JEE Main On 16 April 2018 Question 28
Question: . If the length of the latus rectum of an ellipse is units and the distance between a focus and its nearest vertex on the major axis is $ \frac{3}{2} $ units, then its eccentricity is? [JEE Main 16-4-2018]
Options:
A) $ \frac{1}{2} $
B) $ \frac{2}{3} $
C) $ \frac{1}{9} $
D) $ \frac{1}{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
Focus $ =(ae,0) $ and vertex $ =(a,0) $ Distance between focus and vertex $ =a(1-e)=\frac{3}{2} $
$ \Rightarrow $ $ a-\frac{3}{2}=ae $ Squaring above equation, we get
$ \Rightarrow $ $ a^{2}+\frac{9}{4}-3a=a^{2}e^{2} $ …[1]
Length of latus rectum $ =\frac{2b^{2}}{a}=4 $
$ \Rightarrow $ $ b^{2}=2a $ $ e^{2}=1-\frac{b^{2}}{a^{2}} $ ….(from [2])
$ =1-\frac{2}{a} $ …[3}
Substituting this in equation [1] we get
$ \Rightarrow $ $ a^{2}+\frac{9}{4}-3a=a^{2}(1-\frac{2}{a}) $
$ \Rightarrow $ $ a=\frac{9}{4} $ Therefore, $ e^{2}=1-\frac{2}{a}=1-\frac{8}{9}=\frac{1}{9} $
$ \Rightarrow $ $ e=\frac{1}{3} $ Hence, answer is option D
