Max area of rectangle in an ellipse

Question

I have an ellipse of shape

$\dfrac{(x-y)^2}{2m^2} + \dfrac{(x+y)^2}{2n^2} = 1$ Given that $0<mn \leq 1$.

Find the area of the maximum rectangle that could fit inside the ellipse.

Answer 1

Making a change of variables (rotation) as

$$ \left\{x\to \frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}},y\to \frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}\right\} $$

we get at

$$ \frac{X^2}{n^2}+\frac{Y^2}{m^2} = 1 $$

Now forming the Lagrangian

$$ L(X,Y,\lambda) = 4XY + \lambda\left(\frac{X^2}{n^2}+\frac{Y^2}{m^2} - 1\right) $$

the stationary points are given by

$$ \frac{ \lambda X}{n^2}+2 Y = 0\\ \frac{ \lambda Y}{m^2}+2 X = 0\\ \frac{Y^2}{m^2}+\frac{X^2}{n^2}-1 = 0 $$

and solving we get at

$$ S = 4 X_0Y_0 = 2mn $$