The eccentricity of the hyperbola whose latus rectum is $8$ and conjugate axis is equal to half the distance between the foci is:

Solution

1. Given $2b^2/a=8$, so $b^2=4a$.
2. Given $2b = ae$, so $b=ae/2$.
3. Substitute $b^2=a^2{e}^2/4$ into $b^2=4a$ to get $a^2{e}^2/4=4a$, so $a{e}^2=16$.
4. Use $b^2=a^2(e^2-1)$, so $4a=a^2(e^2-1)$, which means $4=a(e^2-1)$.
5. Divide $a{e}^2=16$ by $a(e^2-1)=4$ to get $e^2/(e^2-1)=4$.
6. Solving $e^2=4e^2-4$ gives $3e^2=4$, so $e=2/\sqrt{3}$.
7. Hence the answer is (B).