and therefore PP’ is a double ordinate to CA. Because 
 the pole O' of KK’ with respect to the ellipse is in CP’Q’, and 
 
KK’‘ and RR’ meet in CAT (Art. 3), therefore OO’ is parallel to 
 the tangent at A and is bisected by CA. Henee the straight 
 line OO" joining the poles of any straight line LL’ with respect 
 to the ellipse and the hyperbola is bisected by CA and is pa- 
 rallel to the tangent at A. 
 
 6. Because in Fig. 2 NT: MT’:: CN: CM:: CT: CT, 
 thereforeCT 'NT=CT“ MT’, and CT2-CA2=CA?-CT”?, 
 and therefore ATT A’ = AT’ T’A’. Henee the ordinates through 
 T and T’ to the diameter ACA’ are equal. 
 
 If the tangents at P’ and Q meet the ordinates QQ’ and 
 PP’ in K’ and E, PN: QM:: CN: CM:: CT GP Reais 
 MT :: EN: K’M’ and therefore PT’ is parallel to QT, and PE: 
 EP’ :: QK’: K’Q’:: CP: CQ, because EP’K’Q is a parallelo- 
 gram and its opposite sides are equal. Similarly the tangents 
 at P and Q’ divide QQ’ and PP’ in the ratio of CP to CQ. 
 
 Because PE: QK’:: CP: CQ, EK’ passes through the 
 center, and because EP’K’Q is a parallelogram,EK’ bisects b’Q. 
 
 7. If anellipse touches an hyperbola and its conjugate, each 
 curve cuts the conjugate diameters of the other proportionally. 
 Let CPP’ (Fig. 4) be any semidiameter. Draw PE’ and 
 P'D’ parallel to the asymptote CO’ meeting the ee and the 
 conjugate hyperbola in E’ and D”. , 
 
 Because P’D’ is parallel to CO’, thoreforad it is bisected by 
 the asymptote CO, and because CA and CB are conjugate semi- 
 diameters, therefore AB is parallel to CO’, and therefore PE’ 
 which is parallel to CO’ or AB is bisected by. CO which bisects 
 AB.. Because PE’ and P’D' are both bisected by CO and are 
 parallel, E’D’ passes through C, and CE’: CD’: : CP; CP". 
 
 If E’D and D’E are double ordinates to CB, DE passes 
 through C, and CD: CE:: CE’ : CD’, and because the tangent 
 at D is parallel to the tangent at D’ (Art. 2) and therefore to 
 CPP’, therefore CD is conjugate to CP. 
 
9 
 
  Itis evident that if BC.is at right angles to AC, CH’ is 
 
 equal to CD which is conjugate to'CP,; and CE is equal to CD’ » 
 which is conjugate to CP’. 
 
 8. Let PNP’ be.a double ordinate to ACA’ (Fig. 5), and 
 
 the tangents at P and P’ meet the tangents at A and A’ in F, E, 
 ¥’, and E’. 
 
 ~ Because PNP’ is parallel to EE’ and FF’ and is bisected by 
 AA’, and the tangents at P and P’ meet in A’A produced, there- _ 
 fore EH’ and FF’ are bisected by AA’. 
 Let EF and E’F’ meet in M, A’P and P’ Ain Q, and A’P’ 
 
 and PA in Q’.. Because MANA’ is harmonically divided, QQ’ | 
 
 passes through M and is parallel to FF’, and EF’ and EF bi- 
 sect PP’inN.. : 
 
 Beeause QM : A’M:: PN: A'N, and QM: AM:: PN: AN, 
 therefore QM?2: AM: A’M:: PN?; AN? A’N:: 
 BC? ; AC*, Hence Q and Q’ are points on the 
 associated conic, and because MANA’ is harmonically divided, 
 CM:CN=CA?,and QN and Q’N touch the hyperbola. 
 
 We have proved that if A’PQ is drawn to meet the curves 
 in P and Q, the tangents at P,Q, and A meet at a point; the 
 tangents at P and Q pass through the feet of the ordinates of 
 Q and P respectively ; and A’PQ is cut harmonically by the 
 ellipse and the tangent at A. It may also be shown that any 
 chord of the conjugate hyperbola passing through A or A’ is 
 
 divided harmonically by the tangents at A and A’ to the 2 f 
 ginal hyperbola. 
 
 Because QMQ’, the polar of N, passes through Q, therefore 
 the polar of Q passes through N, and similarly the polar of P 
 passes through M. Hence the tangent QN at any point Q of 
 the hyperbola is divided harmonically by the ellipse and the 
 diameter of contact AA’, and the tangent PM at any point P 
 
 of the ellipse is divided harmonically by the ba a and the 
 diameter of contact AA’. 
 
10 
 
 We have also proved that EF’ and E’F meet at the point 
 of intersection of PP’ and AA’, touch the hyperbola at Q’ 
 and Q, and meet PA, A’P’, and A’P, P’A, produced at the 
 points of contact. Thus the triangle NQQ’ whose sides are a 
 double ordinate to AA’ and the tangents at its extremities, is a 
 self-conjugate triangle formed by joining the points of inter- 
 section of the opposite sides and diagonals of the quadrilateral 
 APA'P’. It is also formed by the diagonals of the circumscri- 
 bing quadrilateral EE’F’F. It is evident that the reciprocal of 
 a self-conjugate triangle of a conic with respect to an associated 
 conic is a self conjugate triangle of the original conic. 
 
 Because the polars of Q and Q’ with respect to the ellipse 
 are the tangents Q’N and QN (Art. 4), and it is obvious that if 
 QN and Q’'N meet the ellipse in O and 0’, the tangents Q’O and 
 QO’ meet in A’A produced, and similarly the other tangents 
 from Q and Q’ meet in AA’ produced, therefore one of the dia- 
 gonals of the quadrilateral formed by the tangentsfrom Q and 
 Q’ to the ellipse must coincide in direction with PP’. Hence 
 if EN and E’N cut the ellipse ia L and L’, LO and L'O’ meet 
 A’A produced in M, and the triangle formed by the points of 
 intersection of the opposite sides and diagonals of the quadri- 
 lateral formed by the tangents from Q and Q’is the same as 
 
 the self-conjugate triangle NQQ’. Similarly the triangle MPP’ 
 is self-conjugate. 
 
 If CP and CQ meet the tangent at A in K and K’, AK and 
 AK’ are respectively equal to QM and PN, because QM? ; 
 
 PN? :; CMMN: CN.NM :: CM: CN:: CM2; 
 CA2:: QM? : K’A2 3; CA#;: CN?2:; AK?; PN?2, 
 If the angle BCA is right angle, AP and AQ are equally 
 
 inclined to the transverse axis, because the angle PAN=the 
 angle Q’AM=the angle QAM. 
 
 If from any point Tin PP’ produced TR and TR’ are 
 
11 
 
 drawn touching the ellipse, RR’ passes through M,' and the 
 
 straight lines joining the extremities of PP’ and the ordinate 
 through M pass through A and A’. 
 
 9. Ifan ellipse touches a given hyperbola and its con- 
 jugate, the rectangle contained by its principal axes is constant 
 for it is equal to the area of the common circumscribing paral- 
 lelogram which is equal to the rectangle contained by the 
 principal axes of the given hyperbola. 
 
 If a parabola touches an hyperbola and its conjugate at 
 the extremities of conjugate semidiameters and is bounded by 
 an asymptote, its area is constant. 
 
 Let CP and CD (Fig. 6) be any two conjugate semidia- 
 meters. Because PL=PL’, therefore CV=VL', and VO=OL’. 
 Hence CO=3VO, and because CQ? ; PV2 :: CO: 
 
 VO, therefore CQ*=3PV2. Because CL is equal to 
 twice PV, therefore CL bears a constant ratio to CQ, 
 
 Again the area of the parabolic figure QOQ’ is equal totwo- 
 thirds of the rectangle contained by QQ’ and the perpendicular 
 from O to CL, therefore the area of QOQ’ bears a constant ratio: 
 
 to the triangle LL’ and is constant, since LL’ is half the rect- 
 angle contained by the principal axes of the given hyperbola. 
 
 10. If PCP’ and DCD’ (Fig. 7) are any two conjugate 
 diameters of an ellipse or hyperbola, and any ordinate RR’ to 
 DCD’ is cut by the ordinate through Q to PCP’ in M and by 
 QD, QC, and QD’ in V, O, and T respectively, TMVO and 
 TRVR’ and harmonic ranges.. 
 
 Because VN : ND: :QN’ :. N’D,and TN : ND’ :: QN’: N’D' 
 therefore VN. TN: ND.ND’:: QN”?: N’D,N’D?:: 
 RN2 ;:ND.ND’, hence RN*2=VN.TN, 
 
 Again because VN: RN:: RN: TN, 
 therefore RV: RV:: RT: RT. 
 Hence TRVR’ is an harmonic range.. 
 
12 
 
 If the tangent at’D° which is parallel to CP meets D’Q 
 - produced in K, it may be easily shewn that the tangent. at Q 
 
 bisects DK, and therefore T’t=tV’, and according to the proof 
 given above W‘C.T’C=CP2=CM.Ct, 
 therefore Ct2-¢V’2=CM’'.Ct, and C12 CM". Cet? : 
 “2029 Henee Of: ¢V" tv": eM and CT’: CVS: TM’? MoV), 
 _ and therefore T’M’V'C and TMVO are harmonic ranges. 
 From the ‘above it is manifest that ‘if the rectangle con- 
 ~ tained by two parallel sides of a quadrilateral is constant, the 
 - locus of the point of intersection of the diagonals is an ellipse 
 when one of the other sides is fixed and the locus of the point 
 of intersection of the other sides is an hyperbola when one e of : 
 the diagonals. is fixed. 
 
 11. The reciprocal of an ellipse or hyperbola with respect 
 
 to another ellipse or hyperbola which touches it at the extrre- 
 
 mities of a diameter is an ellipse ov hyperbola ae the 
 same diameter. 
 
 Let the ellipse APA’ (Fig. 8) touch the ellipse APA! ab 
 the extremities of the diameter ACA’, and let OT, the polar of 
 
 P, meet CA produced in T. Draw the ordinate PPV, and let 
 it cut OT in P”. 
 
 Because PP’P"R is harmonically divided, » 
 therefore P"V: PV:: PV: PV:: BC: BC : 
 Hence the locus of P” is an ellipse, and. P’T touches it, 
 
 because CV.CT=CA?, 
 
 Therefore the reciprocal of the ellipse APA’ ath reeeeek 
 to the ellipse AP’A’ is the ellipse APA’, Similarly it may be 
 shewn that the reciprocal of the en be AQ with’ hes to 
 the hyperbola AQ’ is the hyperbola AQ”. | 
 
 From the above it follows that the ict of the auxi- 
 liary circle with respect to the ellipse isan ellipse having its 
 minor axis equal to the latus. rectum of the given ellipse, and‘the 
 
183 
 
 “yeciprocal of a rectangular hyperbola with respect to an hyper- 
 bola having the same major axis is an hyperbola whose conju- 
 gate axis is equal to the latus rectum of the latter hyperbola. 
 
 12, The reciprocal of a conic with respect to a concentric, 
 similar, and similarly situated conic is a concentric, suemilar 
 and similarly situated conic. | 
 
 Because (in Fig. 9) CP”: CP’:: CP’: CP:: BIC: BC, 
 
 therefore the locus of P” is a concentric, similar, and simi- 
 
 larly situated ellipse, and because the tangent at P’ is parallel 
 to TT’, therefore TT’ touches the ellipse A”P”B”. Hence the re- 
 ciproeal of the ellipse APB with respect to the ellipse A’P’B’ 
 or its associate A’Q’ is the ellipse A”P’B”. Similarly the reci- 
 procal of an hyperbola AQ with respect to.a concentric, similar, 
 
 and similarly situated hyperbola A’Q! or its associate A’P’B’ 
 is a concentric, similar, and similarly situated hyperbola A”Q”. 
 
 13. ABQ and A’B’R ( Fig. 10 ) are two concentric ellipses 
 having their major axis in the same straight line. Let AB,P 
 be similar to A’B’R. Draw an ordinate QPN. Let the recipro- 
 cal of AB,P with respect to A’B’R be A”BgP’ (Art. 12). P’T’ 
 
 and QT’, the polars of P and Q with respect to A'B/R, meet 
 in the major axis, because CN. CT’=CA’, 
 
 Becanse PN. P’N=RN?-QN. Q’N (Art. 10), 
 therefore QN: PN:: P’N: Q’N:: P'N’: Q’N’: BG: B,O. 
 Hence the locus, of Q’ is an ellipse having the same major 
 
 axis as A”B,P’, and because CN’, CT’=CA”®, there- 
 
 _ fore Q’T’ touches the ellipse A”B”Q’, and the reciprocal of the 
 
 ellipse ABQ with respect to the ellipse A’B’/R is the ellipse 
 
 —A”B'Q’ such that AC.A”C=A’/C2,and BC. B’C=B’C2, 
 
 because BoC: B’C:: P’N’: Q’N’:: BC: B,C, and 
 
 ByC. B,C=BR/C2, ( Art. 12). | 
 
 14, Let PDP’ (Fig. 11) be an ellipse. 
 
 It is required to describe the hyperbola and its conjugate 
 
14 
 
 which touch it and have their conjugate diameters along two 
 given straight lines CA and CB. 
 
 Draw CE’ conjugate to CA, CP parallel. to EP’ and. CD 
 conjugate to CP. Then the hyperbola having CP and CD for its 
 conjugate semidiameters is the hyperbola required. Because 
 the tangent at B is parallel to the tangent at EH’ (Art. 2), it is 
 
 also parallel to CA, and therefore CA and CB are conjugate 
 semidiameters of the hyperbola. 
 
