## Task 22/62

The ellipse and the hyperbola with the following properties are sought:

1. The linear eccentricity is \( e = 20 \).
2. The perpendicular focal rays \( l_{1} \) and \( l_{2} \) are in the ratio \( l_{1}: l_{2} = 4: 3 \).

a) The lengths of the focal rays \( l_{1} \) and \( l_{2} \) are to be determined.
b) The equations of the conic sections are to be established.
a) Let $P_{1}$ and $P_{2}$ be the foci of the two conic sections (due to the equality of the linear eccentricity, the foci also coincide if the axes coincide) and $P_{8}$ one of the points where the rays are perpendicular to each other. The points $P_{1}, P_{2}$, and $P_{8}$ form a right-angled triangle, so $P_{8}$ lies on the Thales circle over $P_{1} P_{2}$. It is immediately apparent that there is exactly one point $P_{8}$ (apart from symmetry about the conic section axes), i.e., the ellipse and hyperbola intersect at $P_{8}$. The following equations now apply:

$$
l_{1}: l_{2}=4: 3 \quad(1) \quad \text { and } \quad P_{1} P_{2}^{2}=4 e^{2}=l_{1}^{2}+l_{2}^{2}
$$

By substituting (1) into (2), we get $4 e^{2}=\frac{25}{16} l_{1}$ and $4 e^{2}=\frac{25}{9} l_{2}$. With $e=20$, this results in $l_{1}=32$ and $l_{2}=24$.

b) The equation of the ellipse can be written in the following form: $x^{2} b^{2}+y^{2} a^{2}=a^{2} b^{2}$ (4). We replace $b^{2}$ with the relation $b^{2}=a^{2}-e^{2}$ and thus obtain

$$
x^{2}\left(a^{2}-e^{2}\right)+y^{2} a^{2}=a^{2}\left(a^{2}-e^{2}\right)
$$

To determine $a$, we calculate the coordinates of $P_{8}$ and substitute them, along with $e=20$, into (4a). By decomposing $y_{8}$, we divide the triangle $P_{1} P_{2} P_{8}$ into two right-angled sub-triangles. Then, according to the Pythagorean theorem, $l_{1}^{2}=y_{8}^{2}+\left(e+x_{8}\right)^{2}$ (5a) and $l_{2}^{2}=y_{8}^{2}+\left(e-x_{8}\right)^{2}$ (5b).

By subtracting one of these two equations from the other and solving for $x^{2}$, we get $x_{s}=5.6$ and thus $y_{s}= \pm 1.2$. Substituting these values into (4a) results in solving the resulting biquadratic equation

$$
a 4-800 a^{2}+12544=0 \rightarrow a_{1}^{2}=784 ; a_{1}=28 \text { and } a_{2}^{2}=16 ; a_{2}=4
$$

From $b^{2}=a^{2}-e^{2}$, it follows further that $b_{1}^{2}=384$, so $b_{1} \approx 19.6$ and $b_{2}^{2}=-384$, so $b_{2} \approx 19.6 i$. Clearly, $a_{2}$ and $b_{2}$ are (in the real numbers) unusable, so the equation of the ellipse is

$$
384 x^{2}+784 y^{2}=301056 \rightarrow \frac{x^{2}}{784}+\frac{y^{2}}{384}=1
$$

Analogously, from the hyperbola equation $x^{2} b^{2}-y^{2} a^{2}=a^{2} b^{2}$ with $b^{2}=e^{2}-a^{2}$, we obtain the values $a_{1}=28, a_{2}=$ 4 and $b_{1} \approx 19.6 i, b_{2} \approx 19.6$. It is evident that in this case, the values $a_{1}$ and $b_{1}$ must be discarded as unusable. The hyperbola equation thus takes the form

$$
384 x^{2}-16 y^{2}=6144 \rightarrow \frac{x^{2}}{16}-\frac{y^{2}}{384}=1
$$