How to find the largest rectangle inside an ellipse I have an ellipse that is defined by center, width and height. The axes of the ellipse parallel to the x and y. I want to find the largest rectangle that completely fits inside this ellipse. Is there an easy way to do this?
And sorry if my terminology is a bit off... it's been a long time since I last tackled with geometry

Top right corner on the ellipse (centered on origin): $x = a \cos \theta$ and $y=b\sin\theta$
where $0\le \theta \le \frac{\pi}{2}, a=\frac{width}{2}$ and $b = \frac{height}{2}$
So area:$$A(\theta) = 4xy=4ab\sin\theta\cos\theta = 2ab\sin2\theta$$
Maximum of sin is at $2\theta=\frac{\pi}{2}$. So, we have $\theta = \frac{\pi}{4}$.
Therefore the 4 corners are: 
$$(a\cos \frac{\pi}{4},b\sin \frac{\pi}{4}),
(a\cos \frac{3\pi}{4},b\sin \frac{3\pi}{4}),
(a\cos \frac{5\pi}{4},b\sin \frac{5\pi}{4}),
(a\cos \frac{7\pi}{4},b\sin \frac{7\pi}{4}),$$
which are of course equal to:
$$(\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}),
(-\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}),
(-\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}),
(\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}),$$
And it's area is: $a\sqrt{2}\cdot b\sqrt{2}=2ab$