 If the hyperbola is given, and it is required to describe 
 the ellipse which touches it and its conjugate and has its con- 
 jugate diameters along CA and CB’, draw CB conjugate to 
 CA and CP parallel to BB’. If CD is conjugate to CP, the 
 
 ellipse which has CP and CD for its conjugate semidiameters 
 is the ellipse required. 
 
 15. Let AB and A,B, (Fig. 12)be any two concentric ellipses. 
 It is required to find the reciprocal of AB with respect 
 
 to A,B,. 
 Describe the hyperbola A,B, and its conjugate 
 
 which touch A,B, and have their ‘tba hile axes along CA 
 and CB ( Art. 14), 
 
 Let A;B, be the reciprocal of AB with respect 
 to A,B,, (Art. 13), and let CD meet the ellipse A Bs 
 in D’. Let CP’ be conjugate to CD’, and let the tangent PO 
 meet CP in O. Produce P’O to P” so that P’O=P'0. Then the 
 ellipse A,B, having CD’ and CP” for its conjugate 
 semidiameters is the reciprocal required. The polar of a point 
 Q with respect to the ellipse A, Bj, is the same as the polar of 
 
 apoint Q’ with respect to the hyperbola A2B,, if QQ’ 
 is parallel to CD and is bisected by CP ( Art. Bi) 
 Because P’O=OP”, therefore VM=MV’, and QV=Q'V’. 
 Because CP’”2-CV2 3: CP’2-CV’2?::: CP”2:; CP’, 
 and Q’V’2; CP’2-CV’? ;: CD’: CP 
 therefore QV?: CD’? :: OP. CV2;: CP”2, 
 
15 
 
 Hence the locus of Q is an ellipse having CD’ and CP” for 
 its conjugate semidiameters. 
 
 16. Let AB (Fig. 13) be an ellipse. 
 
 It is required to find its reciprocal with respect to a con- 
 centric hyperbola A,B). 
 
 Let the ellipse A,B, be the reciprocal of the 
 ellipse AB with respect to the ellipse A,B, (Art. 15). 
 
 Then the reciprocal of AB with respect to the hyperbola 
 A,B, is a concentric ellipse Az;B, which has a point 
 P on it corresponding to a point P’on A,B,, so that 
 PP’ is parallel to the tangent at A, and is bisected by CA}. 
 
 Similarly the reciprocal of an hyperbola with respect to a 
 concentric ellipse or hyperbola may be determined. 
 
 17. Let AR ( Fig. 14) be any conic. 
 
 It is required to find its reciprocal with respect to another 
 conic A,B, whose center S is the focus of AR. 
 
 Draw the circle A,B, whose diameter is a mean, 
 proportional between the principal axes of A,B, and draw the 
 
 rectangular hyperbola A, B, touching the ellipseA, B, 
 at the extremities of its equiconjugate diameters. 
 
 If can be shown that the reciprocal of AR with respect to 
 the circle or rectangular hyperbola A,B, is a circle 
 PQP’ whose center C lies in A,S and_ whose 
 diameter is a third proportional to the diameter of the circle 
 
 A,B, and the latus rectum of AR. 
 
 Hence the reciprocal of AR with respect to A,B, isan el- 
 lipse P, D, P, such that the straight line joining any 
 point Q, on it and the corresponding point Q, on the 
 circle P’QP” is parallel to SD and is bisected by SP. Because 
 the common tangent P’P,is bisected by SP, NM is also bisected 
 by SP, and therefore Q,N=Q,M, and CQ=C,D,. 
 
 Because CP’2-CN2:C,P, 2?-C, M?::CP’2: CP, ? 
 
 therefore Q,.M?:C,P,2-C,M? = C,D,%:C,P,?. 
 
16 
 
 Hence the locus of Q, is an ellipse whose dia- 
 meter D,C,D, parallel to SD is equal to the dimeter : 
 of the se P/QP” and whose diameter P,C, Vg con- 
 jugate to D,C,D, bears to the diameter P’ cp” pb, . 
 the circle a constant ae of CO to OC’, since CO is perpendi- 
 cular to SD and meets PS produced if necessary in O, and CC, 
 is bisected by PS. 
 
 The latus rectum of AR is a fourth proportional to the 3 
 principal axes of ii B, and the diameter of the circle equal i in 
 area to the ellie "p De. 
 
 The reciprocal of a conic AR with respect to an hyperbola, 
 A,B, whose centre is the focus of AR is an ellipse such that the _ 
 straight line joining every point on it and the corresponding 
 point on the eilipse P,D,P, is bisected by SA,, and 
 is parallel to the tangent at ix 1 | 
 
 18. Let A”’R’ (Fig. 15) be any ellipse, and let A’F be any 
 circle, whose center C is in one of the principal axes of A” W’. 
 
 It is required to find the reciprocal of A”R’ with respect — 
 to A'F. | 
 
 Deseribe a circle having A”A” for its diameter, and Jet AP 
 be its reciprocal with respect to the circle A'F. : 
 
 Draw an ordinate R'RM. The tangents at R and R’ meet _ 
 in AA’, and therefore, the straight line joining their poles P | 
 and P’ is perpendicular to AA’. 
 
 The ratio of R’M’ to RM is constant, therefore the ratio 
 of PN to P’N is constant because itis equal to the ratio of QIN | 
 to QN (Art. 10) which is equal to the ratio of R’M to RM. 
 Hence the reciprocal of A”R’ is the conic AP’, and the rect- 
 angle contained by the axes of AP’ and A”R’ perpendicular | 
 to AA’ is equal to the rectangle contained by the axis of AP 
 perpendicular to AA’ and the diameter of the circle A’R. 
 
17 
 
 19. Let ABA’ (Fig. 16) be an hyperbola, and EE’a 
 circle having its center C,; in A’A produced. 
 
 It is required to find the reciprocal of ABA’ with respect 
 to the circle EE’. 
 
 Draw a semidiameter CPQ meeting the associated conic 
 APA’ in P, and draw the ordinate PP’. 
 
 Because the tangents at P’ and Q are parallel ( Art. 2 ), 
 
 therefore their poles P; and Q, lie in a straight line passing 
 through the centre Cj. 
 
 Now because C,T.C,N =C,G.C,P,=C,E?=C,F. 
 CQ) =06,1T.. CM, aid’ 0,A.C,A = CLE 
 =C,A’.C, A, 
 
 therefore C,T: C,A:: C,A,:C,N, and C,T:C, A’ 
 we OPA CAN 
 
 Hence AT: C,A:: A,N:©,N, and A’T: C,A’ 
 aig Nf CIN, 
 
 therefore AT,A’T: C,A.C,A’::A,N.A,N:C,N2, 
 
 Similarly AT’: C,A:: A,M;C,M, and A’T’: C, A’ 
 ;;A,M;C,M, | 
 
 therefore AT’,A’T’-C, A.C, A’:: A,M.A,M:C,M2, 
 but AT,A’T=AT’.A’T’ (Art. 6), 
 
 thereforeA, N.A,N:A,M.A.M::0,N?:C,M?::P,N?2 
 :Q,M?, but P,N?:A,N.A,N is a constant 
 ratio (Art. 18), | 
 
 therefore Q,M*%: A,M.A>M isa constant ratio, 
 
 Hence the locus of Q, is an ellipse, and the hyper. 
 bola A,P,A, and the ellipse A,Q,A, are as- 
 
 sociated conics. 
 
 20. Let the center OC’ of an ellipse A,B, (Fig. 
 17) be in one of the principal axes of another ellipse 
 ABA’. 
 
 __ It is required to find the reciprocal of ABA’ 
 with respect to A,B,. 
 
qi 
 ag NS EEG ISA HRY 
 
 A’ 
 
DR Ne RR ER OER ES 8 se mere 8 Uae A se, 
 
 A 
 
 \ Fux, /f 
 
20 
 
 Describe a rectangular hyperbola A,B, having 
 the equiconjugate diameters of the ellipse A,B, 
 for its conjugate diameters, and let A;B, be its au- 
 xiliary circle, 
 
 Find the reciprocal of ABA’ with respect to 
 the circle A;B,; (Art. 18). Hence find the reciprocal 
 of ABA’ with respect to the rectangular hyperbola 
 A,B, (Art. 5). This reciprocal has a point P on it 
 corresponding to a point P’ on the reciprocal of 
 
 ABA’ with respect to the ellipse A,B, such that 
 PP* is parallel to C’B, and is bisected by C’Ag. 
 
 Similarly the reciprocal of the ellipse ABA’ with respect 
 to the hyperbola A,B, may be determined, and also 
 
 the reciprocal of the hyperbola ABA’ with respect 
 to the ellipse or hyperbola A, B, (Art. 19). 
 
 21. Let AB and A,B, (Fig. 18) be any two 
 
 ellipses. 
 It is required to find the reciprocal of AB with respect to 
 
 + Bat 
 Join the centers C and C’, and let CC’ cut the el- 
 lipse AB in A. Draw AA, _ touching the ellipse 
 AB, and draw C’A, perpendicular to AA,. Bisect 
 AA, in O, and join C’O. Draw C’B; parallel to 
 AAS: 
 jugate touching the ellipse A,B, and having its 
 conjugate diameters along O’O and C’B, (Art. 14). 
 Draw CB, parallel to AA, meeting C/A, produced 
 in C,. Take C,B, equal to CB, and draw the 
 ellipse A,B, having C,A, and C,B, for its con- 
 
 jugate semidiameters. 
 
 The reciprocal of AB with respect to the hy- 
 perbola A;B; is the same as the reciprocal of A,B, 
 
 Describe an hyperbola AB, and_ its econ- 
 
21 
 
 with respect to the ellipse Az;B, (Art. 17), and this 
 reciprocal can be drawn according to Art, 20, and 
 has a point P on it corresponding to a point P’ on 
 the reciprocal of the ellipse AB with respect to the ellipse 
 A,B, such that PP’ is parallel to O’B, and is 
 bisected by C’A,. | 
 
 Similarly the reciprocal of the ellipse AB with respect to 
 the hyperbola A,B, can be drawn, and also the re- 
 ciprocal of the hyperbola AB with respect to the ellipse or 
 hyperbola A,B,. : 
 
 22. The reciprocal of an hyperbola and its conjugate 
 touching an ellipse with respect to another hyperbola and its 
 conjugate touching the ellipse is a third hyperbola and its 
 conjugate touching the same ellipse (Art. 5 ). Also the reci- 
 procal of an ellipse touching an hyperbola and its conjugate 
 with respect to another ellipse touching the hyperbola and its 
 conjugate is a third ellipse touching the same hyperbola and 
 its conjugate. It is also evident that the reciprocals of an hy- 
 perbola and its conjugate touching an ellipse at the extremi- 
 ties of its principal axes with respect to the rectangular hy- 
 perbola and its conjugate touching the ellipse are the same 
 conjugate hyperbola and original hyperbola respectively. 
 
 93. If the rectangular hyperbola AB and the 
 ellipse A,B, (Fig. 19) are associated, the reciprocal 
 of A,B, with respect to the circle AB is an ellipse 
 A, Bghaving its principal axes equal to those ofA, B,, 
 because CA,. CB,=CA?=CA,. CB,=CA,. CB. 
 
 Let CP and CD be any two conjugate semidiameters of 
 A ,B,. If CY is drawn perpendicular to the tangent 
 at P, and P’is the pole of the tangent with respect 
 to the circle AB,CP’.CY=CA?=CA ,.CB,=CY. CD, 
 therefure CP’=CD. 
 
 Tances PN: PN’ :: Bie: Ae %: PN: QN, 
 
 therefore P’N’= QN, 
 
Vp on 
 
 i ne 
 
 Hs 
 
 : caviar ots Pe 
 selgonepenmen 
 
94 
 and because N’K: QN:: CN’: ON: B,C 
 :A,C t: PN: QN, 
 therefore N’K=PN, and PQ=KP’. 
 also PK is equal and parallel to NN’, 
 therefore the angle PQK=the angle PP’K. 
 Again because the triangles CNQ and ON’P’ 
 
 are equal in every respect, 
 therefore OP’=CQ=A,C. 
 Also the triangles CN’K and (QNP are equal in 
 every respect, 
 therefore OP=CK= one and meee ALG: 
 But CK: CQ: :CN :: B,C: AG, 
 ‘hereforo@K 2B. CaOP0'P* andP PA ,C-B,C. 
 Hence the distance of any point P on the ellipse 
 A,B, from the pole of the tangent at the point is - 
 =iiye constant, and K is a point on the circle whose diameter 
 is the minor axis of A,B, or A,B,. 
 Because CP is equal to the semidiameter conjugate to 
 CP’, therefore the pole of the tangent at P’ with 
 respect to the circle AB is the point P. 
 
 Hence CP is perpendicular to the tangent at P’. 
 
 24, The reciprocal of the hyperbola A,B, (Fig. 
 20) with respect to the circle AB is the hy por 
 A.B, such that the reciprocal of the original hyperbola 
 18 cata to its conjugate, and the reciprocal of the conjugate 
 hyperbola is equal to the original hyperbola. It is evident that 
 the hyperbola and its reciprocal are confocal. 
 
 It may be shewn asin Art. 23 that if P’ is the 
 
 pole of the tangent at P, CP and CP’ are perpen- 
 dicular to the tangents at P’ and P, and if P one, 
 the double ordinate to OA., meets the rectaneular 
 hyperbola having A,A-, for its transverse axis in Q, 
 
 QN=P’N’, and P,P’ and CQ bear a constant ratio 
 
25 
 
 and are parallel, Also if P,P’ meets CA, and B,C 
 produced in O and O’, P?O=P,0’=CQ. 
 
