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The eccentricity of the ellipse whose Latusrectum is equal to half of its minor axis is

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$3 / 5$

$2 / 3$

$1 / \sqrt{2}$

$\sqrt{7} / 4$

answer is D.

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Detailed Solution

$\frac{2 b^{2}}{a} = \frac{3}{2} b \Rightarrow 4 b = 3 a \Rightarrow 16 b^{2} = 9 a^{2} \Rightarrow 16 a^{2} (1 - e^{2}) = 9 a^{2} \Rightarrow e = \frac{\sqrt{7}}{4}$

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