Preface: C is an
parabola if and only if C
has exactly one ideal point.
Proof: Suppose $\Delta = 0$.
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, <$B,-2A,0$> .
In any case, C
has exactly one ideal point and so C is a parabola.