 25. Let P (Fig. 21) be one of the points of contact of the 
 ellipse A,B, and the rectangular hyperbola. Let S 
 and S’ be the foci of A, Bay, 
 
 Draw SS, parallel to the tangent at P meeting 
 CB, in 8. , 
 
 Produce CP to meet the directrix XE in EB, and let XE 
 meet the hyperbola AP in Q’. Draw YVQ the 
 double ordinate to CP. 
 
 Because CP and the tangent at P are equally inclined 
 to the asymptote CA,, therefore VEH=VQ=VQ’, and 
 the angle QEQ’ is a right angle. 
 
 Because SS, is parallel to the tangent at P and 
 - jg bisected by CP, and § is the pole of XE with respect to the 
 ellipse AgB,, S, is the pole of XE with respect to 
 the hyperbola AB (Art. 5), and because QQ’ is paral- 
 lel to the tangent at P and is bisected by CP, and XE is the 
 
 polar of S with respect to the ellipse A,B,, there- 
 fore QE is the polar of S with respect to the hyperbola AB. 
 
 Again because S is the pole of QE with respect to the hy- 
 perbola AB, S, is the pole of QE with respect to 
 the ellipse A,B, (Art. 5). 
 
 Hence CX,: CB, :: CB,: CS, 3: CA, : CS: 
 CA,: Cds; | 
 
 therefore CX,. CS,=CB,. CA,=CA’, 
 and X,E is the polar of the focus 8; of the 
 ellipse A,B, with respect to the circle AR. 
 
 Also CS,: CB, :: CA, * CX;, 
 
 therefore CS, . CX,=CB, CA {=CA?, 
 
 and S, is the pole of the directrix X;E., ofthe 
 ellipse A,B, with respect to the circle AB. Take 
 CS,=CS, and CX,=CX,. 
 
28 
 
 We will call S, and S, the conjugate foci, and 
 X,E and X,E, the conjugate directrices of the el- 
 lipse A,B,. 
 
 Because the tangents at Q and Q’ pass through 
 
 S ang S, respectively, therefore CS=2X,Q, and 
 CS, =2XQ’. 
 
 Because SS, is parallel to the tangent at P and 
 is bisected by CP, therefore the pole of any focal chord of the 
 ellipse A,B, with respect to the hyperbola AB or 
 
 the ellipse A,B, is the same as the pole of the cor- 
 
 responding conjugate focal chord with respect to the ellipse 
 A,B, or the hyperbola AB, 
 
 It may be shown that portions of a Jine drawn parallel to 
 an equiconjugate diameter and intercepted by the ellipse or 
 hyperbola and a directrix and a conjugate directrix are equal. 
 
 Also if the tangents to the ellipse from a point in an equi- 
 conjugate diamater cut a directrix anda conjugate directrix 
 which meet in the diameter, the straight lines joining the 
 points of intersection are parallel to the other equiconjugate 
 diameter. 
 
 26, Let CA, and CB, (Fig. 22) cut the hy- 
 perbola RBR’ in R and BF and the ellipse R, B,R 
 in R, and Ry. 
 
 Because RN?: CN*-CB,?:: CA?: CB?, and 
 RN?=CN2, | 
 
 therefore CN? ; CB,2:: CA?: CS*%, and CN; 
 UA tee aloo, 
 
 Hence RR’ is a conjugate directrix of the 
 ellipse A, B,. 
 
 The conjugate foci and the conjugate directrices of the 
 
 hyperbola A,B, are the poles and polars of the direc- 
 
 trices and foci of the hyperbola AB with respect to 
 the circle A, B,. 
 
29 
 
 oie R,N,?: CB,°-CN,? :: A,C*®: B,C?, 
 and R LN,=CN,, 
 therefore CN,?: CB,2 :: A,C*: CS,2, and CN: 
 OAj thy CAVE  y 
 
 Hence R,R, is a conjugate directrix of the hy- 
 perbola A,B. 
 
 If CS, isa third proportional to CN, and CB,, 
 S, is a conjugate focus of the hyperbola Ae a: 
 
 It may be shown as in Art. 25 that the straight shhe join- 
 
 ing the foci and conjugate foci of an hyperbola are parallel to 
 the asymptotes. 
 
 27. Let the reciprocal of the circle FQG ( Fig, 23 ) with 
 respect to the ellipse AB, be the hyperbola whose 
 principal axes are A,A, and B,B,, and let the 
 ‘center C of the circle FQG be in the major axis of 
 AB,. If P and P’ are the poles of the tangent QT 
 with respect to the circle AB and the ellipse AB, 
 Lae PP'N is perpendicular to C’T, because C’N. C’T 
 
 ws 
 Because PN.LN=RN2 ( Art. 10:), and P’N. 
 LN=WN?2, 
 therefore PN: P’N:: RN®?: RYN?:: C’B®: 
 CB 
 Hence if C’'U isa tangent from C’ to the circle FQG, 
 OB, C’U=C’B,2, and A,O: CF:: C’a?: C'U?. 
 If C’, the center of the ellipse A,B, (Fig. 24), 
 is within the circle, let UU’ be its polar with oad to the 
 circle QFG, then the center O of the reciprocal ellipse is the 
 pole of UU’ with respect to the auxiliary ellipse, and AO: 
 CF:: C’A,?: C’V2, and BO.C’V=C’B,2. Also the 
 latus rectum of the reciprocal ellipse is a third propotional to 
 
 the diameter of the reciprocated circle and the latus rectum of 
 the auxiliary ellipse. 
 
32 
 
 28. If one of the principal axes of the ellipse 
 A,B, (Fig. 25) does not pass through C, then the 
 straight line joining P and P’, the poles of the tan- 
 gent QT with respect to the circle A,B and the 
 ellipse A,B,, is perpendicular to A,C’, and P/N: 
 PN ie? COBre OAS. 
 
 Hence if a straight line is drawn through the focus of a conic 
 in any given direction, and a perpendicular is drawn to it from 
 
 any point on the curve, the locus of the point which divides the 
 perpendicular in a constant ratio is a conic. 
 
 If the major axis of the auxiliary ellipse passes through 
 C, the reciprocal is a parabola, and its latus rectum isa third 
 
 proportional to the diameter of the reciprocated circle and the 
 Jatus rectum of the auxiliary ellipse. 
 
 29. Let the vertex of a parabola CQ ( Fig. 26 ) be the 
 
 center of an ellipse AB, and let their axes be coincident and 
 their latera recta equal. 
 
 It is required to find the reciprocal of CQ with respect to AB, 
 Take any point Q on the parabola, and let Q’O be its polar 
 with respect to the ellipse AB, Take CM’=CO, and draw M’Q' 
 
 at right angles to CM’ meeting Q’O in Q’. Let CQ meet the 
 ellipse AB in P, and let CD be conjugate to CP. 
 
 Then DR is to CR in the ratio compounded of 
 BC? to AC? and CN to PN or CM to QM, 
 but QM?; 2BC2;:: CM: AC, 
 therefore DR: CR:: QM: 2AC:: Q’M’: OM’ :: Q’M: 2 CO. 
 Hence QM?; Q'M’2 :3; AC?2: CO?;: CM: 
 CONC M CMa y 
 therefore the point Q’ is on the parabola CQ’ whose latus 
 rectum is equal to that of CQ, and OQ’ is the tangent at Q’. 
 Hence the reciprocal of the parabola CQ with respect to 
 the ellipse AB is the parabola CQ’, and the latera recta of CQ, 
 CQ’, and AB are equal. The reciprocal of the parabola CQ 
 with respect to the hyperbola ABA’ ig also the parabola CQ! 
 
33 
 
 (Art. 5). Also the pole of the tangent at Q with respect to 
 the ellipse AB is Q’. 
 
 It may be shown that the straight lines joining the extre- 
 mities of the ordinates of the parabola through O and M pass 
 through A and A’, and the tangents at those extremities meet 
 on the tangents to the ellipse at A and A’. 
 
 30. Let the ellipse APA’ (Fig, 27) and the parabola CQQ' 
 have equal latera recta, and let their axes be coincident. 
 
 It is required to find the reciprocal of the ellipse APA’ 
 with respect to the parabola CQQ’. 
 
 Let QQ’ be the polar of any point P on the ellipse with 
 respect to the parabola. Draw PQ, parallel to CA 
 meeting QQ’ in Q,, and PN perpendicular to CA. 
 Draw NP, touching the hyperbola AP, at P,, and 
 NG fee ead iealar to CA, Take CN’ equal to 
 
 CN He “and draw N’P” at right angles to AA’ meeting 
 QQ produced i in P’, 
 
 If Q’Q meets AA’ in O, CN=CO, 
 and because CM, CM’=CN®2, therefore QM. 
 Q’M’=4CS.CN, S being the focus of the parabola. 
 Because P,N,2 is to NN,? in the ratio com- 
 pounded of P, N 3 ;. CN, ‘NN, and CN, : NN,, 
 and CN, : NN, sib GN, .CN : : CN,‘CN-CN2 
 >; AC2 ; ACt-CN? 
 33 BC2: PN, 
 _ therefore P,N,? is to NN,® in the ratio com- 
 pounded of BO? ; AC? and BO?: PN?, 
 but 4PN2=4Q, M, 2=(QM+Q’M’)2=4CS ( CM+ 
 CM’+2CN), 
 therefore PN2=CS (OM+0M’)=2CS.0M,. 
 Hence PN?: BC? :: OM, AC, 
 and P,N,?: NN,2:: BOh: AGOM, PN? 
 OR 7) is Q, M,* - OM,2, 
 
RM Ae ve eee ronment 
 q 
 
 sar 
 
 TAs: ' 
 eas z mone 
 Rieder eel 
 
 ne AN FE 
 
36 
 
 therefore NP, is parallel to OQ,, 
 
 and because CN=CO, and CN,=CN’, 
 
 therefore OP’ touches the hyperbola A’P’ at P’. 
 
 Hence the reciprocal of the ellipse ABA’ with respect to 
 the parabola CQQ’ is the associated hyperbola ABA’. 
 
 Because CN‘CN ,=CA2, the tangent at P passes 
 through N,, and the straight limes joining the ex- 
 tremities of the ordinates of the ellipse and its as- 
 sociate through N and N,, pass through A and A’, 
 and the tangents at those extremities meet on the tangents to 
 the ellipse at A and A’ (Art. 8). Also, as stated in the previ- 
 ous Article, the straight lines joining the extremities of the 
 ordinates of the parabola through N and N, pass 
 
 through A and A’, and the tangents at those extremities meet 
 on the tangents to the ellipse at A and A’, 
 
 Hence the -straight lines joining the extremities of the 
 common chords of the parabola, the ellipse and the hyperbola 
 pass through A and A’, and the tangents at those extremities 
 to the parabola or the ellipse and the hyperbola meet on the 
 tangents to the ellipse at A and A’. Also the distance between 
 _ the common chords is equal to the transverse axis of the asso- 
 ciated conics and if the transverse axis is given, the common 
 
 chords are fixed in direction whatever the length of the con- 
 jugate axis may be. 
 
 It is obvious that the chord of contact of the parabola 
 
 and the conjugate hyperbola is the tangent at A, and the com- 
 mon tangents pass through A’. 
 
 The tangents to the parabola at the points where it cuts 
 the ellipse and its associate also touch the hyperbola and its 
 associate, and the chords of contact of the parabola are equal 
 to the chords of contact of the associated conics. 
 
 Since PN, is the tangent at P to the ellipse, 
 the polar of P’ with respect to the parabola is PN,. 
 
37 
 
 31. Let P be any point in the conjugate hyperbola, ( Fig. 
 28 ) and let QQ’ be its polar with respect to the 
 parabola CQQ’. Draw PN perpendicular to AA’, | 
 and through N draw NP, touching the conjugate 
 hyperbola at P,. Draw P,N, perpendicular to BC, 
 and take BN’ equal to N,B’. Draw N’P’ at right 
 angles to BC meeting QQ’ in P’, 
 
 The ratio of P,N,? to N,T,2 is equal to the 
 ratio compounded of P,N,+CN: CN, and CN: CT,, 
 and therefore it is equal to the ratio of AC?4+CN? 
 to BC2, that is, to the ratio compounded of PN? : 
 BC? and AC2: BC?. 
 
 Again 4PN 2=(QM+Q’ M")?=40S (CM+CM’+2CN ) 
 = 8CS.OM.,. 
 
 Hence PN2;BOC*;; OM,: ACG:: OM,?:AC,0M,, 
 
 and therefore P,N,2:; N,T,2:: AC. OM,: BC?:: 
 
 ON esbNGkee OM , 2 2G), M8: 
 
 and because CN=OC, and CN’=CN, therefore OP’ 
 touches the conjugate hyperbola at P’, and because 
 ON P’N’=AC2, P” lies in the polar of N with respect 
 to the original hyperbola. Hence the conjugate hyperbola BP 
 is its own reciprocal with respect to the parabola CQ. As in 
 Art. 8 it may be shown that the straight lines joining the ex- 
 tremities of the chords of the conjugate hyperbola drawn 
 through P and P’ perpendicular to AA’ pass through A and 
 
 A’, and the tangents at those extremities meet on the tangents 
 to the ellipse at A and A’. 
 
