A $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ ellipse, draw a rectangle whose area $t$ is given. Determine the coordinates of the vertices of the rectangle. Furthermore, determine the coordinates of the vertices of the largest area rectangle that can be inscribed in the ellipse.
Let the coordinates of one vertex of the rectangle be $\xi$ and $\eta$, then the coordinates of the other three vertices are $(\xi, -\eta), (-\xi, \eta), (-\xi, -\eta)$,

$$
t=4 \xi \eta
$$

Since $(\xi, \eta)$ is a point on the ellipse, therefore

$$
b^{2} \xi^{2} + a^{2} \eta^{2} = a^{2} b^{2}
$$

From (1) and (2),

$$
\xi = \pm \sqrt{\frac{a^{2}}{2} \pm \frac{a}{4 b} \sqrt{4 a^{2} b^{2} - t^{2}}}
$$

and

$$
\eta = \pm \sqrt{\frac{b^{2}}{2} \pm \frac{b}{4 a} \sqrt{4 a^{2} b^{2} - t^{2}}}
$$

(1) can also be written as

$$
t = 4 \xi \frac{b}{a} \sqrt{a^{2} - \xi^{2}}
$$

or

$$
t^{2} = 16 \frac{b^{2}}{a^{2}} \xi^{2} \left(a^{2} - \xi^{2}\right)
$$

Since the sum of the factors is constant, $t^{2}$, and thus $t$, will be maximum when

$$
\xi^{2} = a^{2} - \xi^{2}
$$

from which

$$
\xi = \pm \frac{a}{2} \sqrt{2} \quad \text{and} \quad \eta = \pm \frac{b}{2} \sqrt{2}
$$

(Imre Csada, Pápa.)

The problem was also solved by: Dömény I., Fodor H., Fuchs I., Haar A., Kiss J., Rassy P., Rosenberg J., Schwarz Gy., Székely J.