Question

An ellipse is inscribed in a rectangle and the angle between the diagonals of the rectangle is ${\tan }^{-1}\left(2\sqrt{2}\right)$. Then the eccentricity of the ellipse is

• A. $\cot {15}^0$
• B. $\cos {45}^0$
• C. $\cot {60}^0$
• D. $\cot {75}^0$

Answer 1

Answer: (B)

$\cos {45}^0$

Let $2a$ be the length of the major axis (rectangle side length) and $2b$ be the length of minor axis (rectangle side width)

So if $\theta$ is angle between the diagonals of the rectangle then, $\tan \frac{\theta }{2}=\frac{b}{a}$

$\Rightarrow {\tan }^{-1}\frac{\frac{b}{a}+\frac{b}{a}}{1-(\frac{b}{a}\times \frac{b}{a})}={\tan }^{-1}2\sqrt{2}$

$\Rightarrow \theta =2{\tan }^{-1}\frac{b}{a}={\tan }^{-1}2\sqrt{2}$

$\Rightarrow {\tan }^{-1}\frac{\frac{2b}{a}}{1-{\left(\frac{b}{a}\right)}^2}={\tan }^{-1}2\sqrt{2}$

$\Rightarrow \frac{\frac{2b}{a}}{1-{\left(\frac{b}{a}\right)}^2}=2\sqrt{2}$

$\Rightarrow \frac{b}{a}=\sqrt{2}(1-\frac{b^2}{a^2})$

Solving above quadratic we get, $\frac{b}{a}=\frac{1}{\sqrt{2}}$

$\therefore e=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{\sqrt{2}}=\cos {45}^o$