JEE MAIN - Mathematics (2016 (Offline) - No. 12)
The eccentricity of the hyperbola whose length of the latus rectum is equal to $$8$$ and the length of its conjugate axis is equal to half of the distance between its foci, is :
$${2 \over {\sqrt 3 }}$$
$${\sqrt 3 }$$
$${{4 \over 3}}$$
$${4 \over {\sqrt 3 }}$$
Explanation
$${{2{b^2}} \over a} = 8$$ and $$2b = {1 \over 2}\left( {2ae} \right)$$
$$ \Rightarrow 4{b^2} = {a^2}{e^2}$$
$$ \Rightarrow 4{a^2}\left( {{e^2} - 1} \right) = {a^2}{e^2}$$
$$ \Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}$$
$$ \Rightarrow 4{b^2} = {a^2}{e^2}$$
$$ \Rightarrow 4{a^2}\left( {{e^2} - 1} \right) = {a^2}{e^2}$$
$$ \Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}$$
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