Let's determine the vertices of a square whose diagonals lie on the coordinate axes and whose sides are tangent to the ellipse

$$
4 x^{2}+9 y^{2}=36
$$

What is the area of the rectangle defined by the points of tangency?
I. solution: The equation of the tangent line at a point $(x_{1}, y_{1})$ of the ellipse (1. the 1st figure) is

$$
4 x_{1} \cdot x + 9 y_{1} \cdot y = 36
$$

or in intercept form,

$$
\frac{x}{\frac{9}{x_{1}}} + \frac{y}{\frac{4}{y_{1}}} = 1
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c4fc7f9b4d3ad0f691e8g-1.jpg?height=710&width=789&top_left_y=477&top_left_x=646)

1. figure

According to the problem, the intercepts of the tangent are equal:

$$
\frac{9}{x_{1}} = \frac{4}{y_{1}}
$$

which means

$$
9 y_{1} = 4 x_{1}
$$

We can also use the fact that the point $(x_{1}, y_{1})$ lies on the ellipse:

$$
4 x_{1}^{2} + 9 y_{1}^{2} = 36
$$

From equations (1) and (2), we can calculate the values of $x_{1}$ and $y_{1}$:

$$
x_{1} = \pm \frac{9 \sqrt{13}}{13}, \quad y_{1} = \pm \frac{4 \sqrt{13}}{13}
$$

This determines 4 points of tangency, symmetric with respect to the origin. The area of the rectangle formed by the points of tangency can be found by taking 4 times the area of a small rectangle:

$$
t = 4 \cdot \frac{9 \sqrt{13}}{13} \cdot \frac{4 \sqrt{13}}{13} = \frac{144}{13} = 11 \frac{1}{13}
$$

Friedlszky Sándor (Bp. II., Rákóczi g. IV. o. t.)

II. solution: Let the length of the diagonal of the desired square be $2d$. The equations of the sides of the square are

$$
\begin{array}{ll}
y = -x + d, & y = -x - d \\
y = x + d, & y = x - d
\end{array}
$$

Let's find the intersection point of the first line and the ellipse. Substituting the value of $y$ from the line equation into the ellipse equation,

$$
\begin{aligned}
& 4 x^{2} + 9 x^{2} - 18 d x + 9 d^{2} = 36 \\
& 13 x^{2} - 18 d x + 9 d^{2} - 36 = 0
\end{aligned}
$$

If the line is tangent to the ellipse, the discriminant of the resulting equation will be 0:

$$
324 d^{2} - 468 d^{2} + 1872 = 0
$$

From this,

$$
d^{2} = 13, \quad d = \sqrt{13}
$$

(As a length, we only need to consider the positive root). Substituting the obtained value of $d$ into (3), we can calculate the $x$-coordinate of the points of tangency:

$$
x = \pm \frac{9 \sqrt{13}}{13}
$$

From the equation of the first tangent, we can calculate the $y$-coordinate of the points of tangency:

$$
y = \pm \frac{4 \sqrt{13}}{13}
$$

The area can be calculated in the same way as in the first solution.

Tusnády Gábor (Sátoraljaújhely, Kossuth g. II. o. t.)

III. solution: We can solve the problem without coordinate geometry. Draw a circle with a radius equal to half the major axis of the ellipse, and draw a tangent that cuts off $d$ units from both the $x$ and $y$ axes (2. figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_c4fc7f9b4d3ad0f691e8g-2.jpg?height=845&width=721&top_left_y=1071&top_left_x=688)

2. figure

The ratio of the affinity is the ratio of half the major axis to half the minor axis: $\frac{3}{2}$. Thus, the tangent to the circle cuts off $\frac{3}{2} d$ from the $y$-axis. The hypotenuse of the $\triangle OAB$ is $\frac{\sqrt{13}}{2} d$. The area of $\triangle OAB$ can be calculated in two ways:

$$
t = \frac{\frac{3}{2} d \cdot d}{2} = \frac{3 \cdot \frac{\sqrt{13}}{2} d}{2}
$$

From this, $d = \sqrt{13}$.

Using the geometric mean theorem for the legs, we can calculate the length of the segment $AC$ from the point of tangency on the circle:

$$
(\sqrt{13})^{2} = AC \cdot \frac{\sqrt{13}}{2} \sqrt{13}
$$

From this,

$$
AC = 2
$$

Using the similarity of triangles $\triangle CDA$ and $\triangle BOA$, it can be seen that the distance $AO = \sqrt{13}$ is divided by point $D$ in the ratio $AC : CB = 2 : \frac{9}{2} = 4 : 9$, so the $x$-coordinate of the point of tangency $E$ is $\frac{9 \sqrt{13}}{13}$. Similarly, in the similar triangles $\triangle B_1OA$ and $\triangle EDA$, the ratio is also $4 : 9$, so the $y$-coordinate of the point of tangency $E$ is $ED = \frac{4 \sqrt{13}}{13}$.

From this, the same area value is obtained as in the previous solutions.