Mathematico | Mathematico

A parabola $P$ is drawn in the $x y$ plane. The vertex of the parabola is $(p, q)$ , where $q < 0$ .

Given that the parabola is convex, prove that the parabola intersects the $x$ axis in two distinct points.

WARNING

You might think that this is obvious, but there are plenty of seemingly obvious statements in mathematics which turn out to be false on closer inspection - an algebraic proof is needed!

Hint

Because the parabola is convex, the complete square form of its equation must be

y = a (x - p)^{2} + q

where $a > 0$ and $q < 0$ .

Solution

Because the parabola is convex, the complete square form of its equation must be

y = a (x - p)^{2} + q

where $a > 0$ and $q < 0$ .

We expand and simplify

\begin{aligned} y & = a (x^{2} - 2 p x + p^{2}) + q \\ = a x^{2} - 2 a p x + a p^{2} + q \end{aligned}

The discriminant is

\begin{aligned} 4 a^{2} p^{2} - 4 a (a p^{2} + q) \\ = & 4 a^{2} p^{2} - 4 a^{2} p^{2} - 4 a q \\ = & - 4 a q \end{aligned}

Since $a > 0$ and $q < 0$ , the expression $- 4 a q > 0$ and therefore there must be two distinct roots.