430 
PROFESSOR J. CASEY ON A NEW 
and, by the method of art. 30, we find the intrinsic equation to be 
■tf 
\/ sin '{ 
If we put/y/ sin— = co s^ this equation becomes 
_ a | oS 
S= ~\/2 J vT— 4 sin 2 3’ 
( 266 ) 
*=^ F (v*- 4 ) < 267 ) 
(4) Find the reciprocal of the parallel to an ellipse. 
Here 
f(<p) =^/a 2 -{-b 2 cot 2 <p, 
A 2 / , . , — * p- 
= sin <p-\-b cos" <p + r 
is the required equation. 
This curve, in Cartesian coordinates, is 
4 try = {a 2 x l + by r 2 % 2 ) 2 (268) 
This curve can by linear transformation be changed into a bicircular quartic. For 
writing the equation in full by putting x 2 -\-y 2 in place of § 2 , and then changing y into 
/ a 2 — r 2 \ i 
y, we get 
A 4 r 2 f r 2 — 2 (ft 2 — r 2 ) (6 2 — r 2 ' 
(x 2 +y 2 y+a*-r*i (a 2 — r 2 ) (6 2 — r 
}(*•+/ 
_| ** ) 7.4 
"T(a 2 -r 2 ^^ 
(269) 
■^3^(5V+aY 2 )| = 0 
128. To find the reciprocal of a bicircular quartic, with respect to one of its circles 
of inversion. 
The following method of generating these curves is given in my memoir on “ Bicir- 
cular Quartics” (see Transactions of the Royal Irish Academy, vol. xxiv. p. 460) : — 
Let + be a conic F, called the focal conic, {x — f) 2J c{y— g) 1 — r 2 =0 a circle J ; 
then if from the centre O of the circle we let fall a perpendicular OT on any tangent to 
F, and take two points, P, P', in opposite directions from T on OT, such that 
OT 2 — TP 2 = OT 2 — TP' 2 = r 2 , 
the locus of the points P, P' is a bicircular quartic. Now, denoting OT by p, and OP 
by g, we get from this construction 
2p%=r 2 -\-f, 
or 
2\\/a 2 cos 2 ct+b 2 sin 2 ct—( fcosa+g sma)\g=?' 2 -\-g 2 . . . . (270) 
