If the distance between the foci of an ellips | vedclass

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(C) Given that the distance between the foci of an ellipse is equal to the length of the latus rectum.The distance between the foci is $2ae$ and the l

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

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(C) Given that the distance between the foci of an ellipse is equal to the length of the latus rectum.The distance between the foci is $2ae$ and the length of the latus rectum is $\frac{2b^2}{a}$.Therefore, $2ae = \frac{2b^2}{a} \implies ae = \frac{b^2}{a} \implies b^2 = a^2e$.We know that for an ellipse, $b^2 = a^2(1 - e^2)$.Substituting $b^2 = a^2e$ into the equation, we get $a^2e = a^2(1 - e^2)$.Dividing by $a^2$ (since $a 
eq 0$), we get $e = 1 - e^2$.Rearranging gives the quadratic equation $e^2 + e - 1 = 0$.Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since eccentricity $e$ must be positive $(0 < e < 1)$, we take the positive root: $e = \frac{\sqrt{5} - 1}{2}$.

If the distance between the foci of an ellipse is equal to the length of the latus rectum, then its eccentricity is

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