Question

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ is inscribed in a rectangle of dimensions $2a$ and $2b$ respectively, If the angle between the diagonals of the rectangle is ${\tan }^{-1}(4\sqrt{3})$, then the eccentricity of that ellipse is

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

Answer 1

Answer: (B)

Now, from the figure $\tan \theta =\frac{b}{a}$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Since, $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$
[where $2\theta$ is angle between the diagonals of the rectangle ]
$\Rightarrow 4\sqrt{3}=\frac{2\frac{b}{a}}{1-\frac{b^2}{a^2}}$
$\Rightarrow 2\sqrt{3}=\frac{\frac{b}{a}}{1-\frac{b^2}{a^2}}\left[\because {\tan }^{-1}(4\sqrt{3})=2\theta \right]$
$\Rightarrow 2\sqrt{3}{\left(\frac{b}{a}\right)}^2+\left(\frac{b}{a}\right)-2\sqrt{3}=0$
$\Rightarrow 2\sqrt{3}{\left(\frac{b}{a}\right)}^2+4\left(\frac{b}{a}\right)-3\left(\frac{b}{a}\right)-2\sqrt{3}=0$
$\Rightarrow 2\left(\frac{b}{a}\right)\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]-\sqrt{3}\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]=0$
$\Rightarrow \left[2\left(\frac{b}{a}\right)-\sqrt{3}\right]\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]=0$
$\Rightarrow \frac{b}{a}=\frac{\sqrt{3}}{2}\left\{\text{ as }\frac{b}{a}>0\right\}$
Now, the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$e=\sqrt{1-{\left(\frac{b}{a}\right)}^2}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$