143 



CONIC SECTIONS. 



Hyper- 

 bola. 



Fir. 40. 



Fig. 41. 



Cor. 2. Straight lines drawn from the intersection of 

 any two tangents to the foci make equal angles with 

 those tangents. 



For produce FH to O any distance: and because the 

 angles FHK, FH& are equal, the angles OHK, OH/c 

 are equal : but OHK=OH/+ 2/HD and OH k = 

 kHg+OHg=fHg + OHg=OHf+2 0Hg,there- 

 fore OH/+2/HD=OH/".f 20Hg; hence 2/"HD= 

 2 OH g and fHD = OH g=QH d (in Fig. 89. No- 1.) 

 or/HD=FH d (in Fig. 39. No. 2.) 



Prop. XII. 



If there be two tangents at the extremities of a chord, 

 and a third be parallel to the chord, the part of this 

 tangent intercepted between the other two is bisected 

 at the point of contact. 



Let HD, Hd be tangents at the extremities of the 

 chord D d, and KP/i another tangent parallel to the 

 chord, meeting the others in K and k ; the line K k is 

 bisected at P the point of contact. From the points 

 of contact D, P, d draw lines to F either of the foci, 

 and from the intersection of each two tangents, draw 

 perpendiculars to straight lines drawn from the focus 

 through their points of contact ; that is, draw HI, H i 

 perpendicular to FD, F d and KM, Km perpendicular 

 to FD and FP ; and k N, k n perpendicular to Yd and 

 FP. The triangles DHI, DKM are manifestly equi- 

 angular, as also tf H i, d kN ; therefore, DH : DK : : 

 HI : KM ; and d H : dk : : Hi : k N, but Kk being pa- 

 rallel to D d, DH : DK : : d H : dk (2. 6. E. ) ; there- 

 fore HI : KM : : H i : k N, now H [ = H i (1. Cor. 11.) 

 therefore KM = k N ; but KM = K m, and k N = k n 

 (1. Cor. 11.) therefore K m == K n ; and since we have 

 manifestly K mi km: KF : k P (4. 6. E.) therefore 

 KP=kp. 



Definitions. 



13. Any chord not passing through the centre which 

 is bisected by a diameter, is called an Ordinate to that 

 diameter. 



14. The segments into which an ordinate divides a 

 diameter are called Abscisses. 



Lemma. 



LetHK'£' be a triangle, having its base K' k' bi- 

 sected at p ; and let K k, any straight line parallel to 

 the base, and terminated by the sides produced, be bi- 

 sected at P, then P, p, the points of bisection, and H, 

 the vertex of the triangle, are in a straight line, which 

 bisects D d any other straight line parallel to K k. 



Join HP, Up. The triangles KHi, and K'H/c' 

 being similar, and K k, K' k' similarly divided at P 

 and p, we have HK :HK'(::K(: K' k') : : KP : K'p; 

 now the angles at K and K' are equal, therefore the 

 triangles HKP, HK'p are similar; and the angle PHK 

 is equal to wHK', and the sum of PHK and pHK 

 is equal to the sum of p H K' and p H K, that is to two 

 right angles (13. 1. E.) therefore HP, Hp lie in the 

 same straight line (14-. I.E.) 



.Again, let D d meet HP in E, then KP : DE 

 ( : : PH : EH) : : P k : E d, but KP s= P *.; therefore 

 DE=Ed. 



Prop. XIII. 



Any chord not passing through the centre, but pa- 

 rallel to a tangent, is bisected by the diameter which 

 passes through the point of contact, or it is an ordinate 

 to that diameter. 



Hyper- 

 bola. 



The chord D d, which is parallel to K k a tangent 

 at P, is bisected at E by the diameter Pp. Draw 

 K' p k' a tangent at p, the other end of the diameter, v ' £? 

 and DPI, d H tangents at D, d, the extremities of the 

 chord meeting the other tangents in K, k, and K', k'. 

 Then KP k and K'pk' are bisected at P and p (12.) 

 therefore the diameter P p when produced must pass 

 through H and bisect D d, which is parallel to K k or 

 K' k', in E ( Lemma. ) 



Cor. 1. Straight lines which touch an hyperbola at 

 the extremities of an ordinate to any transverse diame- 

 ter, intersect each other in that diameter. 



Cor. 2. Every ordinate to a transverse diameter is 

 parallel to a tangent at its vertex. For if not, let a 

 tangent be drawn parallel to the ordinate, then the di- 

 ameter drawn through the point of contact would bi- 

 sect the ordinate, and then the same line would be bi- 

 sected in two different points, which is absurd. 



Cor. 3. All the ordinates to the same transverse di- 

 ameter are parallel to each other. 



Cor. 4. A straight line that bisects two parallel 

 chords, and terminates in the opposite branches, is a 

 transverse diameter. 



Cor. 5. The ordinates to the transverse axis are per- 

 pendicular to it, and no other transverse diameter has 

 its ordinates perpendicular to it. This follows from 

 Cor. 3. to 4. and Cor. 2. to 10. 



Prop. XIV. 



If a tangent to an hyperbola meet a transverse dia- 

 meter, and from the point of contact an ordinate be 

 drawn to that diameter, the semi-diameter will be a 

 mean proportional between the segments of the diame- 

 ter intercepted between the centre and the ordinate, 

 and between the centre and the tangent. 



Let DH, a tangent to the curve at D, meet the dia- p;„._ 42> 

 meter Pp in H ; and let DJLd be an ordinate to that 

 diameter ; then CE : CP : : CP : CH. Through P and 

 p, the vertices of the diameter, draw the tangents PK # 

 p K', meeting DH in K and K' : draw PF, pF, DF to 

 either of the foci, and draw KM, Km perpendicular to 

 FP and FD, and K'N, K'm perpendicular to Fp, FD. 

 The triangles PKM, pK'N are equiangular, for the 

 angles at M and N are right angles, and the angles 

 MFK, NpK' are equal (1. Cor. 10.), therefore PK : 

 pK' : : KM: K'N (4. 6. E.) : : Km : K'n (I. Cor. 11.), 

 but the triangles KmD, K'mD being manifestly equi- 

 angular, Km : K'n : : KD, K'D, therefore PK :pK' : : 

 KD : K'D. But because of the parallel lines KP, DE, 

 K'p, we have PK : pK' : : PH : pH, and KD : K'D : : 

 PE : pE, therefore PH : pH : : PE : pE. Take CG= 

 CE, then pG=PE, and (Cor. Prop. D. 5. E.) PH : 

 Pp : : PE : EG, and taking the halves of the conse- 

 quents, PH : PC : : PE : EC : hence by division PIC : 

 PC : : PC : EC. 



Cor 1. The rectangle contained by PE and Ep is 

 equal to the rectangle contained by HE and CE. For 

 CP 2 = HC.CE=EC 2 — HE.EC (2.2. E.), also CP 2 = 

 EC 2 — PE.Ep (6. 2. E.), therefore EC 2 — HE.EC= 

 EC 2 — PE.Ep, and HE.EC=PE.Ep. 



Cor. 2. The rectangle contained by PH and Hp is 

 equal to the rectangle contained by HE and HC. 



For PIC 2 = CP 2 — PH.Hp (5. 2. E.) also HC 2 = 

 EC.PIC— EH.HC = CP 2 — EH.HC (3. 2. E. and by 

 the Prop.), therefore CP 2 — PH.Hp=CP 2 — EH.HC, 

 and PH.Hp=EH.HC. 





