Rectification of the Conic Sections. 



257 



meeting CH 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 PQR, 

 and if the angle BCD be 

 equal to half the contained 

 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 U and QS equal to QJR ; then 

 PS and PZ7 will be equal to AC and CB, and the angle CDN 

 to TSP ; therefore drawing PT perpendicular to TS, DN and 

 NO will be equal to T8 and TP. But URS being a right angle, 

 Z7# is parallel to PT, and therefore TS : TR : : PS : PU : : AC 

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

 since PT - CN, it follows that P = CE. 



THEOREM. 



Let AT and 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' 



meet in P and P f the circles described * / c F A 

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



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

 B" C") are equal to oFand FA, 

 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- 



A' 



& fj 



Fig. 6. 



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



s 



