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Q.
The eccentric angles of the extremities of the latus rectum of
the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ are given by
Conic Sections
Solution:
Let eq. of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
If the latus rectum is $PQ$ then for the points $P$ and $Q$
$x = ae , \frac{ y ^{2}}{ b ^{2}}=1- e ^{2}$
$\Rightarrow y^{2}=b^{2}\left(1-e^{2}\right)$
$\Rightarrow y =\pm b \sqrt{1- e ^{2}}$
Hence, $P$ is $\left( ae , b \sqrt{1- e ^{2}}\right)$ and $Q$ is $\left( ae ,- b \sqrt{1- e ^{2}}\right)$
If eccentricangle of theextremitiesbe $\theta$,
then, $a \cos \theta=a e$ and $b \sin \theta=\pm b \sqrt{1-e^{2}}$
$\Rightarrow \tan \theta=\pm \frac{\sqrt{1-e^{2}}}{e}=\pm\left(\frac{b}{a e}\right) $
$ \Rightarrow \theta=\tan ^{-1}\left(\pm \frac{ b }{ ae }\right)$
