152 



CONIC SECTIONS. 



Hyper- 



boll. 



Sis. 5*. 



fiv. S3. 



Let Rr be the transverse axis, Yp any other trans- 

 ( verse diameter, draw PE perpendicular toRr; then 

 CE being greater than CIl, and CP greater than CE, 

 much more is CP greater than CR; therefore Yp is 

 greater than Rr. In like manner it is shewn, that if 

 S s be the conjugate axis, and Q q any other second 

 diameter, Q q is greater than S s. 



Prop. XXVI. 



If an ordinate be drawn to any transverse diameter 

 of an hyperbola, the rectangle contained by the abscis- 

 sas of the diameter is to the square of the semiordinate 

 as the diameter to its parameter. 



Let DE be a semiordinate to the transverse diame- 

 ter P p ; let PG be the parameter of the diameter, and 

 O q the conjugate diameter. By the definition of the 

 parameter ( Def. 20. ) P p : Q q : : Q q : PG, therefore 

 Yp : PG : : P^ 2 : Q 9 2 , (2 Cor. 20. 6. E.) But Yf : 

 Oq-i : : PE.E^ : DE 2 , (24.) therefore PE.E_p : DE* : : 

 Yp : PG. 



Cor. Let the parameter PG be perpendicular to the 

 diameter Yp; join pG, and from E draw EM parallel 

 to PG meeting pG in M. The square of DE the se- 

 miordinate is equal to the rectangle contained by PE 

 and EM. For YE.Ep : DE 2 : : Yp : PG, and Yp : PG 

 : : E p : EM : : PE . En : : PE . EM ; therefore DE 2 = 

 PE.EM. 



Scholium. 



If the rectangles PGLp, HGKM be completed, it 

 will appear that the square of ED is equal to the rect- 

 angle MP, which rectangle is greater than the rectangle 

 KP, contained by the abscissa PE and the parameter 

 GP, by a rectangle KH similar, and similarly situated 

 to LP, the rectangle contained by the parameter and 

 diameter. It was on account of the excess of the square 

 of the ordinate above the rectangle contained by the 

 abscissa and parameter that Apollonius gave the curve 

 to which the property belongs the name Hyperbola. 



Prop. XXVII. 



If from the vertices of two conjugate diameters of 

 an hyperbola there be drawn ordinates to any third 

 transverse diameter, the square of the segment of that 

 diameter, intercepted between the ordinate from the 

 vertex of the second diameter, and the centre, is equal 

 to the rectangle contained by the segments between the 

 other ordinate and the vertices of the third transverse 

 diameter. And the square of the segment intercepted 

 between the ordinate from the vertex of the transverse 

 diameter, and the centre, is equal to the square of the 

 segment between the other ordinate and the centre, 

 together with the square of half the third transverse 

 diameter. 



Let Y p, Qq be two conjugate diameters, of which 

 Y p is a transverse, and Qq a second diameter ; let 

 PE, QG be semiordinates to any third transverse di- 

 ameter Rr, then, CG 2 =RE.Er and CE 2 =CG 2 + CR2. 

 Draw the tangents PH, QK, meeting R r in H and K. 

 The rectangles KC.CE and KC.CG are equal, for each 

 is equal to CR 2 (14. and 23.) therefore HC : CK : : CG : 

 CE. But the triangles HPC, CQK are evidently si- 

 milar (Cor. to Def. 19.) and since PE, QG are parallel, 

 their bases CH, KC are similarly divided at E and G, 

 therefore HC : CK : : HE : CG, wherefore CG : CE 

 : : HE : CG, consequently CG 2 =CE.EH = (by 1. Cor. 

 J 4.) RE.E r. Again, from the similar triangles HPC, 



CQK, HC : CK : : CE : KG. Now it was shewn that Hvper. 

