A $9 x^{2}+25 y^{2}=225$ ellipse, we draw tangents at the endpoints of its parameters. Determine the area of the quadrilateral formed by the four tangents.

We draw tangents at the endpoints of the parameters of the ellipse $9 x^{2}+25 y^{2}=225$. Determine the area of the quadrilateral formed by the four tangents.
If the foci of the ellipse are $F_{1}$ and $F_{2}$, and the perpendiculars to the major axis at these points intersect the ellipse at $C_{1}, D_{1}$ and $C_{2}, D_{2}$ respectively, then $C_{1} D_{1}=C_{2} D_{2}=2 p$ is called the parameter. The endpoints of the parameters are thus $C_{1}, C_{2}, D_{1}, D_{2}$. Suppose the tangents drawn at $C_{1}, C_{2}, D_{1}, D_{2}$ intersect each other at points $P, Q, R, S$, then, since the ellipse has a central equation; the points and tangents of the ellipse are symmetric with respect to the axes, i.e., $P$ and $R$ lie on the ordinate axis, $Q$ and $S$ on the abscissa axis, and if $O$ is the origin of the coordinates:

$$
P O Q \Delta \cong Q O R \Delta \cong R O S \Delta \cong S O P \Delta
$$

so:

$$
P Q R S=4 \cdot P O Q=4 \cdot \frac{P O \cdot Q O}{2}=2 \cdot P O \cdot Q O
$$

Therefore, we need to establish the equation of the tangent at point $C_{1}$. The abscissa of $C_{1}$ is:

$$
\xi=c=\sqrt{a^{2}-b^{2}}=\text { eccentricity }
$$

and its ordinate, if the equation of the ellipse is

$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$

is derived from the equation

$$
\frac{a^{2}-b^{2}}{a^{2}}+\frac{\eta^{2}}{b^{2}}=1
$$

We note that both $\xi$ and $\eta$ are taken as positive, because $C_{1}$ is in the first quadrant, so:

$$
\eta=\frac{b^{2}}{a}
$$

Since the equation of the tangent at $C_{1}$ is:

$$
\frac{x \xi}{a^{2}}+\frac{y \eta}{b^{2}}=1
$$

and substituting the values of $\xi$ and $\eta$:

$$
\frac{x \sqrt{a^{2}-b^{2}}}{a^{2}}+\frac{y}{a}=1
$$

or

$$
\frac{x}{\frac{a^{2}}{\sqrt{a^{2}-b^{2}}}}+\frac{y}{a}=1
$$

and thus

$$
\begin{gathered}
P O=\frac{a^{2}}{\sqrt{a^{2}-b^{2}}} \\
Q O=a
\end{gathered}
$$

so

$$
P Q R S=\frac{2 a^{3}}{\sqrt{a^{2}-b^{2}}}
$$

In our case

$$
a=5, \quad b=3
$$

so

$$
P Q R S=62.5
$$

square units.

(Zoltán Harsányi, Eger.)

The problem was also solved by: Bartók I., Braun J., Dömény E., Dömény I., Enyedi B, Haar A., Kiss J., Kürti I., Liebner A., Pivnyik I., Pichler S., Riesz K., Rássy P., Rosenberg J., Riesz M., Szombathy J., Szücs A.