 Hence if the straight lines joining the extremities of two 
 chords of a conjugate hyperbola pass through A and A’, the 
 straight lines joining the points where the chords cut the as- 
 sociated ellipse and the original hyperbola also pass through 
 A and A’, and if the tangents at the first two pairs of points 
 to the conjugate hyperbola meet on the tangents to the ellipse 
 at A and A’, the tangents at the other two pairs of points to 
 
ad 
 EDL, ARAN ORD CONOR AE AR Ee i am 
 
 i 
 { 
 i 
 
 ey 
 
40 
 
 the ellipse and the original hyperbola also meet on the same 
 two tangents. 
 
 It is evident that the polars of B and B’ with respect to 
 the parabola are the asymptotes, and the portions of the asymp- 
 
 totes cut off by the parabola are each equal to the distance 
 between the foci of the hyperbola. 7 
 
 32. We have proved in Art. 29 that the reciprocal of the 
 parabola CQ with respect to the ellipse AB is the parabola 
 CQ’. It is evident that the reciprocal of the parabola CQ with 
 respect to the original hyperbola AE is the parabola CQ’ (Art. 
 5) and the parabola CQ is its own reciprocal with respeet to 
 the conjugate hyperbola BE’ ( Art.5). The following is ano- 
 ther proof. 
 
 Let OQ! ( Fig. 29) be the polar of Q with respect to- 
 the conjugate hyperbola BR, and let CP be conjugate to CQ. 
 Take CM’ equal to CO and draw M’Q’ at right angles to CM’ 
 meeting OQ’ in Q’. 
 
 Q’'M’: OM’ or PN: CN is equal to the ratio CSUM. 
 of BC? ; AC? and CM: QM, but QM2=4C8S:CM 
 
 eneresore QM? ; 2BC2:: CM: AC, 
 
 and Q’M’: OM’ :: QM: 2AC. 
 
 Hence Q’M’? :: QM2 :: CM’2; AC? 3: CM’: OM, 
 
 therefore Q’ is a point on the parabola, 
 
 and because CM’=CO, OQ’ is a tangent. 
 
 Because CM‘CM’=AC2, Q’Q passes through A‘, 
 
 and the tangents at Q and Q’ meet on the tangent at A. 
 
 Also QA and Q’A are equally inclined to the axis.. 
 
 33. If ABA’ (Fig. 26) isa circle, CQ and CQ’ are at 
 right angles to the tangents at Q’ and Q. The focus and direc- 
 trix of the parabola bisect AC and A'C, and its latus rectum is 
 
 equal to the diameter ofthe circle, If in CA and CA’ produced 
 two points X, and 8, are taken such that CX, 
 
 and CS, are each equal to AA’, 8, is the connate 
 focus. of the parabola CQ’, and Xx, is the foot of the 
 
 conjugate directrix. 
 
41 
 
 34. Let the parabolas AP and AQ ( Fig. 30) have their 
 latera recta equal and their axes in the same straight line. 
 , viens is required to find the reciprocal of AQ with respect 
 o AP. 
 
 Draw any straight line QP parallel to the axis MAN. 
 
 Because QM=PN, therefore AM=AN, and PM and QN are 
 tangents and are equally inclined to the axis. 
 
 Also QN is parallel to P’M, and PM is parallel to Q'N. 
 
 Because AN=AM, N lies in the polar of Q’ with respect to 
 the parabola AP. 
 
 Again because the polar of Q’ is parallel to the tangent at 
 P’, QN must be the polar, and since QN touches the parabola 
 AQ at Q, the parabola AQ is its own reciprocal with respect 
 to the parabola AP. 
 
 The above result also follows from Art. 4. 
 
 35. The reciprocal of the parabola CP ( Fig. 31 ) with re- 
 spect to the circle QRQ’ is the parabola CP’, therefore the re- 
 ciprocal of CP with the respect to the rectangular hyperbola 
 QP is the parabola CP,, and the reciprocal of CP 
 with respect to the rectangular hyperbola RP’ is the 
 
 parabola CP,, and these parabolas and _ rectangular 
 hyperbolas have their latera recta each equal to the diameter 
 of the circle QRQ’. : | 
 
 The reciprocal of a parabola CP ( Fig. 42) with respect to 
 a parabola CQ whose axis OS’ is at right angles to ths axis 
 CS of CP may be determined in the following manner. 
 
 Let the tangents PT and QT’ be parallel. Draw OQR 
 parallel to CS’ meeting PT in O. If QR is equal to QO, R is 
 the pole of PT with respect to the parabola CQ. 
 
 Because RQ=QO=LT’, and QH=CT’, 
 
 therefore RH is equal to CLwhich is one-half of PN. 
 
 Again because PT is parallel to QT’, 
 
 therefore PN? : NT2 ;: MT’? ; QM?, 
 and 2AS: NT:: MT’ ; 2AS’. 
 
Py 4 x4 
 os ; / 
 fe W ¥ ‘ i Re ee Tee a SOLAS Os, Ny NES 
 } a ft > 
 ay - ‘ 
 
 ae a: Ae 
 ek ae f 
 
 NS 
 ~ 
 
 ~ 
 
44 
 
 ‘Hence PN 7 2AS :: 24S’: QM, 
 
 and therefore 4AS.AS’=PN.QM=2RH.HC. 
 
 Hence the reciprocal of CP with respect to CP isa rect- 
 angular hyperbola having its aspmptotes along CS and CT’ and 
 
 its latus rectum a mean proportional between the latera recta 
 of CP and CQ. 
 
 36. Let PREF (Fig. 32) be a parabola. IfP,-is a 
 point such that PP, is parallel to the tangent ata 
 given point R and is bisected by another given line CO, it is 
 required to find the locus of P,. 
 
 Let CO meet the tangent at R in C, the diameter 
 through R in O, PP’ in E, and the parabola in F. Produce RC 
 to R, making CR, equal to CR, and join OR, meet. 
 ine) PP in) V,- 
 
 Because PE=EP,, and VE=EV,, 
 
 therefore QV=P,V 
 
 Again the third proportional to RV and PV is constant, 
 
 and RV: R,V, :: RO: R,O=a constant ratio, there- 
 fore the third A doartienal to R,V, and P. 17, is 
 constant, and therefore the locus of P, is @ para- 
 bola passing through F, touching RR, at R,, and 
 having R,O parallel to its axis, 
 If CE is parallel to RV, RV and BR, V, are equal 
 and parallel, and the parabolas are equal, and conversely if the 
 parabolas are equal and their axes parallel, the straight lines 
 
 joining the corresponding points are parallel to one fixed 
 straight line and are bisected by another fixed straight line. — 
 
 37. Let AP ( Fig. 38 ) be a given parabola whose focus is 
 S Let PT be the tangent at P. Produce SP to S’ making S’P 
 equal to SP. Describe a parabola A’P touching PT at P and 
 having S’ for its focus. 
 
 AS is required to find the reciprocal of A’P with respect. 
 to AP 
 
45 
 
 Because the tangent at any point P of a parabola bisects 
 the-angle between the focal distance SP and the perpendicular 
 from P to the directrix, the axis AS is parallel to the axis A'S’. 
 Draw any line QR parallel ‘to either axis, and let QQ’ and RR’ 
 Eee double ordinates through Q and R. Draw the diameter 
 
 Because 4SP.PV=QV2=RV’2=49’P.PV’, 
 therefore PV=PV’. Hence the tangents at Q and Q’ meet 
 at V’, and the tangents at R and R’ meet at V. 
 
 It is obvious that the tangents at Q, R, and P meet at a 
 point, and so do the tangents at Q’, R’, and P. 
 
 Again because RV’ and Q’V are equal and parallel, there- 
 fore RV and Q’V’ are equal and parallel, and so are QV’ and 
 R’V equal and parallel. Also RQ’ and QR’ are bisected in P. 
 
 Because the polar of V with respect to AP passes through 
 RB, the polar of R passes through V, and because the polar of 
 R with respect to AP is parallel to.the tangent at Q, therefore 
 RV must be the polar of R, and since R'V touches the parabola 
 
 A’P at R’, the parabola A’P is its own reciprocal with respect 
 to the parabola AP. 
 
 Hence the tangent at any point on one of the parabolas 
 is divided harmonically by VPV’ and the other curve. 
 
 It is evident that for a point O on the parabola AP there 
 is a point O’ on the parabola A’P such that OO! is parallel to 
 VV’ and is bisected by PT. Hence if from any point O on a 
 
 parabola OO’ is drawn parallel to the axis so that it is bisected 
 by a fixed tangent, the locus of O' is a parabola equal to AP. 
 
 38. Let PR and PQ (Fig. 34) be two equal parabolas 
 having their axes parallel and touching each other at the point 
 P. Take any point O, and through O draw ORR’ parallel to 
 the tangent at P. Draw RQ and VPV’ parallel to the axis, and 
 
 QVQ’ parallel to the tangent at P. Produce OV’ to O’ making 
 O'V’ equal to OV’. 
 Because PV = PV’ (Art. 37), 
 
48 
 
 therefore the polar of V with respect to PR is OV’, 
 
 and therefore the polar of O with respect to PR passes ~ 
 through V, . 
 
 Similarly the polar of O with respect to PQ also passes 
 through V. 
 
 Also the polar of O’ with respect to PQ passes through V, 
 
 therefore the polar of O with respect to PR and the polar 
 of O’ with respect to PQ both pass through V, 
 
 and because they are parallel to the tangents at KE and FE’, 
 
 and the tangents at E and E’ are parallel ( Art. 37 ), 
 
 therefore the same straight line FV is the polar of O with 
 respect to PR and the polar of O’ with respect to PQ. 
 
 Also the polar of O' with respect to PR coincides with the 
 polar of O with respect to PQ, 
 
 Hence the straight line joining the poles of any line with 
 
 respect to the parabolas PQ and PR is parallel to the common 
 tangent at P and is bisected by the diameter through P. 
 
 39. The reciprocal of the parabola CQ (Fig. 35) with 
 respect to the hyperbola PDP’ is the parabola CQ’ ( Art. 29 ), 
 therefore the reciprocal of the parabola CQ with respect to the 
 ellipse PDP’ is the parabola Q,CQ,, because for a 
 point R on the parabola CQ’ there is a point RY on | 
 
 @,CQ, such that RR’ is parallel to CD and is bi- 
 sected by CP (Arts. 5 and 36). 
 
 It is evident that for a point R’ on the parabola 
 Q,CQ, there is a point R’ on the parabola CQ such 
 that R’R” is parallel to CP and is bisected by CD. Also because — 
 
 the parabola CQ’ touches the conjugate hyperbola at Q’ and Q”, 
 the parabola Q,CQ, also touches it at Q, and Q,. 
 
 Because Q’Q, and Q,Q” are parallel to CD and 
 are bisected by CP’, and Q’Q” touches the original hyperbola 
 at A’, therefore Q,Q, touches it at A” so that A’A” 
 
 is parallel to CD and is bisected by CP’. Hence the common 
 
49 
 
 tangents at Q, and Q, meet at the other extremity 
 of the diameter through A”, Also the axis of the 
 parabola Q,CQ, is parallel to CA”, and the tangent 
 at C is parallel to Q,Qo. 
 
 Because the area of the triangle formed by Q,Q, 
 and the tangents at Q, and Q, is constant, the area 
 of the parabolic segment Q,CQ, is also constant, 
 
 The focal chord of the parabola Q,CQ, parallel 
 to Q,Q. is a third proportional to CA” and A’Q,,. 
 It bisects CA” and is double of it if the hyperbola PDP’ is 
 rectangular. 
 
 40. In Fig. 27 if the latus rectum of the parabola is not 
 equal to that of the ellipse, take B,C a mean propor- 
 tional between the semi-latus rectum of the parabola and AC. 
 ‘In N,P, produced if necessary take P, such that 
 
 P,N, is to P,N, as the latus rectum of the para- 
 bola is to the latus rectum of the ellipse. 
 
 Now P,N,?2is to NN,? in the ratio compounded 
 of P,N,?: P,N,? and P,N,?: NN,2, ze. in the 
 ratio compounded of the ratio of the squares on the latera recta 
 and the ratios BC?: AC? and BC2 ; PN? (Art. 30), 
 therefore P,N,? is to NN,? in the ratio compounded 
 of B,C? : AC? and B,C?: PN?2, 
 
 but B,C2: PN? :: B,C?:Q,M,2:: AC:OM, 
 ( Art. 30), 
 
 therefore. P,N,2; NN,2:: B,C? : AC.OM,:: 
 QM. 20M; 2. 
 
 Hence NP, is parallel to OQ’, and therefore Q’/O 
 touches an hyperbola whose latus rectum is a third propor- 
 tional to the latera recta of the ellipse and the parabola, 
 
 Because the reciprocal of an ellipse with respect to a para- 
 bola having the center of the ellipse for its vertex and its axis 
 
50 
 
 coincident with that of the ellipse is an hyperbola whose trans- 
 verse axis is the same as that of the ellipse and whose latus 
 rectum is a third proportional to the latera recta of the ellipse 
 and parabola, therefore the reciprocal of an hyperbola with 
 respect to a parabola having the centre of the hyperbola for its 
 vertex and its axis coincident with that of the hyperbola is an 
 ellipse whose transverse axis is the same as that of the hyper- 
 bola and whose latus rectum is a third proportional to the latera 
 recta of the hyperbola and parabola. 
 
 41. Let CQ be any parabola passing through the centre 
 C ( Fig. 36) of an ellipse PDP’. 
 
 It is required to find the reciprocal of the ellipse PDP’ 
 with respect to the parabola CQ. 
 
