15S 



CONIC SECTIONS. 



#,irabol; 



Pe- 



ng 19. 



tangent PB, and FP a straight line drawn to the point 

 of contact ; let A be the vertex of the axis, and there- 

 fore FA equal to one-fourth of the parameter of the 

 axis ,• FB is a mean proportional between FP and FA. 



Produce FB and FA to meet the directrix in D and 

 C, and join AB. The lines FC, FD, are bisected 

 at A andB (Cor. Def. 1. — 3. and 1. Cor. 6.) ; therefore, 

 (2. 6. E.) AB is parallel to CD, or perpendicular to CF, 

 and consequently a tangent to the curve at A (3. Cor. 2.) ; 

 now BP is a tangent at P, therefore the angle AFB is 

 equal to BFP (f.), and since the angles FAB, FBP, 

 are right angles, the triangles FAB, FBP, are equian- 

 gular ; hence FP : FB : : FB : FA. 



Cor. 1. The common intersection of a tangent and a 

 perpendicular from the focus to the tangent, is in a 

 straight line touching the parabola at its vertex. 



Con. 2. If PH be drawn perpendicular to the tan- 

 gent, meeting the axis in H, and HK be drawn perpen- 

 dicular to PP, PK shall be equal to half the parameter 

 of the axis. For the triangles HPK, PFB are mani- 

 festly equiangular ; therefore HP : PK : : PF : FB : : FB 

 : FA : : FD : FC. But if PD be joined, the line PD is 

 evidently perpendicular to the directrix (6.); there- 

 fore the figure HPDF is a parallelogram, and HP=rFD; 

 therefore PK = FC = half the parameter of the axis. 



Prop. XV. Problem. 



A parabola being given by position, to find its direc- 

 trix and focus. 



Let T>Vd be the given parabola ; draw any two pa- 

 rallel chords Dd, Ee, and bisect them at H and K ; join 

 KH, meeting the parabola in P ; the straight line PHK 

 is a diameter (4- Cor. 9.), the point P is its vertex, and 

 Dd, Ee are ordinates to it. In HP produced take PL, 

 equal to one-fourth part of a third proportional to PH 

 and HD, and draw LN perpendicular to PL, the line 

 LN will evidently be the directrix ( Def. 1 0. & Prop. 1 3. ) 

 Draw PM parallel to the ordinates to the diameter PK, 

 then PM will be a tangent to the curve at P (-2. Cor. 9.) 

 Draw LM perpendicular to PM, and take MF = ML, 

 and the point F will be the focus of the parabola, 

 (I. Cor. 6.) 



SECTION V. 



Of the Curvature of the Conic Sections. 



Definitions. 



1. A circle is said to touch a conic section in any 

 point, when the circle and conic section have a com- 

 mon tangent in that point. 



2. If a circle touch a conic section in any point, so 

 that no other circle touching it in the same point can 

 pass between it and the conic section on either side of 

 the point of contact, it is said to have the same curvature 

 with the conic section in the point of contact, and it is 

 called the Circle of' Curvature. 



Lemma. 



Let PK be any chord in a circle, PX a tangent at 

 one of its extremities, and KL a diameter passing 

 through the other extremity ; draw any chord Hh pa- 

 rallel to ' t the tangent, meeting PK in E, and from its 

 extremities draw HG, h g perpendicular to the diame- 

 ter, meeting PK in M and m ; the square of HE is 

 equal to the rectangle contained by PE and KM, and 



the square of /; E is equal to the rectangle contained by Curvature . 



PE and Km. r-J 



From H and h draw the straight lines HP, hV, HK, 

 hK, and let KN a perpendicular to the diameter, and 

 therefore a tangent to the circle ;:t K, meet PX in N. 

 The triangle MHE is evidently similar to the triangle 

 KNP, and KN=NP, therefore MH= HE ; .hence the 

 angles HMK, HEP are equal. Now, PHE is equal to 

 the alternate angle HPX, that is, to the angle HKM 

 in the alternate segment of the circle (32. 3. E.), there- 

 fore the triangles PHE, HKM are similar, and PE • 

 EH : : HM or EH : MK, therefore HE 2 =PE.MK. In 

 the same way it may be demonstrated, that mh—hE, 

 and that the triangles pfcE, hKm are similar, and 

 therefore PE : Eh : : k m or Eh :'m K, and hence AE ; = 

 YE.mK. 



Prop. I. 



If a circle be described touching a conic section, and 

 cutting off from the diameter that passes through the 

 point of contact a segment greater than the parameter 

 of that diameter, a part of the circumference on each 

 side of the point of contact will be wholly without the 

 conic section ; but if it cuts off from the diameter a 

 segment less than the parameter, a part of the circum- 

 ference on each side of the point of contact will be 

 wholly within the conic section. 



Let Pp be the diameter of a conic section ; let a cir- Fles.SO, 81, 

 cle HP/i touch the section in P, the vertex of the dia- 82, 83. 

 meter, and cut off a segment PK either greater or less 

 than the parameter of the diameter: In the former 

 case, a part HP// of the circumference of the circle on 

 each side of P, the point of contact, will be wholly with- 

 out the conic section, as in Fig, 80. and Fig. 8! ; and 

 in the latter, a part HP/? of the circumference on each 

 side of P will be wholly within the section, as in Fig. 82. 

 and Fig. 83. Through K, draw KL a diameter of the 

 circle ; let DEr/, an ordinate to the diameter of the sec- 

 tion, meet the circle in H and h, so that the points H, 

 P, h may be on the same side of KL, the diameter of 

 the circle, and draw HG, kg, PO perpendicular to KL, 

 the two former lines meeting KPin M and m. Frora 

 K, towards P, place KR in the diameter equal to its 

 parameter : then, in the former case, the point II will 

 fall between K and P, as in Fig. 30. and Fig. 81 ; and 

 in the latter, it will fall in KP produced, as in Fig. 82. 

 and Fig. 83. 



Case 1. First, let the section be a parabola (Figs. 80. 

 and 82.); then DE% also </E 2 :=PE.RK (Prop. 13. of 

 Sect. IV.): Now H£ 2 = PE.KM, and //E 2 =PE.K»» 

 (Lemma): Therefore DE 2 : HE 2 : : KR: KM, and JE l : 

 hE 2 : : KR : Km. Now, if the ordinate D d be suppo- 

 sed to approach to the tangent at the vertex, the points 

 H, h will approach to P, the lines HG, h g to the line 

 PO, and the points M, m to the vertex P, where they 

 will at last coincide ; hence it is evident, that the ordi- 

 nate DEd may be at such a distance from the tangent, 

 that the points M, m, and the vertex P, may be all on 

 the same side of the point R. In this position of the or- 

 dinate, if the segment cut off by the circle be greater 

 than the parameter, as in Fig. 80, then KR will be 

 less than either KM or Km, and therefore DE 2 less 

 than HE 2 , also dE* less than' h E 2 ; so that the points 

 H, h are both without the parabola. If the ordinate 

 approach nearer to the tangent, as the points M, m 

 will also approach nearer to P, the line KR will still be 

 less than either KM, or Km, and therefore DE 2 less 

 than. HE 1 , and c/E 2 less than /«E\ Hence, every point 



