ME. A. CAYLEY ON THE DOUBLE TANOENTS OF A PLANE CUEVE. 107 
y, z) as cm-rent coordinates, is consequently that of a curve, intersecting the given 
curve (^now represented by in the points of contact of the double tangents. Ihe 
process leads to a determinate form nU = 0, of the curve in question, but of course 
any curve whatever, nU+M.U=0, will intersect the curve U=0 in the points of con- 
tact of the double tangents. 
6. I write for the moment 
a=(A,...xx, Y, z)»-^=:0, 
V=|X-f-p?Y4-^:Z -0, 
for the two equations ; the coefficients (A, . . .), as already mentioned, are of the degree 
2(n—2) in {x, y, z) and of the degree 3 in the coefficients of U ; or as we may express it, 
A, y, 
In like manner t], I are of the degree 1) in (^, y, z), and the degree 1 in the coeffi- 
cients of U, or we may write 
I, y!,l^{a,..y{x,y, zf-\ 
7. The equation which expresses that the luie V=0 touches the curve fl=0, is 
FCl = 0, where the facients of the Reciprocant PH are the coefficients (|, pj, 1) of the 
linear function. This equation is of the form 
or attending to the forms of (A, . ..) and (|, >?, 1), it is of the form 
(a . .)6(n-3) + («-2Krt-3)^_^^ ^j4{»-2)(«-3) + (n-l)(n-2){«-3}_Q^ 
or what is the same thing, the form 
viz. the curve through the points of contact of the double tangents is a curve of the 
order fn—2)(n^—9), and its equation contains the coefficients of the equation U=0 of 
the given curve in the degree {n-\-^)(n — 3). And since each double tangent corresponds 
to two points of contact, the number of double tangents is ^n{n—2){n‘--9). This agrees 
with the before-mentioned results. 
8. The whole problem is thus reduced to the demonstration of Mr. Salmon s expres- 
sion for the cm-ve 0 = 0. To fix the ideas, consider the case of a quartic curve 
T=(*XX, Y, Z)^=0, and let the function U=(*X ^5 shortness we may 
wiite it, y, zj-) and certain of its emanants be represented as follows, viz.— - 
a —{x,y,zy 
h =(^,2/,^KX,Y,Z), 
c =(a:,y,z/(X, Y,Z)^ 
d =(x,y,z)(X,Y,Zy, 
e = . . (X,Y,Z)fi 
