Question

An ellipse is inscribed in a rectangle and if the angle between the diagonals is ${\tan }^{-1}2\sqrt{2}$ then $e=$

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{3}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$

$\tan \theta =2\sqrt{2}$

Let $\tan \frac{\theta }{2}=x$.

Then $\frac{2x}{1-x^2}=2\sqrt{2}$

$2x=2\sqrt{2}(1-x^2)$

$x=\sqrt{2}-\sqrt{2}{x}^2$

$\sqrt{2}{x}^2+x-\sqrt{2}=0$

$x=\frac{-1+\sqrt{1+4.2}}{2\sqrt{2}}(\because x>0)$

$=\frac{3-1}{2\sqrt{2}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\tan \frac{\theta }{2}=\frac{h}{a}=\frac{1}{\sqrt{2}}$

squaring, we get $\frac{h^2}{a^2}=\frac{1}{2}$

$\frac{a^2(1-e^2)}{a^2}=\frac{1}{2}$

$1-e^2=\frac{1}{2}\Rightarrow e^2=\frac{1}{2}$

(or) $e=\frac{1}{\sqrt{2}}$