\section*{Problem 17 - V01217}

Given is the ellipse \(9 x^{2}+25 y^{2}=225\) and the point \(P_{1}\left(-1 ; \frac{21}{5}\right)\).

a) Find the equations of the tangents from \(P_{1}\) to the ellipse.

b) Show that the line connecting \(P_{1}\) with the midpoint of the chord of contact passes through the center of the ellipse!

c) The major axis of the ellipse is the axis of a parabola whose vertex is at the center of the ellipse and which passes through \(P_{2}\left(3 ; \frac{12}{5}\right)\).

At what angle do the ellipse and the parabola intersect?

Rearranging the equation of the ellipse yields
}

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

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

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1716.jpg?height=531&width=782&top_left_y=246&top_left_x=637)

a) The equation of the tangent to this ellipse at a point of tangency \(B\) is

\[
\frac{x \cdot x_{B}}{25}+\frac{y \cdot y_{B}}{9}=1
\]

Substituting the point \(P_{1}\), which lies on the tangent, we get

\[
\frac{-x}{25}+\frac{\frac{21}{5}}{9}=1
\]

Rearranging (2) and substituting into (1) yields the coordinates of the two points of tangency

\[
B_{1}\left(-4 ; \frac{9}{5}\right) \quad ; \quad B_{2}\left(3 ; \frac{12}{5}\right)
\]

with the tangents

\[
t_{1}: \quad y=\frac{4}{5} x+5 \quad ; \quad t_{2}: \quad y=-\frac{9}{20} x+\frac{15}{4}
\]

b) The midpoint of the chord of contact \(B_{1} B_{2}\) is \(M_{s}\left(-\frac{1}{2} ; \frac{21}{10}\right)\). It is clearly also the midpoint of the segment \(M P_{1}\) and thus lies on the line from \(P_{1}\) through the center \(M\) of the ellipse.

c) For the parabola, from the equation \(y^{2}=a x\) and the coordinates of the point \(P_{2}\left(3 ; \frac{12}{5}\right)\),

\[
y^{2}=\frac{48}{25} x \quad ; \quad f(x)=y= \pm \sqrt{\frac{48}{25} x}
\]

The tangent to the parabola at \(P_{2}\) has the slope

\[
f^{\prime}(x)=\frac{2}{5} \sqrt{3} \cdot x \Rightarrow f^{\prime}(3)=\frac{2}{5}=m_{2}
\]

The tangent \(t_{2}\) has the slope \(m_{1}=-\frac{9}{20}\). For the angle of intersection of the two lines, i.e., also the angle of intersection of the parabola and the ellipse, we get

\[
\tan \varphi=\frac{m_{2}-m_{1}}{1+m_{1} \cdot m_{2}}=\frac{85}{82} \Rightarrow \varphi=46.03^{\circ}
\]

The ellipse and the parabola intersect at an angle of \(46.03^{\circ}\).