Rectification of the Conic Sections. 



257 



meeting OH and Ch in K and k : then Kk will be ultimately 

 equal to LI 



For LI: Hh : : LT: HT:: CE : CD : : CK : CH : : Kk : Eh. 

 Therefore LI = Kk. 



Lemma 4. If the semiaxes A C and BC of an ellipse be equal 

 to the sum and difference of the sides PQ, QR, of a triangle PQJR, 

 and if the angle BCD be 



equal to half the contained / ~ \ s 



angle PQR (ADa being the 

 semicircle on Aa) ; then, 

 DEN being drawn perpen- 

 dicular to Aa, CE will be 



Fig. 4. 



equal to the base PR. Take Q^and QS equal to QR-, then 

 PS and PU will be equal to AC and (7J5, and the angle CDN 

 to T&P ; therefore drawing PT perpendicular to T8, DN and 

 NC will be equal to TS and TJP. But Z72&S being a right angle, 

 7.ft is parallel to PT, and therefore TS : TR : : PS : PU : : AC 

 :BC \iDN\EN\ but TS = DN, therefore TR = EN', and 

 since PT = CN, it foUows that PR = CE. 



THEOREM. 



Let ^Tand AT' be an ellipse and hyperbola, the semiaxis 

 (CA or C'A) of either being equal 

 to (C'F' or CF) the distance between 

 the focus and centre of the other; and 

 let tangents at the points T and T f 

 meet in P and P f the circles described 

 on the axes, so that FP = FP\ let Fig. 5. 



also a" B" A" be another ellipse whose semiaxes (A" C" and 

 B" C") are equal to ^and .E4, 

 and take in its circumference a 

 point L so that the semidiameter 

 conjugate to that passing through 

 L may be equal to FP or F'P' ; 

 then will the excess of the ellip- 

 tic arc A T above its tangent TP be greater than the excess of 



