To look for ideal points we examine $f(x,y,z)= Ax^2 + Bxy+Cy^2 +Dxz+Eyz+Fz^2 = 0$ when $z = 0$.

So we wish to solve ...

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

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

This is not possible without contradicting $\Delta $<$0$. [$\Delta =B^2-4AC$]

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

Then both $C \ne 0 $**and**$A \ne 0$.

Now we can solve the quadratic equation (*)

either for $x$ in terms of $y$:

$x = \frac{-By \pm \sqrt{B^2y^2 -4 ACy^2}}{2A}
=\frac{-B \pm \sqrt{B^2 -4 AC}}{2A}y $

$y = \frac{-Bx \pm \sqrt{B^2x^2 -4
ACx^2}}{2C} =\frac{-B \pm \sqrt{B^2 -4 AC}}{2C}x $.

However neither approach will yield non-zero real
number solutions because $\Delta = B^2 -4AC$ <
$0$.

or for $y$ in terms of $x$:

Thus