Solution:
Let the equation of the hyperbola be $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$
Length of latus rectum = 8
$\therefore\ \frac{2\text{b}^2}{2}=8$
$\Rightarrow\text{b}^2=4\text{a}$
Conjugate axis = half of the distance between the foci
$\therefore\ 2\text{b}=\text{ae}$
Now, $\text{b}^2=\text{a}^2(\text{e}^2-1)$
From eqs. (i) and (iii), we get
$\frac{\text{a}^2\text{e}^2}{4}=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\text{e}^2=4\text{e}^2-4$
$\Rightarrow\text{e}^2=\frac{4}{3}$
$\Rightarrow\text{e}=\frac{2}{\sqrt{3}}$
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