If the latus rectum of an ellipse subtends a | vedclass

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let $LL^{\prime}$ be the latus rectum,then the coordinates of $L$ are $

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$\frac{\sqrt{5}-1}{2}$

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let $LL^{\prime}$ be the latus rectum,then the coordinates of $L$ are $(ae, \frac{b^2}{a})$.Since $LL^{\prime}$ subtends a right angle $(\pi/2)$ at the centre $C(0,0)$,the angle $\angle LCS = \pi/4$.In the right-angled triangle $	riangle LCS$,we have $	an(\angle LCS) = \frac{LS}{CS}$.$	an(\frac{\pi}{4}) = \frac{b^2/a}{ae}$$1 = \frac{b^2}{a^2e}$$a^2e = b^2$Since $b^2 = a^2(1 - e^2)$,we have $a^2e = a^2(1 - e^2)$.$e = 1 - e^2$$e^2 + e - 1 = 0$Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $e = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.Since eccentricity $e > 0$,we have $e = \frac{\sqrt{5}-1}{2}$.

If the latus rectum of an ellipse subtends a right angle at the centre of that ellipse,then the eccentricity of that ellipse is

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