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An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ is inscribed in a rectangle of dimensions $2 a$ and $2 b$ respectively, If the angle between the diagonals of the rectangle is $\tan ^{-1}(4 \sqrt{3})$, then the eccentricity of that ellipse is
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The correct answer is:
$\frac{1}{2}$
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]
$\begin{aligned} & \Rightarrow 4 \sqrt{3}=\frac{2 \frac{b}{a}}{1-\frac{b^2}{a^2}} \Rightarrow 2 \sqrt{3}=\frac{\frac{b}{a}}{1-\frac{b^2}{a^2}}\left[\because \tan ^{-1}(4 \sqrt{3})=2 \theta\right] \\ & \Rightarrow 2 \sqrt{3}\left(\frac{b}{a}\right)^2+\left(\frac{b}{a}\right)-2 \sqrt{3}=0 \\ & \Rightarrow 2 \sqrt{3}\left(\frac{b}{a}\right)^2+4\left(\frac{b}{a}\right)-3\left(\frac{b}{a}\right)-2 \sqrt{3}=0 \\ & \Rightarrow 2\left(\frac{b}{a}\right)\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]-\sqrt{3}\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]=0\end{aligned}$
$\begin{array}{lll}\Rightarrow\left[2\left(\frac{b}{a}\right)-\sqrt{3}\right]\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]=0 & \\ \Rightarrow & \frac{b}{a}=\frac{\sqrt{3}}{2} & \left\{\text { as } \frac{b}{a}>0\right\}\end{array}$
Now, the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$e=\sqrt{1-\left(\frac{b}{a}\right)^2}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
Hence, option (b) is correct.
Since, $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}$
[where $2 \theta$ is angle between the diagonals of the rectangle]
$\begin{aligned} & \Rightarrow 4 \sqrt{3}=\frac{2 \frac{b}{a}}{1-\frac{b^2}{a^2}} \Rightarrow 2 \sqrt{3}=\frac{\frac{b}{a}}{1-\frac{b^2}{a^2}}\left[\because \tan ^{-1}(4 \sqrt{3})=2 \theta\right] \\ & \Rightarrow 2 \sqrt{3}\left(\frac{b}{a}\right)^2+\left(\frac{b}{a}\right)-2 \sqrt{3}=0 \\ & \Rightarrow 2 \sqrt{3}\left(\frac{b}{a}\right)^2+4\left(\frac{b}{a}\right)-3\left(\frac{b}{a}\right)-2 \sqrt{3}=0 \\ & \Rightarrow 2\left(\frac{b}{a}\right)\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]-\sqrt{3}\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]=0\end{aligned}$
$\begin{array}{lll}\Rightarrow\left[2\left(\frac{b}{a}\right)-\sqrt{3}\right]\left[\sqrt{3}\left(\frac{b}{a}\right)+2\right]=0 & \\ \Rightarrow & \frac{b}{a}=\frac{\sqrt{3}}{2} & \left\{\text { as } \frac{b}{a}>0\right\}\end{array}$
Now, the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$e=\sqrt{1-\left(\frac{b}{a}\right)^2}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
Hence, option (b) is correct.
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