Then $A*C = 0$. So $B \ne 0$. [$\Delta = B^2-4AC \gt 0$.]
The equation becomes $f(x,y,0) = Bxy+Cy^2 = y(Bx+Cy) = 0$
giving two distinct ideal points: <$ 1,0,0$> and <$ -C, B,0$>.