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
 A: 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: $a\sqrt{2}\cdot b\sqrt{2}=2ab$
A: The naive thing is to see the ellipse as a stretched circle.  We know the size of the largest rectangle that fits in a circle of radius $r$ is a square of side ${\sqrt 2}r$.  If the semi-axes are $a$ and $b$ we would expect the rectangle to be ${\sqrt 2}a \times {\sqrt 2}b$.
To prove this, let us put the center at the origin.  The equation of the ellipse is $\left(\frac xa\right)^2+\left(\frac yb\right)^2=1$.  The area of the rectangle is  $4xy,$ where  $(x,y)$ is a point on the ellipse and we choose the point to maximize it.  The area is then $4xb\sqrt{1-\left(\frac xa\right)^2}$.  The derivative of this is $\frac {4b(a^2-2x^2)}{\sqrt{1-\left(\frac xa\right)^2}}$ which is duly zero at $x=\frac a{\sqrt 2}$, giving a side of $\sqrt 2 a$
A: Using the equation $1 = y^2/n + x^2/m$ for the ellipse, write definitions for side lengths as $2(y^2/n)$ and $2(x^2/m)$. Area equals the product of these sides. Using substitution assert that $(1 - x^2/m) = y^2/m$ and in turn the area = $2(y^2/n) \cdot 2(1 - x^2/m)$. Graph this function and find the maximi. These are your coordinate values. Use +/- in front of these values to form a set of coordinates.
