Q35 / 90 · JEE Main 2018 (16 Apr Online) · Mathematics

If the length of the latus rectum 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 is

A
$\frac{2}{3}$
B
$\frac{1}{2}$
C
$\frac{1}{9}$
Correct answer
D
$\frac{1}{3}$

Solution

Given length of latus-rectum of an ellipse is $\frac{2 b^{2}}{a} = 4 \Rightarrow b^{2} = 2 a$ and $a - a e = \frac{3}{2} \Rightarrow a (1 - e) = \frac{3}{2}$ .

Now we know, $b^{2} = a^{2} (1 - e^{2})$

$\Rightarrow b^{2} = a (1 - e) a (1 + e)$

$\Rightarrow 2 a = \frac{3}{2} a (1 + e)$ $(∵ a (1 - e) = \frac{3}{2})$

$\Rightarrow \frac{4}{3} = 1 + e$

$\Rightarrow e = \frac{1}{3}$ .

Hence, the correct answer is option: 'D'.

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Solution

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