Prove that the area of a square inscribed in the ellipse $b^2x^2+a^2y^2=a^2b^2$ is equal to $\frac{4a^2b^2}{a^2+b^2}$ I have this question for a math assignment

Prove that the area of a square inscribed in the ellipse $b^2x^2+a^2y^2=a^2b^2$ is equal to $\frac{4a^2b^2}{a^2+b^2}$

I tried approaching this question the same I would a trig identity, but it doesn't seem to work and the question is different since it's not trig, but conics. 
Does anyone know how to approach this question? I'm having a hard time and would really appreciate it. Thanks!
 A: Symmetry implies that the centre of the square is at the origin, and the sides are parallel to the axes. Such a square's vertex in the first quadrant will lie on the line $y=x$, and the ellipse, of course. Hence $ b^2 x^2 + a^2 y^2 = a^2b^2 $, and inserting $y=x$ and dividing,
$$ x^2 = \frac{a^2b^2}{a^2+b^2}. $$
That's a quarter of the area, so the total is $4$ of this.
A: Let's call $(p,q)$ the vertex of the square then:
$$b^2p^2+a^2q^2=a^2b^2$$
The only way is to put the square with the sides parallel to the axis of the ellipse. Then the other vertex will be $(p,-q),(-p,q),(-p,-q)$.
So the sides will be $2|p|=2|q|$ and the area will be $4p^2=4q^2$. Then
$$b^2p^2+a^2p^2=a^2b^2 \rightarrow p^2=\frac{a^2b^2}{a^2+b^2}\rightarrow 4p^2=\frac{4a^2b^2}{a^2+b^2}$$
A: The vertices of the square has to lie on the ellipose. Furthermore, the vertices has to satisfies, $x=y$ or $x=-y$ as it is a square.
That is $x^2=y^2$.
$$b^2x^2+a^2x^2=a^2b^2$$
$$x^2=\frac{a^2b^2}{a^2+b^2}$$
Now, you just have to find the length of the edge of the square and square them.