 Let S be the focus of CQ. Draw CP parallel to the axis 
 of CQ, and let CD be conjugate to CP. Take CD, a mean pro- 
 portional between CP and 2 CS. Let QO be the polar of a point © 
 R on the ellipse with respect to the parabola CQ. Let RN be 
 the ordinate to CP through R, and draw NP, touch- 
 
 ing the hyperbola PP, at P,. Let P,N, be the 
 ordinate through P,, and take CN, equal to CN,. 
 Draw NR, parallel to CD meeting QO in R, and 
 
 the hyperbola PE Ales In Nie erode take 
 a point R, such that R,N, biree Nie :: 2 CS.CP: 
 CD?2. Draw RQ, and QM, parallel to CP and CD 
 respectively. 
 
 R,N,2 is to NN,2 in the ratio compounded of 
 R Nj? : Pp iN, ? and P,N,?: NN, ?, 7. e. in the ratio 
 Bont peed any N,2: P,N,? and the ratios CD?: 
 CP? and CD?2:;: RN2, ( Art, 30 ) therefore R,N,? is 
 to NN,® in the ratio compounded of 2CS. OP: OP2 
 and 2C8.CP : RN?2, but it may be sane as in 
 Art, 80 that RN2=2C8, OM, and CD, is a mean 
 proportional between CP and 2058, tineistate RN?: 
 Clee OM CP. 
 
aL 
 
 Hence RiN,%: NNj{? 3: 2053 .OMy CD": 
 SOM es OM: 3 OQ) M54: OM. 3, 
 
 and therefore NR, is parallel to OQ,, and be- 
 eause CN=CO, and CN,=CNog, therefore OQ, touches 
 
 the hyperbola whose semidiameter conjugate to CP is a third 
 
 proportional to CD and CD,. 
 
 Hence the reciprocal of the ellipse PDP’ with respect to 
 the parabola CQ is an hyperbola such that its semidiameter 
 conjugate to CP isa fourth proportional to CD, CP, and 2CS. 
 
 42. Let A,P and A,Q ( Fig. 37 ) be two equal 
 parabolas having their axes A,S, and A,S, parallel. 
 
 It is required to describe a parabola which touches them, 
 is equal to each of them, and has its axis parallel to their axes. 
 Let the two given parabolas meet at O. Draw the diame- 
 ters PV and QV’ at distances from O each equal to one-fourth 
 of the distance between A,S, and A,S,. Then the 
 
 parabola AP which touches the given parabolas at P and Q is 
 the parabola required. 
 
 Because O is equidistant from PV and QV’, the straight 
 line drawn through O parallel to PV or QV’bisects PQ and VV’. 
 Hence PQ and VV’ are parallel to the common tangent to the 
 
 equal parabolas A, P and A,Q (Art, 36), and PQW’V 
 
 is a parallelogram, and therefore the diagonals PV’ 
 and QV bisect each other and meet in OV,. Hence the tangents 
 at P,Q, V, and VW meet in OV). : 
 Now the parabola AP must be equal to each of the para- 
 bolas A,P and A,Q. If possible let AP and A,P 
 
 be unequal. Describe a parabola equal to A,P touch- 
 ing it at P and having its axis parallel to A,S,(Art. 
 37). Then P is equidistant from the axis of this para- 
 
 bola and A,S,. Describe a parabola equal to A,Q 
 having its axis parallel to A,S, and touching A,Q 
 at Q (Art. 37). Then Q is equidistant from the axis of this 
 
a2 
 
 parabola and A,S,. Now the axes of these two 
 parabolas must coincide, because the distance between A,Si 
 and A,S, is double the distance between PV and 
 and QV’. Again as the parabolas are equal they cannot cut 
 each other, and since they are each equal to A,P 
 and A,Q and touch them, the tangents at P and Q 
 which meet in OV, bisect the common diameter 
 through O (Art. 37), and therefore the parabolas meet which 
 is impossible. Hence the parabola AP which touches the given 
 parabolas at P and Q is equal to each of them, 
 
 43. Let PD (Fig. 38) be an ellipse, and AP’ a parabola. 
 It is required to find the reciprocal of the ellipse PD with 
 respect to the parabola AP”. 
 
 Draw CP parallel to the axis AS of the parabola AP’, and 
 let CD be conjugate to CP. Describe a parabola A’P’, equal to 
 the parabola AP’, having its axis parallel to AS and touching 
 CD and the parabola AP’. This can be done by describing a 
 parabola having S for its focus, AS for its axis, and its latus 
 rectum double the latus rectum of AP’. The focus S’ of the pa- 
 rabola A’P’ lies in this parabola ( Art. 37 ), and since the ratio 
 of A'S’ to the perpendicular from S’ to the tangent CDQ’ is 
 known, the position of S’ can be determined, Describe a para- 
 bola CR’ equal to AP’, touching CD at C, and having its axis 
 parallel to AS or CP ( Art. 37), and also a parabola RR’ which 
 touches the parabolas CR’ and A’R, is equal to each of them, 
 and has its axis parallel to their axes ( Art. 42 ). 
 
 The reciprocal of the ellipse PD with respect to the para- 
 bola CR’ can be determined (Art. 41). Hence the reciprocal of 
 the ellipse PD with respect to the parabola RR’, A’Q’, or AP’ 
 can be described ( Art. 38 ). Since the reciprocal of the ellipse 
 PD with respect to the parabola CR’ is an hyperbola (Art. 41), 
 
 its reciprocal with respect to the parabola AP’ is also an 
 hyperbola. 
 
23 
 
 44, The reciprocal of a parabola with respect to an el- 
 lipse or hyperbola can be determined as in Art. 21 by taking 
 the center of the reciprocal conic at an infinite distance. 
 
 45. It is evident that if two circles are taken having 
 their centers in the axis of a parabola equidistant from the 
 vertex, the reciprocal of the parabola with respect to one of 
 the circles is equal to the associated conic of the reciprocal of 
 the parabola with respect to the other circle (Arts. 18 and 19), 
 It is also evident (Arts. 18 and 19) that the reciprocals of as- 
 sociated conics having rectangular axes with respect to acircle 
 whose center is in the transverse axis are associated conics. We 
 may also prove it in the following manner. It may be shown 
 thaw (Wig, 39) AO :,CA,9:: CH? : C’A,.C’A,, 
 
 Prob cee Aer OCA | Gea. 3 Bs Ce, 
 
 and C’A.C’A’ : C’E2::: CE? :C’A,.C’Ag, 
 
 therefore CA,: AO:: C'H?; C’A.C’A’, 
 
 mem era 2: A Oper Oy Ai Ns OFA CAA! 
 
 Hencerb, OF 2) A O82 CA CAL (BOs OC 
 —OA?;: BO, 
 
 and therefore if the reciprocal of the ellipse A,B,A, 
 with respect to the circle EE’ is the hyperbola ABA’, the re- 
 ciprocal of the ellipse ABA’ with respect to the circle EE’ is 
 the hyperbola A,B,A,. Hence the reciprocal of the 
 
 hyperbola A,B,Ag, with respect to the circle EE’ 
 is the ellipse ABA’. 
 
 Because BO? : AO.CA,:: C’E?: B,C?, 
 therefore the rectangle contained by the latera recta of 
 A,B,A, and its reciprocal is equal to O’H?, 
 
 46. The rectangle contained by the axes of the reciprocal of 
 the ellipse ABA’(Fig.t7)with respect to the circleA,B, 
 is equal to the rectangle contained by the axes of the reciprocal 
 of the ellipse ABA’ with respect to the hyperbola A;Bz3, 
 
a4 
 
 and also to the rectangle eontained by the axes of the reci- 
 procal of the ellipse ABA’ with respect to the ellipse 
 A,B, (Arts 5 and 9). 
 
 Similarly the rectangle contained by the axes of the reci- 
 procal of the hyperbola ABA’ which is cut by CC’* with re- 
 
 * Let A’AA” ( Fig. 40) be a rectangular hyperbola, and 
 EE’ a circle. Let CC’, the line joining their centers, be at right 
 angles to ‘A’A”, | 
 
 It is required to describe the reciprocal of the hyperbola 
 A’AA” with respect to the circle EE’. 
 
 Let P be the pole of the tangent QTé Draw C'L and C’'L’ 
 touching the hyperbola, and let O be the pole of LL’ Draw 
 C'T’R parallel to the transverse axis meeting the tangent at Q 
 in T’ and the hyperbola in R. Let BK be the polar of R. Draw 
 QM’ and PN’ perpendicular to CC’, and NPM parallel to CC’. 
 
 PN2; NB.NB’:: (C'O-C’N’)? : C’M2-C’K?, 
 Now C/O: CE:: CE: CN,, 
 and because CO’.CN,=CA? = CA’2, 
 therefore C’-A’N, is a right angle, and O’N,.CC= 
 A’O??=AC2+4+CC’?, 
 Also C’N’.C’t=C’F.C’P=C’E2, 
 ; > ,, CA2_ CC’%,.CM’+CA? 
 and C't=CC4 Ct =CO a= 
 Again C’M.C’T’ =C’F.C’P=C’E2, 
 and O7"T": C/t:+-CPs Chis 2CM"?, Ou 
 therefore C’T’2 ; C’t? : : CM’? : QM’2, 
 and OT’? : (CC’ +97)" : : CM’2: CM2+0a2, 
 also C’K2 ; C’E2: : C’E2 : C’R2::C’E2:CC”24+CA2, 
 
 2. Pe ACC i ; 
 Hence PN2:NB.NB’: ( AG24+CC? | CC’.CMG ) 
 CM's + CA2 U 
 
 ‘ (OC.CMW’+CA)2 ~CO?+CA2 
 | OA Oe as te 
 (CA2+C0'2) (CC’“.CM’+CA?) 
 
 ‘ ( CC’—CM’ )2CA2 ae AQ. , 9 
 i (CC’.CM’+CA? )?(0C?2+CA%) * ‘ CA2;CC 24CA > 
 
 ee 
 
a) 
 
 spect to the circle A,B, is equal to the rectangle 
 
 contained by the axes of the reciprocal of the hyperbola ABA’ 
 with respect to the ellipse or hyperbola A,B,, 
 
 and therefore the duplicate ratio of the axes of the reciprocal of 
 the hyperbola A’A A” with respect to the circle EE’ 
 is equal to CA2: CC’2+CA2, whilst it can be easily 
 shown that the duplicate ratio of the axes of the reciprocal of 
 the circle A’AA” with respect to the circle EE’ is equal to 
 CA? : CC?2—CA2, 
 
 Because OA ,2: OB? :: CA?: CCO’2+CA2, 
 
 and OB2: C’E2;: C’E?: CC’2+-CA2, 
 therefore OB? C’H?;:.; OAL :CA. 
 
 Hence the reciprocal of the hyperbola A’AA” with respect 
 to the circle EK’ is an hyperbola having its latus rectum equal 
 to the latus rectum of the reciprocal of the circle A’AA” with 
 respect to EE’ and its asymptotes perpendicular to the tangents 
 C’L and O'L’. ’ 
 
 Also OA,. OB: CA.C’E :: C’H? ; C’A’3, whilst 
 it can be easily shown that the rectangle contained by the axes 
 of the reciprocal of the circle A’AA” with respect to EE’ is to 
 CA.C’E as C’E? is to C’U®, where C’U is a tan- 
 gent from C’ to the circle A’AA”. Hence the rectangles con- 
 tained hy the axes of the reciprocals are in the triplicate ratio of 
 C’'U to C’A’, and therefore the rectangle contained by the axes 
 of the reciprocal of the rectangular hyperbola A’AA” which CC’ 
 does not cut is not the same as the rectangle contained by the 
 axes of the reciprocal of its conjugate which is cut by CC’ 
 
 If B’AB” ( Fig. 41 ) is any hyperbola, and the straight line 
 joining its center and the center of the circle HE’ is at right 
 angles to the transverse axis B’B”, we may proceed thus. 
 
 Let P be the pole of the tangent Q’t. Draw C’L and C'L’ 
 touching the rectangular hyperbola, A’AA”, and O’L, 
 and O’L, touching the hyperbola BPAB”%. Ifa point 
 N, is taken in AC produced such that CN,.CC’ 
 =AC?2, LL’ and L,L, both pass through N,. 
 
 Let O be the pole of LL,N with respect to the circle | 
 EH’. Draw C’R’R parallel to the transverse axis 
 
56 
 
 The reciprocal of the ellipse AB (Fig. 18) with respect 
 to the hyperbola A,B; is the same as the reciprocal. 
 of the ellipse A,B, with respect to the ellipse 
 A;B, (Art. 17), and the rectangle contained by the 
 axes of this reciprocal is equal to the rectangle contained by 
 the axes of the reciprocal of the ellipse AB with respect to the 
 ellipse A,B, (Art. 21). 
 
 meeting the hyperbolas in R’ and R and the tangents Qt and | 
 
 Q’t in T and TI’. Let BK, be the polar of R’. Since the tan- 
 gents at Q and Q’ meet 4 ACt, QQ'M’ is. perpendicular to 
 CC’. Draw PN’ perpendicular to CC’ and NPM, parallel to CC’. 
 
