For an arbitrary point $P$ on an ellipse, the projection of $P$ onto the major axis is $Q$, and the midpoint of the segment $O Q$ is $R$, where $O$ is the center of the ellipse. What can we say about the ellipse if the line perpendicular to $P R$ at $P$ is exactly the tangent at $P$?
I. solution. Let the canonical equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the coordinates of point $P$ be $x_{1}$ and $y_{1}$. Then the coordinates of the other points are:

$$
Q\left(x_{1}, 0\right), \quad R\left(\frac{x_{1}}{2}, 0\right), \quad O(0,0)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_7a247e76e863029235feg-1.jpg?height=566&width=865&top_left_y=411&top_left_x=619)

Since $P$ is a point on the ellipse: $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1$. The equation of the tangent line to the ellipse at point $P$ - as is well known -

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

its normal vector is $\mathbf{n}\left(\frac{x_{1}}{a^{2}}, \frac{y_{1}}{b^{2}}\right)$. According to the conditions of the problem, $\overrightarrow{R P}$ is perpendicular to this tangent, so $\overrightarrow{R P} \| \mathbf{n}$. Therefore, there is a real number $\lambda$ such that $\overrightarrow{R P}=\lambda \mathbf{n}$.

Since $\overrightarrow{R P}=\overrightarrow{O P}-\overrightarrow{O R}$, we have $\overrightarrow{R P}=\frac{x_{1}}{2} \mathbf{i}+y_{1} \mathbf{j}$, and thus

$$
\frac{x_{1}}{2}=\lambda \frac{x_{1}}{a^{2}} \quad \text { and } \quad y_{1}=\lambda \frac{y_{1}}{b^{2}}
$$

Expressing $\lambda$, we get $\lambda=\frac{a^{2}}{2}$ and $\lambda=b^{2}$, so $a^{2}=2 b^{2}$. Since $a$ and $b$ are positive, this means that $a=\sqrt{2 b}$.

Accordingly, there is indeed an ellipse for which the assumption holds for every point - even for the endpoints of the major axis - and every such ellipse has a major axis that is $\sqrt{2}$ times the length of the minor axis. An interesting consequence of this is that the distance between the two foci is exactly equal to the length of the minor axis.

Every step we have taken is reversible, so this condition is also sufficient.

II. solution. Draw the circle centered at $O$ with a radius equal to the semi-major axis of the ellipse! The ellipse is the image of this circle under a perpendicular affinity, if we choose the line of the major axis as the axis of the affinity. Let $P^{\prime}$ be the point on the circle corresponding to $P$. The tangent line to the circle at this point and the tangent line to the ellipse at $P$ intersect on the major axis at a point $S$.

The right triangle $S P R$ is divided into two similar triangles by the altitude $P Q$, and for the sides of these triangles, we have:

$$
P Q: S Q=R Q: P Q, \quad \text { or } \quad P Q^{2}=S Q \cdot Q R
$$

Similarly, in the right triangle $S P^{\prime} O$:

$$
P^{\prime} Q: S Q=O Q: P^{\prime} Q
$$

and since $O Q=2 R Q$, we have $P^{\prime} Q^{2}=2 S Q \cdot Q R$. Comparing these:

$$
P^{\prime} Q^{2}=2 P Q^{2}, \quad \text { or } \quad \frac{P Q}{P^{\prime} Q}=1 / \sqrt{2}
$$

so the ratio of the affinity is $1: \sqrt{2}$.

Since the minor axis is the image of the circle's diameter under the affinity, and the major axis is the circle's diameter, in this ellipse, the major axis is $\sqrt{2}$ times the length of the minor axis.