CoNRAN — Some Theorems on the Ttvisted Cubic. 

 13. The osculating plane is 



■X 6+ d, y . 6 + 6 z . - (b 6 - (b 



- cos -— -- + V sm — —^ — sm — —-^ = cos — ~ ; 

 a 2 b 2 c 2 2 



257 



or, 





2 



2 



3« - 36 

 2 



Differentiating twice and solving, we find the point on the twisted cubic 

 itself given by 



oa - 



6 sm — - — 



X 2 



3a -0 



^ y 



a . 3a — 36 h 



sm — 2 — 



o cos 



z 3a- 39 



~ = cot — - — . 



c 2 



14. Some interesting results may be noticed by comparing the 

 coordinates of the points on the two cubics. 



Let xyz and xy'z' be two correspoiiding points on the given cubic, and 

 XYZ ?a\di X'Y'Z' the coordinates of the points of contact of their 

 osculating planes with the hyperboloid 



Also let 



X' y z- 

 a^ b' c' 



= g. 



Eccentric 

 angle. 



X 



a 



h 



z 

 c 



X 



a 



Y 



b 



Z 



q 



^ 3 sin (a 4 5) 

 sin 35 



3 cos (a + 5) 

 sin 35 



cot3S 



cos (a + 5) 

 cos 35 



sin (a + 8) 

 cos 35 



- tan 35 





x' 

 a 



y 



h 



c 



a 



I" 



T 



c 



e + 7r 



- 3 cos {a + S) 

 cos 35 



- 3 sin (a + 5) 

 cos 35 



- tan 35 



sin (a + 5) 

 sin 35 



cos (a + 5) 

 sin 35 



cot 35 



15. Therefore 



X' = - 3X' and 

 y = -3Y' 



z^Z' 



K. I. A. PROC, VOL. XXVII., SECT. A. 



x' = - 3X. 

 z' ^Z. 



[B71 