 Because QM’? =CM’.M’z, 
 and Q’M’2; CM’.M’t :: B’C2;AC2, 
 therefore Q’M’2 :QM’2:; BC? : AC? 
 Again C'T: C/T’ :: QM’: O’M’:; ACE BCs: 
 CM, : OM, : 
 and CR: CR’ :: A@: BIC .::. CK ~CK 
 Hence:C’M, + G0 Mies: (Cok Oi 
 and (M,?—C’K,2 : (M2—C’K? :: OM,? :0’M? 
 >: AC? : By Oe and tharepore 2N# tas NB, or PN? 
 OM, 2—C’/K, Sethe ratio compounded of PN2: (’M?2 
 —C’K2 and OM2—CK? : C’M , 2—C’K , 2=the ratio 
 compounded of CA?: CO’24CA2 and B’C?: AC2= the 
 ratio pomponadcd of OA,2 : C’K® and C’K?: C’K,? 
 SQ Aj 2aO' Kate =BO? 0074042, 
 Hence PN?2; NB, NB: OA, * Pee 
 
 and therefore ane reciprocals 7 the hyperbole B ‘AB" and 
 
 A’AA” with respect to the circle EE’ have the same conjugate 
 axis, and their transverse axes are inversely proportional to the 
 transverse axes of the reciprocated conics. Hence the reciprocal 
 of the hyperbola B’AB” with respect to the circle EE’ is an 
 hyperbola having its asymptotes perpendicular to the tangents 
 C’L, and C’L,, and it axes proportional to A’O’ 
 and BC. 
 
 We have proved that 
 
 OA,: CA:: CH?2: CC’2+CA2, 
 
 and ‘CK? : CE? :: OA,:CA 
 therefore OB? : CH?:: OA, CA: BO?, 
 
 Ww 
 
57 
 
 Similarly the rectangle contained by the axes of the reci- 
 procal of the hyperbola AB with respect to the ellipse or hy- 
 perbola A,B, is equal to the rectangle contained by 
 the axes of the reciprocal of the hyperbola A,B, with 
 respect to the ellipse or hyperbola A,bB,, and there- 
 
 fore the rectangle contained by the axes of the reciprocal of 
 
 the ellipse AB with respect to the ellipse or hyperbola 
 A,B, is equal to the rectangle contained by the 
 
 axes of the reciprocal of the hyperbola AB with respect to the 
 ellipse or hyperbola A,B). 
 
 Hence the rectangle contained by the latera recta of the 
 hyperbola B’AB” and its reciprocal. with respect to EE’/=C’H2= 
 the rectangle contained by the latera recta of the ellipse AB’ 
 and its reciprocal with respect to EE’, and this rectangle is con- 
 _ gtant if the centre of the given auxiliary circle is in one of the 
 
 principal axes of the reciprocated conic. 
 
 It is evident that the reciprocal of the hyperbola conjugate 
 to BA,B, with respect to the circle EE’ is an ellipse 
 which touches the hyperbola B’/AB’ at L, and L,. 
 
 Because the rectangle contained by the axes of the reci- 
 procal of the hyperbola BAB’ (Fig.17) with respect to the 
 ellipse or hyperbola A,B, is the same as the rec- 
 tangle contained by the axes of the reciprocal of the hyperbola 
 BAB’ with respect to the circle A,B;, and OA,‘OB 
 ( Fig. 41) is to CACE in the ratio compounded of 
 C’H®: C’A’S and AC: BC, and also the rectangle 
 
 contained by the semi-axes of the reciprocal of the ellipse or 
 
 hyperbola AB/’A, with respect to EH’ is to CA.C’E 
 in the ratio compounded of C’H? : C’U3 and AC:B’C, 
 where C’U isthe tangent drawn from C’ to the circle 
 AA’A,, therefore the rectangle contained by the 
 axes of the reciprocal of the conic AB’A, which CC’ 
 
 cuts with respect toa conic directly or indirectly associated 
 with EE’ is to the rectangle contained by the axes of the reci- 
 procal of the hyperbola B’AB” which CC’ does not cut, in the 
 triplicate ratio of C’A’ to C’U. 
 
60 
 
 47, To illustrate the preceding articles we shall recipro- 
 cate the following theorems :— 
 
 (1). The straight lines joining the extremities of two 
 focal chords of a conic intersect in the directrix. 
 
 (2). The straight line drawn from the focus of a conic 
 to the point in which the tangent meets the directrix, is at 
 
 right angles to the straight line drawn from the focus to the 
 point of contact. . 
 
 (3). The straight line joining the feet of the perpendi- 
 culars drawn from the focus to the tangents at the extremities 
 of any focal chord of a conic passes through a fixed point. 
 
 (4). The locus of the middle point of a chord of a conic 
 passing through a given point is a similar conic. 
 
 (5). The tangents at the ends of a focal chord of a para- 
 bola intersect at right angles in the directrix. 
 
 (6). The foot of the perpendicular from the focus on the 
 
 tangent at any point of a parabola lies on the tangent at 
 the vertex. 
 
 (7). The quadrilateral formed by joining the points of 
 intersection of the diameters drawn from the angular points of 
 a quadrilateral inscribed in a parabola and the diagonals has 
 its sides parallel to those of the inscribed quadrilateral. 
 
 (8). The tangent at any point of an ellipse or hyperbola 
 is equally inclined to the focal distances of that point. 
 
 (9). The tangents drawn to an ellipse or hyperbola from 
 
 an external point are equally inclined to the focal distances of 
 that point. 
 
 (10). Pairs of tangents to an ellipse or hyperbola at right 
 angles to each other intersect on a fixed circle. | 
 
 (11). The locus of the point of intersection of the focal 
 chords drawn parallel to any conjugate diameters of an ellipse 
 
 , or hyperbola is a concentric, Boeke: and similarly situated 
 ~ ellipse or hyperbola. 
 
61 
 
 (12). The intercept of the tangents at the ends of a focal 
 chord on either asymptote subtends a right angle at the focus. 
 
 (18). Conics inscribed in a parallelogram are concentric, 
 and the quadrilateral described about concentric conics is a 
 parallelogram. —_ 
 
 (14). Tfa rectangular hyperbola circumscribes a tri- 
 angle, it passes through the orthocenter. 
 
 (15). The locus of the point at which the opposite sides 
 of a parallelogram subtend equal angles is the rectangular hy- 
 perbola circumscribing the parallelogram. 
 
 (16). Any chord of a rectangular hyperbola drawn paral- 
 Jel to the transverse axis subtends a right angle at either vertex. 
 
 The second theorem is the reciprocal of the theorem, ‘ The 
 tangent at any point of a circle is at right angles to the dia- 
 meter through the point of contact’, with respect to a circle. 
 
 The fifth and sixth theorems are obtained by reciprocating 
 the theerem, ‘ The angle in a semicircle is a right angle ’ with 
 respect to a point on the circumference. The sixth theorem can 
 also be obtained by reciprocating the theorem, ‘The straight 
 line joining the vertex of a parabola and one of the extremities 
 of a focal chord, and the diameter through the other extremity, 
 meet in the directrix,’ with respect to the parabola itself. 
 
 The reciprocals’ of the theorems I, 2, 3, 4, 7, 8, 9, 10, 11, 
 
 12, 13, 14, 15 and 16 with respect to the conic itself or its as- 
 sociate, are, 
 
 (17). If one of the diagonals of a quadrilateral circum- 
 scribing a conic is the directrix, the other two diagonals pass 
 through the focus. 
 
 (18). The straight line joining the focus of a conic and 
 the point of intersection of a diameter and the directrix is per- 
 pendicular to the tangents at the ends of the diameter. 
 
 (19). The straight lines joining the extremities of a 
 
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64 
 
 focal chord and the points of intersection of the directrix and 
 the focal chords parallel to the tangents at those extremities, 
 meet in a fixed straight line. 
 
 (20). The envelope of the straight line drawn through 
 a point ina given straight line parallel to the tangents to a 
 
 conic at the ends of the diameter passing through the point 1s 
 a conic. 
 
 (21). If a quadrilateral circumscribes a parabola, the 
 straight lines drawn parallel to any two sides from the points 
 of intersection of the alternate tangents meet in the diameter 
 through the point of intersection of the other two sides. 
 
 (22). The tangent at any point of an ellipse or hyper- 
 bola is equally inclined to the diameters conjugate to the dia- 
 meters through the points of intersection of the tangent and 
 the directrices. 
 
 (23). Ifachord PQ meets the directrices in E and F, 
 the angle between the diameters conjugate to CP and CE is 
 
 equal to the angle between the diameters conjugate to CQ 
 and OF. 7 
 
 (24). If two tangents to an ellipse or hyperbola are at 
 right angles to each other, the envelope of the chord of contact 
 is a concentric conic having its axes in the duplicate ratio of 
 the axes of the given conic. 
 
 (25), The envelope of the straight line joining the points 
 of intersection of the directrices and conjugate diameters of an 
 
 ellipse or hyperbola is a concentric, similar, and similarly situ- 
 ated ellipse or hyperbola. 
 
 (26). If thestraight lines drawn parallel to an asymptote 
 from the ends of a focal chord meet the directrix in E and EF, 
 the diameters conjugate to CE and CE’ are at right angles, 
 
 (27). Conies described about a parallelogram are con- 
 centric, and the quadrilateral formed by joining the points of 
 intersection of concentric conics is a parallelogram. 
 
65 
 
 (28). Ifa triangle circumscribes a rectangular hyperbola, 
 and straight lines are drawn from the center at right angles to 
 
 the diameters through the angular points to meet the opposite 
 sides, the points of intersection lie in a tangent. 
 
 (29). If a parallelogram circumscribes a rectangular 
 
 hyperbola, the portions of a tangent intercepted by its adjacent 
 sides subtend equal angles at the.center. 
 
 (30). The portion of the tangent at the vertex of a rec- 
 tangular hyperbola intercepted by the tangents from any point 
 in the conjugate axis subtends a right angle at the center. 
 
 The theorems 20 and 27 can also be obtained by recipro- 
 
 cating the theorems 4 and 13 with respect to the auxiliary circle 
 or the associated circle ( Figs. 19 and 20 ), 
 
 The following are the reciprocals of the theorems 1, 2, 3, 
 
 8, 9, 10, 11 and 12, with respect to the auxiliary circle or the 
 associated circle ( Figs. 19 and 20 ). 
 
 (31). If one of the diagonals of a quadrilateral circum- 
 
 scribing a conic is the conjugate directrix, the other two diago- 
 nals pass through the conjugate focus. 
 
 (32). The portion of the conjugate directrix of an ellipse 
 
 or hyperbola between any point in it and its polar subtends a 
 right angle at the center. 
 
 (33). The straight lines joining each extremity of a con- 
 jugate focal chord of an ellipse or hyperbola and the point of 
 intersection of the conjugate directrix and the diameter at right 
 
 angles to the diameter through the extremity meet in a fixed 
 straight line. 
 (34). The intercept of the conjugate directrices of an 
 
 ellipse on a tangent is divided at the point of contact into two 
 parts wnich subtend equal angles at the center. 
 
 (35). Ifa chord of an ellipse is produced to meet the 
 conjugate directrices, the portions intercepted between the con- 
 
  jugate directrices and the curve subtend equal angles at the 
 center. 
 
66 
 
 (36). The envelope of a chord of an ellipse or hyperbola 
 which subtends a right angle at the center is a concentric circle, 
 ( 37). The envelopeof the straight line joining the poimts 
 of intersection of the conjugate directrices and conjugate dia- 
 
 meters of an ellipse or hyperbola is a concentric, similar, and 
 stnilarly situated ellipse or hyperbola. 
 
 (38). The portion of the conjugate directrix intercepted — 
 by the straight lines drawn parallel to an asymptote from the 
 
 extremities of a conjugate focal chord subtends a right angle 
 at the center. 
 
 The following are the reciprocals of the theorems 2, 5, 6, 
 and 7 with respect to the circle having the vertex of the para- 
 bola for its center and its diameter equal to the latus rectum 
 of the parabola ( Art. 33). The reciprocals of the theorems 1 
 and 3 with respect to the circle are the 31st and 41st theorems, 
 
 (39). The portion of the conjugate directrix of a para- 
 
 bola included between any point in it and its polar subtends a 
 right angle at the vertex. 
 
 (40). A conjugate focal chord of a parabola subtends a 
 right angle at the vertex, 
 
 (41). Any diameter intercepted between the parabola and | 
 the conjugate directrix subtends a right angle at the vertex. 
 
 (42). Ifa quadrilateral circumscribes a parabola, the ~ 
 straight lines drawn from the points of intersection of the al- 
 ternate tangents to the points of intersection of the tangent at 
 the vertex and any two sides, meet in the line joining the ver- ° 
 tex and the point of intersection of the other two sides. 
 
 1f the 41st theorem is slightly modified thus, ‘The locus © 
 of the point of intersection of a chord of a parabola and the 
 diameter which bisects another chord at right angles to the first, 
 
 is a straight line’, and then reciprocated with respect to the 
 circle, we get the following theorem :— 
 
 (43). The straight line drawn perpendicular to a chord 
 
67 
 
 of a parabola passing through the vertex from:its pole passes 
 through a fixed point. 
 
 The reciprocals of the theorems 31, 32, 33, 34, 35, 36, nye 
 
 38, 39, 40 and 42 with. respect to the conic itself. or its asso- 
 ciate, are, 
 
 (44). The straight lines joining the ends of: two conju- 
 
 gate focal chords of an ellipse or hyperbola intersect in the con-- 
 jugate directrix. 
 
 (45). Ifa conjugate focal chord of an ellipse or hyper-- 
 bola passes through the pole of another conjugate focal chord, 
 the diameters bisecting them are at right angles to each other. 
 
 (46). 1£ the tangent at each extremity of a conjugate 
 focal chord of an ellipse or hyperbola meets the conjugate focal 
 chord which is bisected by the diameter at right angles to the 
 diameter through the extremity, the straight line joining the- 
 points of intersection passes through a fixed point. 
 
 (47): Phe diameters bisecting the conjugate focal chords 
 drawn from a point on: an ellipse make equal angles with the 
 diameter passing through:.the point. 
 
 (48); Hf conjugate foeal-chords are drawn through the 
 point of intersection of two tangents to an ellipse, the diameters- 
 
 through the-points of contact make equal angles with the dia- 
 meters bisecting the conjugate focal chords. 
 
