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If the length of the latusrectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is $\frac{3}{2}$ units, then its eccentricity =

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a

$\frac{2}{3}$

b

$\frac{1}{9}$

c

$\frac{1}{2}$

d

$\frac{1}{3}$

answer is D.

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Detailed Solution

$2 \frac{b^{2}}{a} = 4 \Rightarrow b^{2} = 2 a SS' = 2 ae = \frac{3}{2} ae = \frac{3}{4} \Rightarrow a^{2} e^{2} = \frac{9}{16} a^{2} - b^{2} = \frac{9}{16} a^{2} - 2 a + 1 = \frac{9}{16} + 1 {(a - 1)}^{2} = \frac{25}{16} \Rightarrow a - 1 = \frac{5}{4} a = \frac{9}{4} N o w ae = \frac{3}{4} \Rightarrow e = \frac{1}{3}$

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