Proof for Ellipse Case

Preface: C is an ellipse if and only if C has no ideal points.

Proof: Suppose $\Delta$ < $ 0$.
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 $

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

$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 $\Delta = B^2 -4AC$ < $0$.

Thus C has no ideal points and C is an ellipse.