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 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.