390 Mr DAVIES on the Equations of Loci 



From this also we get the value of the differential coefficient, viz. 



d(j> 



sn 



dtr _ sin 



d(j> = >l,hi*~c) sin 2 



where d <r is the corresponding element of the curve. By means of (3.) we 

 can eliminate the differentials, and thus obtain the equation of the curve, as 

 in piano, in finite relations between (p and 4-. 



We may illustrate this by one or two examples. 



1. To find the perpendicular from the pole upon the tangent cf a sphe- 

 rical conic section, referred to the focus and major axis. 



The equation of a conic section referred to the said co-ordinates, is 



cos 2 a cos 2 6 

 n0 = : -Mnr2aq=sm26cos0 ................................. <*> 



Differentiating, we get 



d (f) (cos 2a cos 2e) sin 2e sin 6 d& 

 c^*0 = (sin 2a q= sin 26 cos 0)* ' 



d <p _ sin* (p sin 26 sin 6 

 d6~ cos2a cos2e. 



Again, from (4) we find 



A _ ( cos ^a cos ^ e ) ~*~ s i n ^ a tan (R\ 



sin 2e tan 



From this 



, ., sin* 2f tan 2 sin 2 2a tan* 0n=2 (cos Sa^os 2e) sin 2a tan (cos 2a cos Se) 1 



Sill- C7=l - COS 2 = - - - - - r .-7; - 5-; - - - - - 



sin 2 2e tan 2 9 



(cos 2 2a cos 2 2e) sin 2 + 2 (cos 2a cos 2e) sin 2a sin cos (cos 2a cos 2e) 2 cos* 



sin 2 2e sin 2 



_ (cos 2a cos 2e) {(cos 2a+cos 2e) sin 2 + 2 sin 2a sin cos (cos 2a cos 2e)cos 2 0} 



sin 2 26 sin 2 



_ cos 2a cos 2e cos 2a (sin 2 cos s 0)+cos 2e (sin* 0+cos z 0)zp2 sin 2a sin cos 

 sin 2 2e sin 2 



_ cos 2 a cos 1 e cos 2a cos 20+sin 2a sin 20+cos 26 

 sin 2 26 sin 2 



_ cos 2a cos 26 cos 26 cos 2 (azp0) . 



sin* 26 sin 2 ^ 



