372 
PKOFESSOK J. CASEY ON A NEW 
The curve A, intersects the line CN in the points where the pole-conic of CL meets it , 
and it touches CN in the point whose polar conic passes through C. 
14. The polar line of the point C with respect to U will cut CN at its point of contact 
with A,. The same polar line will be a tangent to the pole-conic of CL, and will be 
the polar of the point C with respect to the polar conic of C. Hence it will with C 
divide harmonically the segment of CL included between the points of contact with A x . 
15. We can get the equation of the line of which A l is the envelope as follows : — Since 
y=(v—x)t we have vt—(y-\~xt); and substituting in equation (13) we get 
a 3 xt 3 — (2a 2 x— a 3 y— b 3 )tf -{-(ape— 2a 2 y— 2b 2 )t-\-a i y-{-b, .... (15) 
which is the required line, and the discriminant with respect to t will be Cartesian 
equation of A x . This discriminant is 
2 7 a\x\ ay+bf-p 4 a 3 x( a i x—2a 2 y—2 b 2 ) 3 
— Hpip-\-b^{2a 2 x—a 3 y—bf 
— ( 2a 2 x — a 3 y — bf(a l x — 2a yy —2bf 
+ 1 %a 3 x(2a 2 x—a 3 y—b 3 )(a l x—2a 2 y—2b 2 )(a l y + b x ) = 0. 
. . . (16) 
This equation is of the fourth degree, as it ought, since the curve has a double tangent. 
16. If we denote the equation (15) by T, and since a cusp is a point at which three 
consecutive tangents intersect, the conditions that there shall be a cusp are that 
1 u ’ 
dfr 
dr 2 '' 
: 0 ; 
and eliminating x and y from these equations,- we get the following determinant : — 
a 3 t—2a 2 , a 3 , b 3 
a l —2a 2 t, a 3 t—2a. 2 , b 3 t—2b 2 , 
ad , a l —2a 2 t, b l —2b 2 t, 
This determinant is a cubic in t, showing that there are three cusps. The values of 
t, got from this equation, if substituted in equation (15), will give us the three cuspidal 
tangents. 
17. If we denote the singularities by the following notation — 
Class v, Degree p. 
Double tangents r, Double points b, 
Cusps x, Points of inflection i, 
we have the singularities of the curve A, as follows : — 
^ = 4, r— 1, | 
<= o , a=o, J 
v=6, 
x=3, 
( 18 ) 
