400 Mr DAVIES on the Equations of Loci 



and DAK=a. Then we shall have 



/ 2 =:a* ZabcosO a+b' 2 ........................ (1.) 



Now join EB, and denote the angle EBA by <p : and since EAB is a right 

 angle, we shall have 



Eliminating r from (1), (2), we have the polar equation of the cone, re- 

 ferred to origin B, and lines AB, AK ; viz. 



c 2 tan 2 < = a* ZabcosO a + & 2 (3.) 



Also since all parallel sections of the cone whose vertex is B and base 

 are similar, we have the following quantities constant in all the sec- 

 tions ; viz. 



a 2 + 6 2 a b 



and 



c* c 2 



Put these equal to tan 2 a, and tan* u a : then the equation of the cone be- 

 comes 



tan 2 <p = tan 2 a, 2 tan 2 a,, cos B a, or 



n tan 2 a, tan 2 d> . 



cos6 a= -^ j I- = cot 2 a,, (tan 2 a, tan 2 0) (4.) 



,- 13,11 ft f/ 



Now since every cone of the second degree admits of circular sections, it 

 is quite obvious that all the equations of the cone, whatever form they may 

 assume under their respective circumstances of data, may be transformed 

 into (4). Nevertheless, it is often extremely troublesome to effect these 

 transformations, and almost always unnecessary ; and we shall at once seek 

 the most general form that the equation of the cone referred to its vertex can 

 assume. The method is extremely simple, and precisely similar to that 

 which we have already employed. 



We may take as the equation of a line of the second order referred to 

 polar co-ordinates situated in its own plane, 



r 2 (tan 2 a cos *6 + 2 tan a tan /3 tan 7 cos 6 sin 6 + tan 2 (3 sin* 6) 

 -\-%a,r tan a, cos 6 

 + 2 b, r tan /3, sin Q 



(5.) 



