Question

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. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{9}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

Focus $=(ae,0)$ and vertex $=(a,0)$

Distance between focus and vertex $=a(1-e)=\frac{3}{2}$

$⟹a-\frac{3}{2}=ae$

Squaring above equation, we get

$⟹a^2+\frac{9}{4}-3a=a^2{e}^2$ ...[1]

Length of latus rectum $=\frac{2b^2}{a}=4$

$⟹b^2=2a$ ...[2]

$e^2=1-\frac{b^2}{a^2}$

$=1-\frac{2a}{a^2}$ ...(from [2])

$=1-\frac{2}{a}$ ...[3}

Substituting this in equation[1] we get

$⟹a^2+\frac{9}{4}-3a=a^2(1-\frac{2}{a})$

$⟹a=\frac{9}{4}$

Therefore, $e^2=1-\frac{2}{a}=1-\frac{8}{9}=\frac{1}{9}$

$⟹e=\frac{1}{3}$

Hence, answer is option (D)