I tried the following:
I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $\Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. Letting $b^2-4ac=0$ makes $\Delta_x(y)$ a linear function of $y$ and if $y \ge \frac{4ah -d^2}{2bd − 4ae}$ then for any of those $y$'s there is only one corresponding $x$. So IF for the $P(x, y)$ there are only four possible shapes: circle, ellipse, parabola and hyperbola then the only choice is parabola.
But the proof is not only complete but also doesn't look so rigorous. Are there any better proofs?