\section*{Problem 19 - V01219}

Draw the ellipse

\[
9 x^{2}+25 y^{2}=225
\]

and determine graphically and analytically the points where the focal rays are perpendicular to each other.
}

Rearranging the equation of the ellipse yields

\[
\frac{x^{2}}{25}+\frac{y^{2}}{9}=1
\]

i.e., the semi-axes \(a=5\) and \(b=3\). The linear eccentricity is thus \(e=\sqrt{a^{2}-b^{2}}=4\). The foci have the coordinates \(F_{1}(-4,0)\) and \(F_{2}(4,0)\).

The construction is carried out with compass and straightedge using the classical method:

We choose a point \(H\) on \(A B\), draw a circular arc around \(F_{1}\) with radius \(H A\) and a circular arc around \(F_{2}\) with radius \(H B\). The intersection points of these circles are (symmetric to the major axis) points \(X_{1}\) and \(X_{2}\) of the ellipse.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1718.jpg?height=537&width=740&top_left_y=405&top_left_x=658)

The points on the ellipse where the focal rays intersect perpendicularly must lie on the Thales circle over the segment \(F_{1} F_{2}\).

For the construction, draw a circle around \(M\) with radius \(M F_{1}\). The intersection points with the ellipse are the four points \(P_{1}, P_{2}, P_{3}, P_{4}\) with the required property.

The described Thales circle has the equation \(x^{2}+y^{2}=16\). Solving for \(y^{2}\) and substituting into (1) yields

\[
9 x^{2}+25\left(16-x^{2}\right)=225 \quad \Rightarrow \quad x_{1 ; 2}= \pm \frac{5}{4} \sqrt{7} \quad ; \quad y_{1,2}= \pm \frac{9}{4}
\]

The desired points thus have the coordinates

\[
P_{1}\left(\frac{5}{4} \sqrt{7} ; \frac{9}{4}\right) \quad ; \quad P_{2}\left(-\frac{5}{4} \sqrt{7} ; \frac{9}{4}\right) \quad ; \quad P_{3}\left(-\frac{5}{4} \sqrt{7} ;-\frac{9}{4}\right) \quad ; \quad P_{4}\left(\frac{5}{4} \sqrt{7} ;-\frac{9}{4}\right)
\]