# Show that $P(x,y)=0$ is a parabola if $b^2-4ac=0$.

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?

• 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 $AC=0$. – mathlove Nov 27 '18 at 5:26
• @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. – 72D Nov 27 '18 at 5:50