FORM OF TANGENTIAL EQUATION. 
429 
127. The reciprocal of the parallel curve is found at once from the tangential equa- 
tion. For it is evident that the polar equation of the reciprocal of the parallel to the 
curve 
*=m 
is 
^=/(p)sin<p+r, (262) 
where r is the distance between the curve and its parallel, and k is the radius of the 
circle of reciprocation. 
Examples. 
(1) To find the reciprocal of the parallel to a parabola. 
We have 
f{<p) =a tan <p, 
A: 2 a sin 2 <p . 
q COS <p — 
is the required equation. This, expressed in Cartesian coordinates, is 
(k 2 x—ay 2 ) 2 =r 2 x\x i -\-y 2 ) (263) 
This curve has three double points, namely, the origin and the points where the conic 
k 2 x—ay 2 meets the line at infinity. 
Again, the curve is evidently the envelope of the conic 
x 2 +y 2 +2ih(k 2 x— (264) 
The discriminant of this conic is 
(1 — 2 a{i,)[i?k\ 
This shows that there are two values of p>, for which the conic breaks up into a pair 
of lines; hence the curve has four double tangents. Therefore the characteristics of 
the curve are 
S=3, t=4, 
v=6, <=6, z=0. 
(2) Find the reciprocal of the parallel to the curve 
% m = a m sin mQ. 
From equation (37), art. 24 we have at once the polar equation of the reciprocal 
A 2 
S' 
n+ 1 j 
±r. 
(3) Find the intrinsic equation of the lemniscate. 
The tangential equation is (see art. 25) 
v — a ( sin 2 cosec < p, 
( 265 ) 
