Question

The eccentric angle of a point $P$ lying in the first quadrant on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\theta .$ If $OP$ makes an angle $\varphi$ with x-axis, then $\theta -\varphi$ will be maximum when $\theta =$

• A. ${\tan }^{-1}\sqrt{\frac{a}{b}}$
• B. ${\tan }^{-1}\sqrt{\frac{b}{a}}$
• C. $\frac{\pi }{4}$
• D. none

Answer 1

Answer: (A)

${\tan }^{-1}\sqrt{\frac{a}{b}}$

$\tan \varphi =$ slope of OP$=\frac{b\sin \theta -0}{a\cos \theta -0}$
$\therefore$ $\tan \varphi =\frac{b}{a}\tan \theta$

We have to find max. value of $\theta -\varphi .$
It will be max. when $\tan (\theta -\varphi )$ is max.

$\therefore y=\tan (\theta -\varphi )=\frac{\tan \theta -\tan \varphi }{1+\tan \theta \tan \varphi }$ = $\frac{\tan \theta (1-\frac{b}{a})}{1+\frac{b}{a}{\tan }^2\theta }$ or $y=\frac{a-b}{a\cot \theta +b\tan \theta }$

y will be maximum if $z=a\cot \theta +b\tan \theta$ is min.
$\frac{dz}{d\theta }=-a\cos {\sec }^2\theta +b{\sec }^2\theta =0$
${\tan }^2\theta =\frac{a}{b}$ or $\tan \theta =\sqrt{\frac{a}{b}}$
Clearly $\frac{d^{2}z}{d{\theta }^2}=+$ $ive$, hence $Z$ is min. so that $y$ is max.