If the eccentricity of an ellipse is 1/sqrt{2 | vedclass

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(D) Given eccentricity $e = \frac{1}{\sqrt{2}}$.The formula for the latus rectum of an ellipse is $L = \frac{2b^2}{a}$.We know that for an ellipse,$b^

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Semi-major axis

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(D) Given eccentricity $e = \frac{1}{\sqrt{2}}$.The formula for the latus rectum of an ellipse is $L = \frac{2b^2}{a}$.We know that for an ellipse,$b^2 = a^2(1 - e^2)$.Substituting the value of $e^2 = \frac{1}{2}$ into the equation:$b^2 = a^2(1 - \frac{1}{2}) = a^2(\frac{1}{2}) = \frac{a^2}{2}$.Now,substitute $b^2$ into the latus rectum formula:$L = \frac{2(\frac{a^2}{2})}{a} = \frac{a^2}{a} = a$.Since $a$ represents the semi-major axis,the latus rectum is equal to its semi-major axis.

If the eccentricity of an ellipse is $1/\sqrt{2}$,then its latus rectum is equal to its

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