256 Proceedings of the Royal Irish Academy. 



This meets the conic in 



[t, i) and (i. t^. 



Hence (fig. 2) \i it, - j be the coordinates of P, and i - t, - -j those of Q, 



(-, f'\ will be the coordinates of R. 

 \f I 



Let Q and be the eccentric angles of P and R : then since 



.-. t = e*^^-'^), 

 and - = e*('^-"); 



.-. (p + 26 = 2mr + 3a. 



12. The discussion being still confined to the case of the cubical hyperbola, 

 we take the principal axes of the locus of centres and the chord joining the 

 (imaginary) points toi, «<»2, as axes of coordinates. 



The equation of the hyperboloid referred to in paragraph 9 may be 

 written 



a- 0' c 

 The point of contact of an osculating plane is given by 

 X - a cos y - ^ sin B z 



and 



where 





Solving we get 







X 





a 



a sin d - h cos 9 c' 



X - a cos (j) 1/ - h sin z 



- a sin (p b cos c 



29 + = 2mr + 3a. 



9+6 f a- 9 



cos ---- cos O 4 - Q - 



r^ " 3 (o - 9) 



cos — -- cos ' 



. e + (b . ( a- 9 



sm ^ sm a + 



2/_ 2 ^ V 2 



b~ 9 - <h~ S(a- 9) ' 



COS — ~- cos — ^ 



s ^ 0-0 ^ 3 (o - 0) 

 - = tan — ^ = - tan -_, • 



c 2 2 



Hence the point of contact lies on a twisted cubic. 



