The length of the major axis of an ellipse is $12 \mathrm{~cm}$, and the minor axis is $10 \mathrm{~cm}$. If we use the major axis as the x-axis and one of its vertices as the origin, we need to find the abscissa corresponding to an ordinate of $4 \mathrm{~cm}$. Furthermore, determine the equation of the tangent line to the ellipse at the endpoint of the positive ordinate, its angle of inclination, and the area of the triangle formed by the ordinate axis, the abscissa axis, and the tangent line.
We choose one of the vertices of the ellipse as the origin of the coordinate system, so the equation of the ellipse is:

$$
y^{2}=2 p x-\frac{p}{a} x^{2}
$$

Given that

$$
p=\frac{b^{2}}{a}=\frac{25}{6}, \quad a=6, \quad y=4
$$

it follows that

$$
x^{2}-12 x=-\frac{576}{25}
$$

from which

$$
x_{1}=2 \frac{2}{5}, \quad x_{1}^{\prime}=9 \frac{3}{5}
$$

The equations of the tangents at these points are:

$$
y=\frac{5}{8} x+2 \frac{1}{2}, \quad y=-\frac{5}{8} x+10
$$

The slopes of the tangents are determined by the following equations:

$$
\operatorname{tg} \alpha=\frac{5}{8}, \quad \operatorname{tg} \alpha^{\prime}=-\frac{5}{8}
$$

from which

$$
\alpha=32^{\circ} 19^{\prime} \quad \text { and } \quad \alpha^{\prime}=147^{\circ} 41^{\prime}
$$

The areas of the corresponding triangles are:

$$
t=5 \mathrm{~cm}^{2}, \quad t^{\prime}=80 \mathrm{~cm}^{2}
$$

(Grofffits Gábor, Pozsony.)

The problem was also solved by: Prohászka J., Schiffer H., Szabó I.