 HC : CK : : CG : CE, therefore CG : CE : : CE : KG. bola. 

 consequently CE 2 =CG.GK=(3.2.E.) CG^-f-GC CK '*" 

 But GC.CK=CR 2 (23.) therefore CE 2 =CG e -f-CRA 



Cor. 1. Let S s be the diameter that is conjugate to 

 R r, then R r is to S * as CG to PE, or as CE to QG. 

 For R r s : S * 2 : : RE.E r, or CG* : PE 2 , therefore 

 R r : S s : : CG : PE. In like manner R r : S .? : : CE 

 QG. 



Cor. 2. The difference between the squares of CE, 

 CG, the segments of the transverse diameter to which 

 the semiordinates PE, OG are drawn, is equal to the 

 square of CR, the semidiameter. For it has been 

 shewn that CE'^=CG 2 -}-CR 2 ; therefore CE 2 — -CG 2 = 

 CR*. 



Cor. 3. The difference of the squares of any two 

 conjugate diameters is equal to the difference of the 

 squares oi the axes. Let Rr, Ss be the axes, and Yp, 

 Q q any two conjugate diameters ; draw PE, QG per- 

 pendicular to R r, and PL, QM perpendicular to S s. 

 ThenCE 2 — CG 2 z=CR 2 , and CM* — CL 2 , or GQ 2 — 

 PE= = CS 2 , therefore CE 2 + PE 2 — (CG 2 + GQ 2 ) = 

 CR 2 — CS 2 , that is (47. 1. E) CP 2 — CQ 2 =CR 2 — 

 CS 1 , therefore Pp 2 — Q 5 2 = R r 2 — S s 2 . 



Prop. XXVIII. 



If four straight lines touching conjugate hyperbolas 

 at the vertices of any two conjugate diameters, the pai 

 rallelogram formed by them is equal to the rectangle 

 contained by the transverse and conjugate axis. 



Let Y p, Qq be any two conjugate diameters, a pa- p;^ 5 $ 

 rallelogram DEGH formed by tangents to the conju- °" 

 gate hyperbolas at their vertices is equal to the rect- 

 angle contained by A a, B b, the two axes. __ 



Let A a, one of the axes, meet the tangent PE in K; 

 join QK, and draw PL, QM perpendicular to A a. 

 Because CK : CA : CA : CL (14.) and CA : CB : : 

 CL : QM ( 1. Cor. 26.) ex stq. CK : CB : : CA : QM ; 

 therefore CK.QM=CB.CA. But CK.QMrr twice the 

 trian. CKQrrparal. CPEQ ; therefore the paral. CPEQ 

 rrCB.CA ; and, taking the quadruples, the parallelo- 

 gram DEGH is equal to the rectangle contained by 

 Aa and B6. 



Prop. XXIX. 



If two tangents at the vertices of any transverse dia- 

 meter of an hyperbola meet a third tangent, the rect- 

 angle contained by their segments between the points 

 of contact, and the points of intersection, is equal to 

 the square of the semidiameter to which they are pa- 

 rallel. And the rectangle contained by the segments 

 of the third tangent between its point of contact and 

 the parallel tangents, is equal to the square of the se- 

 midiameter to which it is parallel. 



Let PH, p h, tangents at the vertices of a transverse p; ?> 5 g a 

 diameter P;;, meet DH/i, a tangent to the curve, at 

 any point D, in H and h ; let CQ be the semidiameter 

 to which the tangents PH, ph are parallel, and CR 

 that to which H^ is parallel ; then PH.p/i=;CQ*, and 

 DH.D/; = CR 2 ; let ti'h meet the semidiameters CP, 

 CQ in L and K. Draw DE, RM parallel to CQ, and 

 DG parallel to CP. Because LP.LpzrLE.LC (2. 

 Cor. 14.), LP : LE : : LC : Lp; hence, and because 

 of the parallels PH, ED, CK, ph, PH : ED : : CK : 

 p h, wherefore YH.p h = ED.CK. But ED.CK = 

 CG.CK=CQ 2 (23.), therefore YH.ph = CQ*. Again, 

 the triangles LED, CMR are evidently similar, and 



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