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Q.

${LL}^{1}$ is the latus rectum of an ellipse and ${ΔS}^{1} {LL}^{1}$ is an equilateral triangle . Then e =

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a

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

b

$\frac{1}{\sqrt{4}}$

c

$\frac{1}{\sqrt{3}}$

d

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

answer is B.

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Detailed Solution

$S' L = LL' = S' L' I n ∆ S^{'} L L^{'} \Rightarrow \tan 30^{0} = \frac{(\frac{b^{2}}{a})}{2 a e} \Rightarrow \frac{1}{\sqrt{3}} = \frac{b^{2}}{2 a^{2} e} \frac{a^{2} (1 - e^{2})}{2 a^{2} e} = \frac{1}{\sqrt{3}} \Rightarrow \frac{1}{e} - e = \frac{2}{\sqrt{3}} \Rightarrow \frac{1}{e} - e = \frac{3 - 1}{\sqrt{3}} = \sqrt{3} - \frac{1}{\sqrt{3}} \Rightarrow e = \frac{1}{\sqrt{3}}$

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