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If the latusrectum of an ellipse subtends a right angle at the centre of that ellipse, then the eccentricity of that ellipse is
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Verified Answer
The correct answer is:
$\frac{\sqrt{5}-1}{2}$
Let equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $L L^{\prime}$ is latusrectum, then co-ordinate of
$$
L\left(a e, \frac{b^2}{a}\right)
$$
$L L^{\prime}$ subtend $\pi / 2$ Angle at the centre, so angle
$$
L C S=\pi / 4
$$
$$
\begin{array}{rlrl}
\Rightarrow & \tan \frac{\pi}{4} & =\frac{\frac{b^2}{a}}{a e} \\
\Rightarrow & 1 & =\frac{b^2}{a^2 e} \\
\Rightarrow & a^2 e & =b^2=a^2\left(1-e^2\right) \\
\Rightarrow & e & =1-e^2 \\
\Rightarrow & e^2+e-1 & =0 \\
\Rightarrow & & e & =\frac{-1 \pm \sqrt{5}}{2} \Rightarrow e=\frac{\sqrt{5}-1}{2} .
\end{array}
$$
Let $L L^{\prime}$ is latusrectum, then co-ordinate of
$$
L\left(a e, \frac{b^2}{a}\right)
$$
$L L^{\prime}$ subtend $\pi / 2$ Angle at the centre, so angle
$$
L C S=\pi / 4
$$
$$
\begin{array}{rlrl}
\Rightarrow & \tan \frac{\pi}{4} & =\frac{\frac{b^2}{a}}{a e} \\
\Rightarrow & 1 & =\frac{b^2}{a^2 e} \\
\Rightarrow & a^2 e & =b^2=a^2\left(1-e^2\right) \\
\Rightarrow & e & =1-e^2 \\
\Rightarrow & e^2+e-1 & =0 \\
\Rightarrow & & e & =\frac{-1 \pm \sqrt{5}}{2} \Rightarrow e=\frac{\sqrt{5}-1}{2} .
\end{array}
$$
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