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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?

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  • $\begingroup$ Strictly speaking, $P(x,y)=0$ is not necessarily a parabola when $b^2-4ac=0$. Take $a=b=c=d=0,e=h=1$. But as I've written in the answer to your another question, we can say that by a suitable rotation, $P(x,y)=0$ becomes an equation of the from $AX^2+CY^2+DX+EY+F=0$ with $AC=0$. $\endgroup$ – mathlove Nov 27 '18 at 5:26
  • $\begingroup$ @mathlove, evaluation of P(x,y)=0 by rotation method is much easier, but the book didn't introduced it (yet?). But after learning about it in your answer I could evaluated different cases too. $\endgroup$ – 72D Nov 27 '18 at 5:50

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