Question

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

• A. ${\tan }^{-1}\left(\pm \frac{ae}{b}\right)$
• B. ${\tan }^{-1}\left(\pm \frac{be}{a}\right)$
• C. ${\tan }^{-1}\left(\pm \frac{b}{ae}\right)$
• D. ${\tan }^{-1}\left(\pm \frac{a}{be}\right)$

Answer 1

Answer: (C)

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, $a\cos \theta =ae$ and $b\sin \theta =\pm b\sqrt{1-e^2}$
$\Rightarrow \tan \theta =\pm \frac{\sqrt{1-e^2}}{e}=\pm \left(\frac{b}{ae}\right)$
$\Rightarrow \theta ={\tan }^{-1}\left(\pm \frac{b}{ae}\right)$