Mathematico | Mathematico

The two parabolas shown have equations

y = x^{2} + 2 a x + p

and

y = - x^{2} + 2 b x + q

The vertex of each parabola lies upon the other.

Intersecting parabolas

Given that the area of the triangle $A B C$ is $1000,$ find the value of $a + b .$

Hint

Complete the square to find the coordinates of $A$ and $B .$ You don't actually need the coordinates of $C,$ you can find the base and height of the triangle just by using $A$ and $B$ and symmetry.

Hint

The vertex $B$ lies on $y = x^{2} + 2 a x + p,$ so the coordinates of $B$ must satisfy this equation. This helps you get rid of $p$ and $q$ from the problem.

Hint

Don't expand out,

(a + b) (a^{2} + b^{2} + 2 a b)

but notice that

a^{2} + 2 a b + b^{2} = (a + b)^{2}

Solution

Let's complete the square to find an expression for $A$

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

so we have $A (- a, p - a^{2})$ .

Let's do simimlar for $B$

\begin{aligned} y & = - [x^{2} - 2 b x] + q \\ = - [{(x - b)}^{2} - b^{2}] + q \\ = b^{2} + q - {(x - b)}^{2} \end{aligned}

so we have $B (b, q + b^{2})$ .

Considering the symmetry of the triangle, its base can be given by

2 (b - (- a)) = 2 (a + b)

It's height is given by

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

so the area is

(a + b) (q - p + a^{2} + b^{2}) = 1000 (i)

The vertex $B$ lies on $y = x^{2} + 2 a x + p,$ so the coordinates of $B$ must satisfy this equation

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

We substitute this into the area of the triangle to get

\begin{aligned} (a + b) (q - p + a^{2} + b^{2}) & = 1000 \\ (a + b) (2 a b + a^{2} + b^{2}) & = 1000 \\ (a + b) (a + b)^{2} & = 1000 \\ (a + b)^{3} & = 1000 \\ a + b & = 10 \end{aligned}