NIMCET 2021 Question Paper with solutions

Given, Ellipse $\frac{{\mathit{x}}^2}{{\mathit{a}}^2}-\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1$Let $\mathit{P}$ be the end point of Lotus Rectum
and $\mathit{P}=\left(\mathit{a}\cos \mathrm{\theta },\mathit{b}\sin \mathrm{\theta }\right)......$ (1)
Now to Calculate, eccentric angle $\mathrm{\theta }$ at the end of LR i.e.
at $\mathit{P}\left(\mathit{a}\cos \mathrm{\theta },\mathit{b}\sin \mathrm{\theta }\right)$
We know that from above figure
$\mathit{P}=\left(\mathit{a}\mathit{e},\frac{{\mathit{b}}^2}{\mathit{a}}\right).......\left(2\right)$
We will compare coordinates
$\mathit{a}\cos \mathrm{\theta }=\mathit{a}\mathit{e}$
$\mathit{b}\sin \mathrm{\theta }={\mathit{b}}^2∕\mathit{a}$
$\cos \mathrm{\theta }=\mathit{e}$
$\sin \mathrm{\theta }=\mathit{b}∕\mathit{a}$We get $\left|\tan \mathrm{\theta }\right|=\frac{\mathit{b}}{\mathit{a}\mathit{e}}$
OR
$\tan \mathrm{\theta }=\pm \frac{\mathit{b}}{\mathit{a}\mathit{e}}$
$\mathrm{\theta }={\tan }^{-1}\left(\pm \frac{\mathit{b}}{\mathit{a}\mathit{e}}\right)$