 (49). The loeus of the pole of a side of a rhombus in- 
 scribed in an ellipse or hyperbola is a concentric conic whose 
 axes are in the duplicate ratio of the axes of the given conic. 
 
 (50): The locus of the point of intersection of conjugate 
 focal chords of an ellipse or hyperbola parallel to conjugate: 
 
 diamaters is.@ concentric, similar, and similarly situated ellipse 
 or hyperbola. 
 
 «51 ): The diameters which bisect the conjugate focal 
 chords passing through the points of intersection of an asymp- 
 
 tote and: the tangents drawn froma point. in. the conjugate: 
 directrix ave at right angles to each other. 
 
68 
 
 (52). The straight line joining the extremities of the dia- 
 meters drawn through the points where the tangent at the ver- 
 tex of a parabola meets a conjugate focal chord and the conju- 
 gate focal chord through its pole, passes through the focus. 
 
 (53). The tangents drawn from the points of intersection 
 of the tangent at the vertex of a parabola and the diameters 
 through the ends of a focal chord meet in the conjugate directrix. 
 
 (54). One of the sides of a quadrilateral inscribed in a 
 parabola, and the straight line joining the points where the dia- 
 gonals meet the lines joining the vertex and the angular points 
 opposite to the side, meet in the tangent at the vertex, | 
 
 The following are the reciprocals of the theorems 1, 2, 4, 
 10,11, 12,14, 15 and 16, with respect to the parabola which 
 has the centre of the conic for its vertex, and its latus rectum 
 equal and parallel to the Jatus rectum of the conic ( Fig. 27 ). 
 
 (55). If one of the diagonals of a quadrilateral circum- 
 scribing an ellipse or hyperbola coincides in direction with the 
 latus rectum of the associated hyperbola or ellipse, * the other 
 two diagonals pass through the foot of the directrix of the latter. 
 
 (56). The semi-latus rectum of an ellipse or hyperbola is 
 a mean proportional between the focal distances of two points 
 in the latus rectum, one of which lies in the polar of the other 
 with respect to the associated hyperbola or ellipse. 
 
 (57). The envelope of the straight line joining a point 
 in the conjugate axis of an ellipse or hyperbola and the point 
 of intersection of its polar and a given straight line is a conic. 
 
 (58). If the semi-latus rectum of an ellipse or hyper- 
 bola is a mean proportional between the ordinates of the 
 
 extremities of a chord which cuts the transverse axis, the 
 envelope of the chord is a concentric conic. 
 
 *The associated conics referred to in this article sae ae 
 angular axes. 
 
69 
 
 (59). If from the point of intersection of a tangent to 
 an hyperbola or ellipse and the conjugate axis, a perpendicular 
 is drawn to a latus rectum of the associated ellipse or hyper- 
 bola, and from the point of contact a perpendicular is drawn 
 to the other latus rectum, the envelope of the straight line 
 
 joining the feet of the perpendiculars is a concentric hyperbola 
 or ellipse. 
 
 (60). The semi-latus rectum of an hyperbola is a mean 
 proportional between the focal distances of the points of inter- 
 section of the latus rectum and the straight lines joining an 
 extremity of the conjugate axis and the extremities of a chord 
 
 of the associated ellipse passing through the foot of the directrix 
 of the hyperbola. 
 
 (61). Points taken in the sides of a triangle circumscribing 
 a circle such that the perpendiculars from them to any diameter 
 are in the opposite direction to those from the opposite angular 
 
 points and are third proportionals to them and the radius of 
 the circle, lie in a tangent. 
 
 (62). If one of the diagonals of a quadrilateral described 
 about a circle passes through the center, the portions of the 
 tangent parallel to this diagonal intercepted by the perpendi- 
 
 culars from the points of intersection of any other tangent and 
 the sides of the quadrilateral subtend equal angles at the center. 
 
 (63). Parallel tangents to a circle intercept on any other 
 tangent a length which subtends a right angle at the center. 
 
 The reciprocals of the theorems 55, 57, 58, 59, 60, 61, 62, 
 and 63, with respect to the conic itself, are, 
 
 (64). The straight lines joining the ends of pairs of 
 chords of an ellipse or hyperbola which pass through the foot 
 
 of the directrix of the associated hyperbola or Ree intersect 
 in the latus rectum of the latter. 
 
 (65). The locus of the point of intersection of a chord 
 
 parallel to the transverse axis of an ellipse or hyperbola and 
 the line joining its pole and a given point is a conic. 
 
70 
 
 (66). If the semi-latus rectum of an ellipse or hyperbola 
 ‘is a mean proportional between the ordinates of the extremities 
 
 of achord which cuts the transverse axis, the locus of the pole 
 of the chord is a concentric conic. 
 
 (67). The straight lines joining the feet of the direetrices 
 of an ellipse or hyperbola and two points in the conjugate axis 
 one of which lies in the polar of the other with respect to 
 
 the associated hyperbola or ellipse, meet on a concentric 
 hyperbola or ellipse. 
 
 (68). The semi-latus rectum of an . hyperbelas is a mean 
 proportional between the focal distances of the points of inter- 
 section of the latus rectum and the straight lines joining the 
 foot of the directrix and the points where the tangent parallel 
 to the transverse axis meets the tangents drawn from: a_ point. 
 ~ in the latus rectum to the associated ellipse: | 
 
 (69). If the angular points of a triangle described about 
 a circle and points in a diameter are equidistant from: another 
 diameter at right angles to the first and on opposite sides of it,, 
 
 the straight lines joining these points and the opposite points: 
 of contact meet on the circumference. 
 
 (70). Ifa quadrilateral inscribed in a eirele has two 
 sides parallel, the angles contained by two pairs of straight. 
 lines drawn from an extremity of the diameter parallel to. 
 these sides to the points of intersection of diameter perpendi- 
 cular to the sides and the lines joining any point in the cir-. 
 cumference to the angular points of the quadrilateral are equal.. 
 
 (71). The angle in a semicircle is a right angle. 
 
 The following are the reciprocals of the theorems 1, 6, and’ 
 
 7, with respect to the parabola turned through a right angle 
 about the vertex (Fig. 42 ). 
 
 (72). If the tangent to the auxiliary circle of a rect- 
 angular hyperbola at the point where it meets an asymptote, 
 cuts the curve and its conjugate in O and O’, and one of the 
 
71 
 
 diagonals of a quadrilateral circumscribing the original hyper- 
 
 bola coincides in direction with CO’, the other two diagonals are 
 parallel te CO. 
 
 (73). If the straight line joining the vertex and an ex- 
 tremity of the latus rectum of a parabola meets the diameter 
 through one of the extremities of a focal chord in O, the locus 
 of the point of intersection of the diameter through the other 
 extremity and the line drawn through O perpendicular to the 
 axis, is a rectangular hyperbola. 
 
 (74). The straight lines drawn from the points of inter- 
 section of an asymptote of a rectangular hyperbola and any two 
 sides of a circumscribing quadrilateral to the points of inter- 
 section of the alternate tangents meet in the line drawn parallel 
 
 to the asymptote from the point of intersection of the other 
 two tangents. 
 
 48, If an ordinate RQON to the diameter PCP’ 
 ( Fig. 43.) meets the ellipse in Q,an asymptote in O, and the 
 conjugate hyperbola in R, then ON24+QN2=CD32, 
 and RN2+QN2=2CD2. 
 
 Let CQ!’ be conjugate to CQ. 
 
 Because CN: Q'N’:: CP: CD :: CN:ON, 
 
 therefore Q’N’=ON. 
 
 Hence QN2+ON2 =QN24Q/’N’2=CD2, 
 Again because RN?—ON2=CD2, 
 therefore RN2+QN2=2CD2, 
 and RN2—QN2=20N32, 
 
 If any chord EFE’E’ is drawn parallel to one of the lines 
 joining the points of contact meeting the hyperbola and _ its 
 conjugate in E and E’and the ellipse in F and F’, then EF=H’F’. 
 
 Because the asymptote CO bisects PD, it also bisects EE’ 
 and FE” in V, and therefore EF=E’F’. 
 
 Similarly if RQ is produced to meet the ellipse and the 
 
72 
 
 conjugate A ie in Q, and R,, it may be shown 
 that RQ=R, 
 
 49. The eects FP and FQ ( Fig. 5) are proportional 
 to the diameters parallel to them, because FP is to FA as the 
 diameter parallel to FP is to BCB’, and FA is to FQ as BCB' 
 is to the diameter parallel to FQ. It may also be shown that 
 if any tangents FP and FQ are proportional to the diameters 
 parallel to them, F is a point in one of the common tangents. 
 Also the rectangles contained by the segments of chords drawn 
 through a point are in the duplicate ratio of the diameters 
 
 parallel to them only if the point lies in one of the common 
 tangents or diameters of contact. 
 
 Let CF meet the ellipse in P,, the hyperbola 
 in Q,, AP in V, and AQ in V’, and let CD and 
 CD” be the semi-diameters conjugate to CP, and CQ,. 
 
 It is evident that CF bisects AP and AQ and is parallel to A’PQ. 
 Because AP? : 4CV.VF:: CD?: CP,?2, 
 and AQ?; 4CV“V’E:: ; ‘CD’?: CQ, 2, 
 and VE: Vos: CVs CN 3 3 ge 2 oe. 
 therefore CV.VF: CP,?2:: CV.VF: CQ; 8 
 and CV’.WE:; CQ,? 3; CV.WE: Cre 
 Hence AP2: 40D? :: CV.WF: CQ,3, 
 and AQ?: 4CD’23; CV. WE: CP 2 
 and therefore AP: AQ::CD.CP,: CD’.CQ,. 
 50. The following are some of the properties of the 
 ellipse and its associate. ; 
 Let the tangent PF ‘(fFig. 44) meet the hyperbola in O 
 and O’, and let the tangent QF meet the ellipse in E and E’. 
 (1),..,.Because AT.TA’: CA?: : OT Te Ta 
 and. AT’. Al: Ome: : (14.20 ioe 
 and AT.TA’=AT.T’A’ (Art. 6), 
 therefore OT,TO’= Q’T’?. Similarly HT Pie Pas, 
 
_ 
 aod 
 
 73 
 
 ek Se he QT: s NT MET! 42 "ORY CQ, 
 Similarly PT: Q'T) »: CP: CQ. 
 (3). Because KK’ is parallal to QQ’, 
 
 therefore KN=NK’, 
 and PN=NP’, Preperore PK=P’K’, and PF=K’'F”. 
 
 Also KF=P’F’, and PK’, FF’, and QL’ are all equal and 
 parallel. Similarly FL=Q’F’, and KL and K’L' are parallel to 
 CQ’ and CQ respectively. 
 
 (4)78 Because HRIP ER” :-PF2% ¢ PT? + PT? :. 3 
 REAY PH, 
 
 therefore EF. FE’=KF2=P’F’?, 
 
 Similarly FO.FO’/=FL72=Q’E’2, 
 
 Also if Q’E’ meets the ellipsein E, and E,, and 
 P’E’” meets the hyperbola in O, and apy HE, WE,= 
 E’K’2=PF2, and K’YO,.”O =F Q?. 
 
 - (5). Let QPC and YPC be produced to meet the ellipse 
 in P, and P” and the hyperbola in Q, and Q’. 
 
 BOO POOP: >, CI2 ...CP23; CD40, 
 mre EW) i bey PQ. 3's CD’? ; CQ’2, 
 
 boee Gd QP yas CO ss (PQ P4Q")s CO’? gincd 
 
 PP’ is parallel to QQ’, 
 
 therefore EQ.QE’=P’0,.P’O,. 
 
 Similarly PO.PO=Q’H,.Q’Eo. 
 
 (6). If the tangents PT and Q’T’ meet CQ’ and CQ in R 
 and V respectively, PR? —OR?=Q/V2—HE, V2, and 
 PR2-Q’V2=OR?2—E, V2=FR?—F’Q’2 -F’V24PF2, 
 
 Hence PR?—FR?—PF2=Q/V2—W’Q2—F'V2, 
 and therefore (PR+FR-PF)PF=(Q’V+F’Q’-F’V)F'V, 
 
 Hence FR.PF=F’Q’.F’V. 
 HD: Let the tangents QT’ and P’'T meet CQ’ and CQ in 
 V' and R’. 
 
 Because CV: CP :: CP: CQ :: CQ: CR’:: CV’: CP’:: 
 
 CP’: CQ’:: CQ’: CR 
 therefore VV’ and RR’ are parallel to PP’. 
 
74 
 
 (8). Because FV’2—EV’? : V2—E,V? :: 
 FK2; E’K’2 :; FV2; V2, 
 therefore EE, is parallel to FF’ or PP’. 
 Similarly E7E,,OO,, and O’O, are parallel to PP’. 
 (9) Because FW ?2EV’2 ; FR2-OR? :; P/E’? ;: 
 Ey? os WV ee. 
 therefore EO and HO’ are parallel to CQ’ 
 Similarly E,O., and E,QO, are parallel to CQ. 
 (10). Because FE: EO:: P’F’: P’Q’, and FE’: E’0":; P’F:P’Q’, 
 therefore FE,FH’ : EO.H’O’: : P’F’? : PQ’, but 
 FE, FE’=P’F’2, 
 therefore sey Mee Similarly E,O,.E,0, 
 - POQ2 
 (ll). PE: FQ’: PP: FL:: PK: Qis: 2h QU. 
 Q’L’ :; CP2: CQ?. 
 Similarly P’E?: FO :: CP?2* CO] 
 (12). Because F’K’: E,K’:: FK: EK:: FL: OL, 
 therefore EH, K’: OL:: F’K’: FL:: PK’: QL. 
 Hence P’E, is parallel to QO. Similarly PH, 
 is parallel to QO’. 
 Also PE and PE’ are parallel to Q’O, and Q’O, 
 
 respectively. 
 (18). PF is to FQ inthe ratio compounded of PF : FQ! and — 
 HQ’: FQ, 2 ¢°CP*:; CQ?2-and TO: Tay 
 therefore PF is to FQ in the ratio compounded of 
 CD2: CD’? and CD,: CD’, 
 Hence PF: FQ:: CD,CD,; CD’?:: CD,?; 
 CDeODe 
 
 Also P’R"'s F’Q":* CD.CD, + CD52 :aeGp? 5 
 
 CD’.CD,:: FE: FO, | 
 
 therefore PF.FO: QF.FE:; CD,%: CD’? :: 
 CD,?2: CD. 
 
