If the latus rectum of an ellipse is equal to | vedclass

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(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Length of the latus rectum $= \frac{2b^2}{a}$. Length of

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$\frac{\sqrt{3}}{2}$

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(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Length of the latus rectum $= \frac{2b^2}{a}$. Length of the minor axis $= 2b$. According to the given condition,the latus rectum is equal to half of the minor axis: $\frac{2b^2}{a} = \frac{1}{2} (2b) = b$. Dividing both sides by $b$ (since $b 
eq 0$),we get $\frac{2b}{a} = 1$,which implies $\frac{b}{a} = \frac{1}{2}$. Squaring both sides,we get $\frac{b^2}{a^2} = \frac{1}{4}$. The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting the value of $\frac{b^2}{a^2}$: $e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

If the latus rectum of an ellipse is equal to half of its minor axis,then its eccentricity is

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