160 



CONIC SECTIONS. 



Curvature. 



Tig. 89. 



Pig. 90. 



Let the ordinate DJLd now meet the arch PLI any- 

 where as at h, draw hg perpendicular to KL, meeting 

 KP in in, then it will appear as before that d E * : h E * 

 : : KP : K m : : KO : K g ; but KO is less than K g, and 

 therefore rfE 1 is less than h E 2 , thus the arch PALI 

 falls wholly without the parabola. 



Case 2. Let the section be either an ellipse or hyper- 

 bola, of which P p is a diameter, and PKL the circle 

 of curvature at its vertex, cutting off PK equal to its 

 parameter. Draw KL the diameter of the circle, and 

 KQ perpendicular to KL, and let pQ, a tangent to the 

 conic section in p, meet KQ in Q. Join PQ ; this line 

 will necessarily meet the circle again ; let it meet the 

 circle in I; and draw IS, IT parallel to Qp, QK meet- 

 ing PK in S, T. Then because of the parallels, p P : 

 f S : : QP : QI : : KP : KT, hence pV : KP : : p S : KT : : 

 p S.SP : KT.SP ; but KT.SP=IS 2 (Lemma) therefore 

 p P : KP : : p S.SP : SP, hence I is a point in the ellipse 

 or hyperbola. (Prop. 18. Sect. II. and 26. Sect. III.) 



Let DEd an ordinate to the diameter P p meet the 

 arch PLKI any where in H, if the point K is between 

 P and p, or the arch PHI, if K is in pP produced. Let 

 D d meet PI in Y, draw HG perpendicular to KL, 

 meeting PK in M, and PI in Z, and draw YV paral- 

 lel to HM, meeting KP in V. Because EY, p Q are 

 parallel, also VY, KQ; Pp : pE : : (QP : QY: :) KP: KV; 

 now KP being the parameter, we have, as in Case 2. 

 Prop. 1. DE 2 : HE 1 : : KV : KM : : QY : QZ; but where- 

 ever the point H be taken in the arch PHI, QY is great- 

 er than QZ, therefore also DE 2 is greater than HE* ; 

 thus the arch PHI falls wholly within the conic sec- 

 tion. 



Let the ordinate DE d now meet the other arch P k I 

 any where in h ; draw h g pei-pendicular to KL meet- 

 ing KP in m, and IP in z, then it will in like manner 

 appear that d E 2 ■: h E 2 : : KV : K m : : QY : Q z ; and 

 since in this case Q Y is less than Q z, therefore d E l is 

 less than liE 1 ; hence the arch P h I is wholly without 

 the conic section. 



Prop. IV. 



The chord of the circle of curvature which is drawn 

 from the point of contact through the focus of a para- 

 bola, is equal to the chord which the circle cuts off from 

 the diameter of the parabola ; and half the radius of the 

 circle is a third proportional to the perpendicular from 

 the focus upon the tangent, and the distance of the point 

 of contact from the focus. 



Let PK be the chord cut off from the diameter, and 

 PFG the chord passing through F, the focus ; draw 

 PN the diameter of the circle, join GK, GN, and draw 

 FL perpendicular to the tangent at P. Because the 

 lines PFG, PK make equal aagles with the tangent at 

 P (6. Sect. IV.) the angles PGK, PKG are equal (32. 8: 

 E.) ; hence PG=PK. 



Again, the triangles FLP, PGN being manifest- 

 ly similar, FL : FP : : PG or 4PF : PN ; hence FL : FP 

 :; FP : ± PN or i the radius. 



2 FP 1 

 Cor. 1. Hence the radius is equal to--^r — 



2 FL3 

 Cor. 2. The radius is also equal to ■ j.^ , where AF 



ig the distance of the focus from the vertex of the pa- 



FL 2 

 -rabola; for FP = -^- (14. of Sect. IV.) 



Cor. 3. Hence also the radius is equal to 

 where L denotes the parameter of the axis, for 



^L.FP 3 Curvature. 



FL 3 ' v— - 



2 FP* 



FL 



2 AF.FP 3 



J L.FP3 



~ AF.FP.FL~ FL' ' 



Prop. V. 



The radius of the circle of curvature at the vertex 

 of any diameter of an ellipse or hyperbola is a third 

 proportional to the perpendicular drawn from the cen- 

 tre upon the tangent and half the conjugate diameter ; 

 and the chord which is drawn from the point of con- 

 tact through the focus, is a third proportional to the 

 transverse axis and conjugate diameter. 



Let PK be the chord cut off from the diameter, and p; 

 PFG the chord passing through F, the focus ; draw 

 PN the diameter of the circle, and from the centre O 

 draw OR perpendicular to PK, which will bisect PK 

 in B ; join GN, and draw the conjugate diameter QC</, 

 meeting PG in M, and PN in S, then PS is equal to 

 the perpendicular from the centre C upon the tangent. 

 The triangles PSC, PRO are similar, therefore PS: PC 

 : : PR : PO, but PC : CQ : : CQ : PR (Def. of parame- 

 ter), ; therefore PS : CQ : : CQ : PO. 



Again, the triangles PSM, PGN are similar; 

 therefore PM : PS : : PN : PG, but PS : CQ : : (CQ : 

 PO : :) Q q : PN, therefore, PM : CQ : : Qq : PG, or 

 since PM=AC (Cor. 22. Sect. II. and Cor. 30. Sect. III.) 

 Aa:Qq::Qq:PCr. 



Cor. 1. Hence the radius of curvature is equal to 



CO 1 



-~-, and the chord passing through the focus is equal 



2CQ* 

 to — ^ 



The radius of curvature is also equal to 

 for PS = ^~ -• (Prop. 20. Sect. II. and 



X^L. 



91-. 



AC 

 Cor. 2. 



AC.BC 



Prop. 28. Sect. III.) 



Cor. 3. Draw FL from the focus perpendicular to 

 the tangent, and let L denote the parameter of the 

 transverse axis ; the radius of curvature is also equal to 

 i L FP 3 

 ~^= — . For the triangles PFL, MPS are manifest- 



FL 5 



ly similar, therefore FL : FP : : (PS : PM or AC : :) BC : 



FP CQ 3 FP 3 



CQ; hence CQ=-pj- X BC, and A( J^ C = ^ X 



BC 1 _FP 3 

 AC~ _ FL 3 



This expression for the radius of curvature is the 

 same for all the conic sections. 



SECTION VI. 

 Or the Sections of a Cone. 



Definitions. 



1. If ADB be a circle, and V a point out of its plane, ^ 

 both having a fixed position ; and if a straight line '&' 

 a VA, indefinitely extended both ways, pass through 

 V, and move in the circumference of the circle ; when 

 it has made a complete revolution, it will have gene- 