75 
 
 (14). Because EE, and OO, are ordinates to CA,. 
 therefore PE. and P’E,, and QO and Q’O, meet 
 in CA ; 
 
 let them. meet CA in H and H’, 
 
 Because QO is parallel to F'E,,. 
 
 therefore QM: MH’:: PN: NH:: PN: NH, 
 
 and therefore the straight: lines bisecting PE and QO, and: 
 
 P’E, and Q’O,, pass through the center. 
 
 Similarly the straight lines. bisecting: PE’ and QO’, and’ 
 
 P’E, and Q’O,, pass through the center, 
 (15), TH? s: 00%: FE: FO’: PR’ 2’"Q' 3 CD? :- 
 €D’.CD,, 
 
 ang HR 0,07 8 PPR. BOL: PEs EQ: 
 
 CD,2:CD‘.CD,, 
 therefore EH/.E, BE, -O07.0,0, 3: PF? : FQ”, 
 and EHH’.0,0,,: OO%.E,E,::; CD*: CD,*:: 
 CD.CD’: CD,.CD,. 
 (16). FE" bisects TT’ and divides NT and MT” similarly,. 
 because PF: FT:: FT’: FQ. 
 (17). Because FK: KT’:: PK: KK’:: QU’ :LL’:: FT: TL. 
 therefore KT is parallel to T’L. ‘Similarly K’T is parallel. 
 to TL’. 
 (18). Because QF’: ET’: QT ; PT’ s::CQ@.::CP, 
 therefore P,T’ is parallel to CF, 
 and because PF: FT:: PT’: QT :: CP: CQ,. 
 therefore QT is parallel to CF. 
 
 Hence P,T’ and Q,T meet the ellipse and the 
 hyperbola respectively in points which are the points. of con- 
 tact of tangents drawn from F. 
 
 (19). Produce OK and O,E, to meet: the ellipse: 
 in Ez and EH, and the hyperbola in O, and O,. 
 
76 
 
 Because PE : Q’0, ::PK : QL! :: CP: CQ®:: 
 CV :'\ CY: % EER, : Q’Q”, and PH Ge: 
 CR Chis: PRA Oro. 
 therefore PE, and EP, are parallel to 0,Q” 
 and Q’O,; respectively, and PEP,EH; is similar 
 to Q’0,Q’03. 
 Similarly it may be shown that P’H,P"H, is 
 similar to OQO,Q,. 
 Because CV’: CP’:; CQ’: CR, therefore EE. 
 OO0,=P’P”.YQ”. Also E, E,.0,03=PP,.QQ). 
 (20). Because EV? : CD,? 3 GRl = yw 
 CP2, and EV’: CD? ;: CP’2—CV?: CP, 
 therefore E,V : EV’::CD, :CD::H,E,: EH, 
 Similarly OO’: 0,0, :: CD,: CD’ :: CD, : CD, 
 Hence’ HE’ .OO@E E,.0,0;. 
 (21).) Because:H, FY; PE: PE: EP 
 therefore P’E, is parallel to PE. 
 Similarly P’E,, QO, and Q’O, are parallel to 
 PH’ Q’O,, and QO’ respectively. 
 (22). Because the triangle PEO is similar to the triangle 
 He PQs 
 and the triangle PQO is similar to the triangle 
 203’, 
 therefore EPQO is similar to P/H,0,Q. 
 Similarly EZ, P’Q’O, is similar to E’PQO’. 
 (23). Because HE, EH, and E,E, are parallel to 
 conjugate diameters, E, HK, passes through the center, 
 Similarly HE’; passes through the center, 
 (24). Because FK: FT’:: FT: FL:: QL’: QL:: CP: CQ, 
 and FP: PT:: FT’; FQ:: PP: OT + 2CP ean) 
 therefore FP: FT:: FT: FL, and FK; FT’: FT’: FQ. 
 
77 
 
 (25). Because the triangles PKE’ and PKE are equal to 
 the triangles P’K’E, and P’K’H,, therefore the tri- 
 angles EPEY and E ,P’E, are equal, Similarly the 
 triangles PE,EH, and ae are equal, 
 
 (26). PE’: Q0,::E Ese Claes 2? b, Haat GG), 
 lee Coen at oO. Les CC)2, 
 
 (27). Because the polars of Q’ and T pass through K, 
 therefore Q’T is the polar of K. Also the polar of K passes 
 through the point of intersection of QV’ and P”P ( Art. 10 ), 
 therefore Q'T and P”P meet on the tangent QV’. Similarly P’I’ 
 and QQ” meet on the tangent PR. Also QT and P,P’ meet on 
 the tangent Q/V, and PT’ and Q’Q, meet on the 
 
 tangent P’R’. Also L, K’, and L’ are the poles of PT’, QT, and 
 P'T’ respectively with respect to the ellipse. 
 
 (28). Because PK: KP’::CP: CQ, 
 therefore CK is parallel to QP’, 
 and because CV’: CP’::CP,:CQ,, 
 therefore Q,P’ is parallel to P,V’. 
 Similarly CK’L is parallel to P,Q’ or Q,R 
 (29). Draw QT” parallel to PT meeting CA produced in T”, 
 Because QM: MT’:: PN: NT::QM: MT", 
 therefore MT’=MT", 
 and because CT: CT” ::CP:CQ:: CT’: CT, 
 therefore CT?=CT’.CT”=CM2-MT’2, 
 Similarly CT’??=CN?2-NT?, 
 (80). Itis evident that the semiordinate through T is to 
 TP as BC is to CD,. 
 (Sly. Because TP; TQ’ 2; PN Q'M:: CP: CQ:: TR: 
 mene: bh Veo bib ss Ves ER 
 therefore TP2=T’Q’. T’V, and TP’2=T’Q,T’V’, 
 (32). The straight lines drawn through T’ and M parallel 
 to the asymptotes meet on the hyperbola, and the straight 
 lines drawn through N and T parallel to the asymptotes 
 
 also meet on the hyperbola, because the ordinates through 
 
78 
 
 the points where they meet bisect MI’ and NT, and 
 CT’. CM=CA2=CT.CN. 
 
 (38). If the tangents at P and Q meet BCB’ in ¢ and 1’, it 
 may be shown that Ct?-Ci’2=2zBC?, and therefore 
 the ordinates through ¢ and ?’ of the conjugate and the original 
 
 hyperbola respectively are equal. 
 
 (34). Because Pt is to the semiordinate of the con- 
 jugate hyperbola through ¢ as CD, is to AC, and PT.Pt 
 =OD,’, j 
 
 therefore the rectangle contained by the ordinates of the: 
 conjugate and original hyperbolas through ¢ and T respectively 
 is equal to AA’ BB’ which is also equal to the reetangle con- 
 tained by the ordinates through ?’ and T’ of the original 
 hyperbola and the ellipse respectively. 
 
 (35). Since a concentric, similar, and similarly situated 
 ellipse passing throught any point Q, passes through Q,, D’, 
 and Dg, therefore the common diameters of this ellipse and 
 the hyperbola and its conjugate are conjugate, and the 
 other common chords are proportional to AC and BC, and have: 
 
 therefore a constant ratio. Because QT” is parallel to PT, QT” 
 touches the ellipse at Q. 
 
 (36). It can be easily shown that CKL’ PQ’, and FE” 
 meet at a point, and P’T” is parallel to the line -joining F and 
 the point of intersection of PQ’ and P’Q. 
 
 51. If the axes of the ellipse and its associate are rect-. 
 angular,Fig.44) PH: QE:: CP?2: CQ?, FE.FER= 
 FP2, and FO.FO’=FQ?. (1). 
 
 Because FR PF=F’Q’ F’'V=QF.FV’, 
 
 therefore P,Q,R,V’ are concyclic, and the angle PQF=the. 
 angle FRV’=the angle FOE, and therefore P,E,O,Q are also: 
 concyclic. (2). 
 
 Because EO is parallel to EO’, P,E’,0’.Q are also con- 
 
 cyclic. (3).. 
 
79 
 
 Because PE: QO:: PF: QF:: CP2: CQ?:: 
 Pan GR. ti PP): O00, 
 
 and the angle EPP,=the angle QOE, therefore 
 the triangles EPP, and "Q00, are similar. (4). 
 
 Because PE is parallel to QO’ or Q,0,, there- 
 fore Q,0,Q,,0, are concyclic, Also because QO is 
 parallel to PE’ or P 18 therefore P,H,P,,Ei; are 
 coneyclic, and P,,E;,Q,,0 4 are concyclie, ap 
 
 Because PY’ is parallel to PQ, and PV’ and QR are equally 
 inclined to the axis, therefore QQ’ biseets the angle P'QR. (6). 
 Because HH’: OO7:; CP?: CQ’: ET’.T’E’:: OT.TO’, 
 
 therefore the harmonic mean between ET’ and T’E’ ig 
 equal to the harmonic mean between OT and TO. (7). 
 
 Because EO,.HO: EQ? :: CQ’? : CD’ 
 and KO OL, : OP? pcan CD 
 
 therefore HO, : OE, (3 EQ2; OPS, (8). 
 It is evident that the HAI lines joining the foci of the 
 hyperbola and its conjugate touch the ellipse. (9). 
 
 If an ellipse and an hyperbola jhave the same transverse 
 axis, it may be shown that the tangents to each curve from the 
 focus of the other are at right angles. (10). 
 
 Similar properties of the parabola ( Fig. 30 ) ean be easily 
 deduced. 
 
 52. If an ellipse and its associate have rectangular axes, 
 and through any point O (Fig. 45) chords LOL’ and ROR’ 
 are drawn so as to be bisected in O, the sum of the squares on 
 
 the chords is equal to the sum of the squares on the diameters 
 parallel to them, 
 
 Because LO? : CRs-COs a3 pes OE 
 >: CF? : CQ? :: OK? : CO?:: LO?+OK2: CP2, 
 therefore LO? +0K2=CE®, 
 and because RO2—~OK *=CF2, 
 therefore LO?+RO2=CKE?2+CF32, 
 53. We have proved that EO (Art. 50) is parallel to E’O’. 
 The reciprocal of this theorem with respect to the conic itself 
 
80 
 
 is, ‘The third diagonal of the quadrilateral formed by the tan¢ | 
 
 gents from P to the hyperbola and from Q’ to the ellipse coin: 
 
 cides with CD’. The reciprocal of the theorem with respect to 
 the parabola ( Fig. 27 ) is, ‘The third diagonal of the quadri« 
 lateral formed by the tangents from the extremities of equal 
 
 ordinates of an ellipse and its associate on opposite sides of the — 
 
 axis 1s parallel to the axis. | 
 
 The reciprocal of the theorem ‘PE and QO are equally ins 
 
 clined to the axis, and PE’ and QO’ are also equally inclined to 
 the axis’ (Art, 51), with respect to the parabola ( Fig. 27), is, 
 ‘The tangents at the extremities of equal ordinates of a conic 
 and its associate, and the tangents from these points to the as- 
 sociated conic meet in points equidistant from the axis’. 
 
 The reciprocal of the theorem ‘P’T’ is parallel to the line 
 joining F and the point of intersection of PQ’ and P’Q’ (Art.50) 
 
 with respect to the conic itself, is, ‘PQ’ and FE’ intersect in 
 
 CL’,’ and the reciprocal of the theorem with respect to the ue 
 
 parabola (Fig. 27) is, ‘If K,K, and L,L, are any 
 
 two equal ordinates of a conic and its associate, the point of — 
 
 intersection of K,K, and the ordinate through the 
 
 point where the tangents at K, and L, meet, and — 
 
 the point of intersection of K, K,and the tangent at L, 
 lie on a straight line parallel to the, axis’. 
 
 Since F (Fig. 44)'is the pole of PQ’, and F” is the pole of 
 
 P’Q, FF’ is the polar of the point of intersection of PQ’ and P’'Q. _ 
 
 Because KT is parallel to T’L, therefore the straight line 
 
 joining the point of intersection of PP’ and Q'T and the point — 
 
 of intersection of QQ! and PT’ passes through the center. 
 
 If the tangents at P, andQ, meet QQ’ and PP’ 
 in G, and Gg respectivdy, FG, and FG, are tangents 
 because P,1’ and Q,T meet the curves in points 
 which are the points of contact of the tangents drawn from F, 
 
 Because P,T’ and Q,T are parallel to CF, 
 
 therefore UG, and CG, are conjugate to CF, 
 
 and because PE is parallel to P’E,, and PH’ is 
 
 parallel to P’E,, 
 
 therefore the tangents at P and P’ meet the tangents from — 
 
 Q’ and Q on two diameters. 
 
 SOReRAdaasacesascansscadanracaccasacddcnaahsednasascanncanAeRaases. 
 
 yn ite