Question

If the eccentric angle of a point lying in the first quadrant on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be $\alpha$ and the line joining the center to the point makes an angle $\beta$ with x-axis then $\alpha -\beta$ will be maximum when $\alpha =$

• A. $0$
• B. ${\cot }^{-1}\sqrt{\frac{a}{b}}$
• C. ${\tan }^{-1}\sqrt{\frac{a}{b}}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (C)

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

Now, $\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$

$(\alpha -\beta )$ will be ${max}^m$, when $\alpha -\beta =\pi /2$ and $\alpha =(\beta +\pi /2)$

$\tan \left(\pi /2\right)=\frac{\tan \alpha +\tan \alpha }{1-{\tan }^2\alpha }$

$1-{\tan }^2\alpha =0$

${\tan }^2\alpha =1$

And $\frac{a}{b}=1\left(for\alpha \right)$

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