Task B-4.1. Calculate the area of a square whose two vertices are at the foci, and the other two at the vertices of the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{b^{2}}=1$. The foci of the ellipse are on the x-axis.
Solution. Sketch

![](https://cdn.mathpix.com/cropped/2024_05_30_7c0a9615fb5cd46ea1c5g-18.jpg?height=602&width=723&top_left_y=905&top_left_x=663)

Since it is a square, it must hold that $e=b$,

or $a^{2}=2 b^{2}, b^{2}=4, b=2$.

The area is $P=2 b^{2}=8$ square units.