81 



For tangents at H and L will meet in a point T in Aa produced ; 

 the triangles HMT and DNC will be similar, and DN and HM will 

 be similarly divided in E and L, therefore CE and TL will be paral- 

 lel, and consequently CE and CL will be conjugate semidiameters. 



Lemma 3. Take in the elUpsea point I indefinitely near to L, and 

 draw through it mh parallel to MH ; join Ch, and with the centre C 

 and a radius equal to CE describe the arc K/f meeting CH and CA 

 in K and k : then KA; will be ultimately equal to L/. 



For U : Bh :: LT : HT :; CE : CD :: CK ; CH :: Kk : Uh. 

 Thei'efore hi =; Kk. 



Lemma 4. If the semiaxes AC and 

 BC of an ellipse be equal to the sum 

 and difference of the sides PQ, QR, of /^,^rVs~~\ \ ^-^ 



a triangle PQR, and if the angle BCP 

 be equal to half the contained angle 

 PQR (ADa being the semicircle on 



Aa); then, DEN being drawn perpendicular to Aa, CE will be equal 

 to the base PR. Take QU and QS equal to QR ; then PS and PU 

 will be equal to AC and CB, and the angle CDN to TSP ; therefore 

 drawing PT perpendicular to TS, DN and NC will be equal to TS 

 and TP. But URS being a right angle, UR is parallel to PT, and 

 therefore TS : TR :: PS : PU :: AC : BC :: DN : EN ; but TS = DN, 

 therefore TR w EN, and since PT = CN, it follows that PR = CE. 



THEOREM. 



Let AT and AT be an ellipse and hyperbola, the semiaxis (CA or 

 C'AO 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 the circles described on the axes, so that FP == FP'; 



VOL. XVI. If 



