Let's construct an ellipse - with a given semi-major axis $a$ - which touches two given parallel lines in such a way that the two focal radii from the foci to any point of tangency form angles in the ratio $1:3$. (It is sufficient to construct the axes of the ellipse and the points of tangency.)





翻译完成，如需进一步帮助，请告知。
Let's consider the task as solved. The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_88bd1d2de4cbdcc6cb2fg-1.jpg?height=686&width=816&top_left_y=248&top_left_x=644)

The major axis $AB$ as a diameter, with the main circle drawn over it, contains the feet of the perpendiculars from the foci $e_{1}$ and $e_{2}$ to the tangents $T_{1}, T_{1}^{\prime}$ and $T_{2}, T_{2}^{\prime}$, respectively.

Due to central symmetry, the quadrilateral $F_{1} E_{2} F_{2} E_{1}$ is a parallelogram. The reflection of $F_{2}$ over $e_{2}$ is denoted by $F_{2}^{\prime}$. Thus, $F_{2}^{\prime} T_{2} \| F_{1} T_{1}$, and due to central symmetry and reflection, $F_{1} T_{1} = F_{2} T_{2} = F_{2}^{\prime} T_{2}$, so $F_{1} F_{2}^{\prime} \parallel T_{1} T_{2}$, meaning that $T_{1} T_{2}$, and similarly $T_{1}^{\prime} T_{2}^{\prime}$, are the midlines of the parallelogram $F_{1} E_{2} F_{2} E_{1}$.

Since, according to the problem, $\angle F_{1} E_{2} F_{2} = 180^{\circ} - 4 \alpha$, and as is known, the tangent bisects the angle between the radius vectors, therefore $\angle F_{2}^{\prime} E_{2} T_{2} = \angle F_{1} E_{2} T_{2}^{\prime} = \angle T_{2} E_{2} F_{2} = 2 \alpha$, and as corresponding angles, $\angle T_{2} O X = 2 \alpha$. But as corresponding angles,

$$
\angle T_{2} O B = \angle E_{2} F_{1} B = \alpha
$$

and thus

$$
\angle B O X = 2 \alpha - \alpha = \alpha,
$$

which means that $OB$ bisects the angle $\angle T_{2} O X$.

Accordingly, the construction procedure is as follows: A circle of radius $a$ drawn around a point $O$ on the midline of the given parallel tangents $e_{1}$ and $e_{2}$ intersects the lines $e_{1}$ and $e_{2}$ at the points $T_{1}, T_{1}^{\prime}$ and $T_{2}, T_{2}^{\prime}$, respectively, and the midline at the point $X$, from which the acute angle $\angle T_{2} O X$ is observed. The bisector of this acute angle is the line of the major axis. The construction of points $A, B, F_{1}, F_{2}$, $E_{1}, E_{2}, C, D$ is then straightforward.

The solvability condition is that $2a > d$, where $d$ is the distance between $e_{1}$ and $e_{2}$.

If $T_{1}^{\prime} T_{2}^{\prime}$ is considered instead of $T_{1} T_{2}$, then we arrive at the second ellipse corresponding to the chosen point $O$, which is the reflection of the first ellipse over the line passing through $O$ and perpendicular to the tangents.