CONIC SECTIONS. 



153 



bola. 



LE, LD are similarly divided at P and H, also at p 

 and h, therefore PE : HD : : (LE : LD) CM : CR, also 

 pE : fiQ : : (LE : LD : :) CM : CR, hence taking the rect- 

 angles of the corresponding terms, FE.pE : HD.hD : : 

 CM : : CR 2 . But if C, D be joined, the points D and 

 R are evidently the vertices of two conjugate diame- 

 ters (Cor. to Def. 19-)> and therefore PE.pE=CM* 

 (27.), therefore HD.M>=CR*. 



Cor. The rectangle contained by LD and DK, the 

 segments of a tangent intercepted between D, the point 

 of contact, and Yp, Qq, any two conjugate diameters, 

 is equal to the square of CR, the semidiameter to which 

 the tangent is parallel. 



Let the parallel tangents PH, p h, meet LK in H 

 and h, and draw DE a semiordinate to Pp. Because 

 of the parallels ED, PH, CK, ph, LE : LD : : EP : DH, 

 and EC : DK : : Ep : T>h, therefore LE.EC : LD.DK : : 

 EP.Ep : DH.D/;. But LE.EC=EP.E w (l. Cor. 14.), 

 therefore LD.DK=DH.D/*= (by this Prop.) CR 2 . 



Prop. XXX. 



If two straight lines be drawn from the foci of an hy- 

 perbola perpendicular to a tangent, straight lines drawn 

 from the centre, to the points in which they meet 

 the tangents, will each be equal to half the transverse 

 axis. 



Let Vd D be a tangent to the curve at P, and ED,fd 

 perpendiculars to the tangent from the foci, the straight 

 lines joining the points C, D and C, (/ are each equal 

 to AC, half the transverse axis. Join FP,_/P, and pro- 

 duce FD, P/' till they intersect in E. The triangles 

 FDP, EDP have the angles at D right angles, and the 

 angles FPD, EPD equal (8. ), and the side DP com- 

 mon to both ; they are therefore equal, and consequent- 

 ly have ED=DF, and EP=PF, wherefore E/=FP— 

 TfzzAa. Now, the straight lines FE, F/ being bisect- 

 ed at D and C, the line DC is parallel to E/, (2. 6. E.) 

 and thus the triangles F fE, FCD are similar, therefore 

 F/:/'E, or Art : : FC : CD : but FC is half Wf, there- 

 fore CD is half of Aa. 



Cor. If a straight line Qq be drawn through the 

 centre parallel to the tangent Dd, it will cut off from 

 PF, P / the segments PG, Pg, each equal to AC, half 

 the transverse axis. For CVPG, CDPg are parallelo- 

 grams, therefore PG=rfC=AC, and Pg=DC=AC. 



Prop. XXXI. 



The rectangle contained by perpendiculars drawn 

 from the foci of an hyperbola to a tangent, is equal to 

 the square of half the conjugate axis. 



Let PrfDbe a tangent, and FD, fd perpendiculars 

 from the foci, the rectangle contained by FD and yd 

 is equal to the square of BC, half the conjugate axis. 



It is evident from the last Proposition, that the points 

 D, d are in the circumference of a circle, having the 

 same centre as the hyperbola, and radius CA half the 

 transverse axis. Now, FDrf being a right angle, if dC 

 be joined, and produced, it will meet DF in H, a point 

 in the circumference (31. 3. E.); and since FCzr/C, 

 and CH = Cd, and the angles FCH, fCd are equal, 

 FH is equal to/,/, therefore DF.ajf=bF.FH=AF.aF 

 (Cor. 36. 3. E.)=CB* (3.) 



Cor. If PP, P/ be drawn from the point of contact 

 to the foci, the square of FD is a fourth proportional to 

 /P, FP, and CB*. For the angles fVd, FPD are equal 

 48.), and FDP, fd P are right angles, therefore the tri- 



VOL. VII. part l 



Hyper- 

 br.la. 



angles FDP, fdV are similar, and/P : FP : :/d: FD : ; 

 fd.ED or BC ! : FD*. 



Prop. XXXII. 



If from C, the centre of an hyperbola, a straight line pig. £* 

 CL be drawn perpendicular to a tangent LD, and from 

 D, the point of contact, a perpendicular be dfawn to 

 the tangent, meeting the transverse axis in H, and the 

 conjugate axis in h, the rectangle contained by CL and 

 DH is equal to the square of CB, the semiconjugate 

 axis; and the rectangle contained by CL and Dh is 

 equal to the square of CA, the semitransverse axis. 



Let the axes meet the tangent in M and m, and from 

 D draw the semiordinates DE, De, which will be per- 

 pendicular to the axes. The triangles DEH, CLw are 

 evidently equiangular, therefore DH : DE : : Cm : CL, 

 hence CL.DH=DE.Cw, but DE.Cw or Ce.Cm=BC J 

 (15.), therefore CL.DH=BC\ In the same way it is 

 shewn that CL.D/i^AC. 



Cor. 1 . If a perpendicular be drawn to a tangent at 

 the point of contact, the segments intercepted between 

 the point of contact and the axes are to each other re- 

 ciprocally as the squares of the axes by which they are 

 terminated. For AC 2 : BC 2 : CL.Dh : CL.DH ::Dk: 

 DH. 



Cor. 2. If DF be drawn to either focus, and HK be 

 drawn perpendicular to DF, the straight line DK shall 

 be equal to half the parameter of the transverse axis. 

 Draw CG parallel to the tangent at D, meeting DH in 

 N, and DF in G. The triangles GDN, HDK are si- 

 milar, therefore GD : DN : : HD : DK ; and hence 

 GD.DK = HD.DN. But GD = AC (Cor. 30.) and 

 ND=CL, therefore AC.DK=HD.CL= (by the Prop.) 

 CB 2 , wherefore AC : BC : : BC : DK, hence DK is half 

 the parameter of Art, (Def. 20.) 



Prop. XXXIII. 



If through P and Q, the vertices of two semidiame- 

 ters of an hyperbola, there be drawn straight lines PD, 

 QE parallel to one of the asymptotes CM, meeting the 

 other asymptote in D and E, the hyperbolic sector 

 PCQ is equal to the hyperbolic trapezium PDEQ; 



Let CQ meet PD in T, the triangles CDP, CEQ are 

 equal (1. Cor. 19-); therefore, taking the triangle CDT 

 from both, the triangle CTP is equal to the quadrilate- 

 ral DEQT : to these add the figure PTQ, and the hy- 

 perbolic sector PCQ is equal to the hyperbolic trape- 

 zium PDEQ. 



Prop. XXXIV. 



If from the centre of an hyperbola the segments CD, Fig. f9. 

 CE, CH be taken in continued proportion in one of the 

 asymptotes, and the straight lines DP, EQ, HR be 

 drawn parallel to the other asymptote, meeting the hy- 

 perbola in P, Q, R ; the hyperbolic areas PDEQ, QEH R 

 are equal. 



Through Q draw a tangent to the curve, meeting 

 the asymptotes in K and L ; join PR, meetmg the 

 asymptotes in M and N ; draw the semidiameters CP, 

 CQ, CR; let CQ meet PR in G. Beeause QE is paral- 

 lel to CM, and KQ is equal to QL (2. Cor. 18.), CE is 

 equal to EL ; and because MC, PD, RH are parallel, 

 and MP is equal to RN (1. Cor. 18.), CD is equal to 

 HN. Now, by hypothesis, CD : CE : : CE : CH, there- 

 fore NH : LE : : CE : CH ; but CE : CH : : HR : EQ 

 u 



Fig. 5.0. 



