The equation of a certain ellipse is $\left(\frac{x}{5}\right)^{2}+\left(\frac{y}{3}\right)^{2}=1$. At the right focus of this ellipse, we draw the positive direction of the ordinate and construct a tangent to the ellipse at the endpoint of this ordinate. What is the area of the triangle formed by this tangent and the two coordinate axes?
Let's determine the coordinates of the point of tangency $P$; its abscissa is the eccentricity:

$$
e=\sqrt{a^{2}-b^{2}}=4
$$

Substituting this value of the abscissa into the equation of the ellipse, we obtain the ordinate of the point of tangency:

$$
\frac{16}{25}+\frac{y^{2}}{9}=1
$$

from which

$$
y= \pm \frac{9}{5}
$$

According to the condition, only the positive value can be taken. The equation of the tangent line passing through point $P$ is:

$$
\frac{4 x}{25}+\frac{9 y}{45}=1
$$

or

$$
4 x+5 y=25
$$

To calculate the area of the triangle in question, we still need to determine the intercepts of the tangent on the axes; if we substitute zero for $x$ in the equation of the tangent, we get $y=5$, and if $y=0$, then $x=6.25$. Therefore, the area of the triangle is

$$
t=6.25 \times 2.5=15.625
$$

(Jakobovits Daniel, forealisk. VIII. o. t., Körmöczbánya.)

The problem was also solved by: Friedmann Bernát, Hönigsfeld Tivadar, Klein Mór, Riesz Frigyes.