Maximum Parallelogram Area within an Ellipse

Let the coordinates of the four vertices of the parallelogram be $(x_1, y_1)$ , $(x_2, -y_2)$ , $(-x_1, -y_1)$ and $(-x_2, y_2)$ , so that all $x_1$ , $x_2$ , $y_1$ and $y_2$ are positive.

The area of parallelogram is given by:

$\begin{aligned} A & = \frac 12 \left((x_2-x_1)(-y_2+y_1) + (-x_1-x_2)(-y_1-y_2) + (-x_2+x_1)(y_2-y_1) + (x_1+x_2)(y_1+y_2) \right) \\ & = 2(x_1y_2+x_2y_1) \end{aligned}$

Using Cauchy-Schwarz inequality on the following.

$\begin{aligned} \left(\frac {x_1}a \cdot \frac {y_2}b + \frac {x_2}a \cdot \frac {y_1}b\right)^2 & \le \left( \frac {x_1^2}{a^2} + \frac {y_1^2}{b^2} \right) \left( \frac {x_2^2}{a^2} + \frac {y_2^2}{b^2} \right) = 1 \\ \implies x_1y_2+x_2y_1 & \le ab \\ \implies A & \le \boxed{2ab} \end{aligned}$

Equality occurs when $x_1=x_2=\dfrac a{\sqrt 2}$ and $y_1=y_2=\dfrac b{\sqrt 2}$ , a rectangle.