The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is (axes are coordinate axes)

We have 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2{\mathrm{b}}^2}{\mathrm{a}}=8\\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{b}=\frac{1}{2}\left(2\mathrm{ae}\right)\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2}{\mathrm{a}}{\left(\frac{\mathrm{ae}}{2}\right)}^2=8\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{ae}}^2=16----\left(1\right)\\ \text{ Also, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\times \frac{{\mathrm{b}}^2}{\mathrm{a}}=8\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{b}}^2=4\mathrm{a}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}^2\left({\mathrm{e}}^2-1\right)=4\mathrm{a}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{ae}}^2-\mathrm{a}=4----\left(2\right)\end{array}$

From (1) and (2), we have 

$16-\frac{16}{e^2}=4$

$\begin{array}{r}\mathrm{or} \frac{16}{{\mathrm{e}}^2}=12\\ \mathrm{or} \mathrm{e}=\frac{2}{\sqrt{3}}\end{array}$