We examine $f(x,y,z)= Ax^2 + Bxy+Cy^2 +Dxz+Eyz+Fz^2 = 0$ when $z = 0$.

So ...

$f(x,y,0) = Ax^2 + Bxy+Cy^2 = 0$ (*).

**Case 1:**$A = 0$.

Then $B = 0$ [$\Delta = B^2-4AC= 0$] and $C \ne 0$, so $y = 0$.

Thus C has only the one ideal point < $1,0,0$ >.

**Case 2:**$A \ne 0$.

Then solving the quadratic (*) for $x$ in terms of $y$ we obtain $x= \frac{-B}{2A}y$. [$\Delta = 0$.].

Thus C has exactly the single ideal point with homogeneous coordinate, <$B,-2A,0$> .