PROFESSOR CAYLEY’S SUPPLEMENTARY MEMOIR ON CAUSTICS. 
11 
paring with the second of the two expressions of U, it appears that we in fact have 
V=27U. 
9. Taking as parameter tan ^ 0, or if we please cos 0+\/— 1 sin 0, the foregoing values 
of (x,y) in terms of 0 give (x,y, 1 ) proportional to rational and integral functions of the 
parameter of the degree 6 ; so that not only the curve is a sextic curve, but it is a uni- 
cursal sextic, or curve of the order 6 with the maximum number, = 10 , of nodes and 
cusps ; that is, if § be the number of nodes and k the number of cusps, we have &+«= 10 . 
Moreover, introducing the same parameter into the equation of the tangent, this equation 
is seen to be of the degree 4 in the parameter ; that is, the class of the curve is = 4 : this 
implies 2d+3*=26, and we have therefore S=4, «= 6 . To verify these numbers, it is 
to be remarked that it appears by the equation of the curve that there is at each of the 
circular points at infinity a triple point in the nature of the point #= 0 , j )=0 on the 
curve y a =x 4 ; such a point is in fact equivalent to two cusps and a node, and we have 
thus the two circular points at infinity counting together as 2 nodes and 4 cusps ; there 
should therefore besides be 2 nodes and 2 cusps, and I proceed to establish the existence 
of these by means of the expressions for ( x , y) in terms of 0 . 
10. To find the cusps, we have 
^0 =— sin 20(3 cos 0— 2m) =0, 
cos 20(3 cos 0— 2m) =0, 
which are each of them satisfied if only 3 cos0 — 2m=0, or cos 0="— ; the corresponding 
values of (x, y) are found to be 
8m 3 i f -i 
X— m gy-, 2 /=±( 1 ' 
4m l \ 2 
'“9" / ’ 
and we have thus two cusps situate symmetrically in regard to the axis of x; the cusps 
are real if m< f, imaginary if m> f ; for m=f, the two cusps unite together at the 
point x=-\ on the axis of x, giving rise to a higher singularity, which will be further 
examined, post, No. 12. 
11. The curve is symmetrical in regard to the axis of x, and hence any intersection 
with the axis of x, not being a point where the curve cuts the axis at right angles, will 
be a node. Hence, in order to find the nodes, writing y= 0, this is 
sin 0(1 — 2 m cos 0+2 cos 2 0 ) = 0 , 
giving sin 0 = 0 , that is, 
or 
0 = 0 , x= 2 — m; 
0=7r, #= — 2— m ; 
but these are each of them ordinary points on the axis of x ; or else giving 
1 — 2 m cos 0 + 2 cos 2 0 = 0 , 
cos 0 =i(m+>/ m 2 — 2 ). 
c 2 
that is, 
