256 



Geometrical Theorems on the 



intersect PT in q : then since FP and Fp are perpendicular to 

 the tangents PT and^tf, it follows (by the preceding lemma) that 

 Pq is ultimately equal to the increment 

 of the difference between the arc AT 

 and the tangent TP. With the centre 

 F and semiaxis FA describe the equi- 

 lateral hyperbola ^GAH, and bisect 

 the angles AFP, AFp by the straight 

 lines FL, Fl. Then (since the square 

 of the radius vector of an equilateral 

 hyperbola is inversely as the cosine of 

 twice the angle which it makes with 

 the axis) the square of LF will be 

 equal to AF x PF\ and because the 

 angle LFl is half the angle PFq, there- 



Fig. 2. 



fore the area LFl will be equal to one-fourth of the rectangle 

 under AF and Pq. 



Hence the hyperbolic sector A FL is equal to one-fourth of 

 the rectangle under A F and the difference between the para- 

 bolic arc AT and the tangent TP. 



Lemma 2. If on either axis of an ellipse a semicircle be 

 described, of which CD and CH 

 are two radii at right angles to 

 each other, and if DN and HM 

 be drawn perpendicular to the 

 axis aA, and meeting the ellipse 

 in E and L ; then CE and CL 

 will be conjugate semidiameters. 



For tangents at H and L will meet in a point T in Aa pro- 

 duced ; 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 parallel, and consequently CE and CL will be 

 conjugate semidiameters. 



Lemma 3. Take in the ellipse a point / indefinitely near to 

 L, and draw through it mh parallel to MS\ join Ch, and with 

 the centre C and a radius equal to CE describe the arc Kk 



Fig. 3. 



