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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Aicharya · Accepted Answer

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} $

If the latusrectum of an ellipse subtends a right angle at the centre of that ellipse, then the eccentricity…

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