374 
PE0FE3S0E J. CASEY ON A NEW 
22. The curve A m is of the degree 2 m(n—m). 
Demonstration. — If O' be any point on the line CN, then since the (n—m) th polar of 
O' cuts CL in [n — m) points, the lines drawn from O' to these points will make up 
{n—m) tangents, and the line CL itself counts for m tangents. Hence the n tangents 
which can he drawn from O' are accounted for. Now if the point O' itself be on the 
curve A m , only (n— 1) tangents can be drawn from it, and two of the points in which the 
line CL is intersected by the polar curve of the (n — m)th degree must coincide, that is 
the polar curve must touch CL. Hence we have to find the points on CN whose polar 
curves of the (n— m)th degree, with respect to U, will touch CL. In order to find the 
number of solutions of this problem, we will use trilinear coordinates. Let ( a , b , c) 
(a', b', d) be the coordinates of two fixed points on CN, then the coordinates of any 
variable point on it are a-\-ka', b-\-kb ] , c-\-kd, and the polar curve of this point of the 
{n — m) th degree, with respect to U, is 
{(«+*«') S+( i + W ')|+( c + fc, )s}’” U = 0 ' (22) 
Now this equation contains the variables in the degree n—m, and its coefficients 
contain k in the mth degree. Hence the condition that it will touch any given line will 
contain k in the degree 2m(n— m— 1); and this is the number of points in which the 
curve A m intersects the line CN, but it touches CN in m points ; .*. the total number 
of points in which the curve meets CN is 2 m{n — m). 
Hence the proposition is proved. 
23. The following are the singularities for the curve A m : — 
v—n, (j.=2m{n—m), i= 0. 
2 r=n 2 —n— 2mn -j- 2m 2 , 
l = 2m 2 (n—m) 2 — 10m(n — to) + 4%, 
x, — <omn — 6to 2 — 3 n. 
Cor. yj-\-2r=n 2 —n, and is therefore the same for the curves A„ A 2 , &c. ; that is, it is 
independent of to. 
Cor. 2. The curves A OT , A n _ m have the same singularities. 
Examples. 
(1) Find the tangential equation of the cuspidal cubic ay 2 =x 3 . 
Eliminating y between this and the equation y=(v — x)t, we get 
x 3 — at 2 x 2 +2avt 2 x— av 2 t 2 =Q 
The discriminant of this is 
which is the required tangential equation. 
(24) 
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