Now we can solve the quadratic equation (*) for $x$ in terms
of $y$ :
$x = \frac{-B\ y \pm \sqrt{B^2 y^2 -4 AC y^2}}{2A} = \frac{-B
\pm \sqrt{B^2 -4 AC}}{2A}y $
giving ideal points : < $ -B \pm \sqrt{B^2 -4
AC},2A,0$ >
So C has exactly two
distinct [but not very pretty] ideal points .