66 
INVESTIGATIONS IN CONNEXION WITH CASEY ? S EQUATION. 
[395 
If however the curves U=0, V=0, TP = 0 have a common intersection, then the 
curve in question has a node at this point, and besides touches each of the three 
curves in n 2 — 1 points; and similarly, if the curves 17 = 0, V = 0, W — 0 have k common 
intersections, then the curve in question has a node at each of these points, and besides 
touches each of the three curves in n? — k points. 
In particular, if U 0, F = 0, W = 0 are conics having two common intersections, 
then the curve is a quartic having a node at each of the common intersections, and 
besides touching each of the given conics in two points; whence, if the coefficients 
f, g, h (that is, their ratios) are so determined that the quartic may have two more 
nodes, then the quartic, having in all four nodes, will break up into a pair of conics, 
each passing through the common intersections, and the pair touching each of the 
given conics in two points; that is, the component conics will each of them touch 
each of the given conics once. Taking the circular points at infinity for the common 
intersections, the conics will be circles, and Ave thus see that Casey’s theorem is in 
effect a determination of the coefficients f, g, h, in such wise that the curve 
\/(fU) + J(gV) + s /(hW) = 0, 
(which when U= 0, 1^=0, IP = 0 are circles, is by what precedes a bicircular quartic) 
shall have two more nodes, and so break up into a pair of circles. 
The question arises, given TJ = 0, V — 0, TP = 0, curves of the same order n, it is 
required to determine the ratios f : g : h in such wise that the curve 
>J(fU) + J(gV) + ^(hW) = o, 
may have two nodes; or we may simply inquire as to the number of the sets of 
values of (/ : g : h), which give a binodal curve, V (/!/") + V (gV) + p {liW) = 0. 
I had heard of Mr Casey’s theorem from Dr Salmon, and communicated it together 
with the foregoing considerations to Prof. Cremona, who, in a letter dated Bologna, 
March 3, 1866, sent me an elegant solution of the question as to the number of the 
binodal curves. This solution is in effect as follows : 
Lemma. Given the curves Z7=0, V = 0, TP = 0 of the same order n; consider the 
point (/, g, h), and corresponding thereto the curve fU + g V + li TP = 0. As long as the 
point (/, g, h) is arbitrary, the curve fU + g V + h W = 0, will not have any node, and 
in order that this curve may have a node, it is necessary that the point (f, g, h) 
shall lie on a certain curve 2; this being so, the node will lie on a curve J, the 
Jacobian of the curves U, V, W\ and the curves J and 2 will correspond to each 
other, point to point; viz. taking for (f g, h) any point whatever on the curve 2, 
the curve fJJ + gV+liW = 0 will be a curve having a node at some one point on the 
curve J; and conversely, in order that the curve fU+gV+ hW = 0 may be a curve 
having a node at a given point on the curve J, it is necessary that the point (/, g, h) 
shall be at some one point of the curve 2. The curve 2 has however nodes and 
cusps; each node of 2 corresponds to two points of J, viz. the point (/, g, h) being 
at a node of 2, the curve fU + gV + hW = 0, is a binodal curve having a node at