Answer
$$\eqalign{
& {\bf{e}} = \sqrt 5 \cr
& {\bf{e}} \approx 2.2 \cr} $$
Work Step by Step
$$\eqalign{
& 8{y^2} - 2{x^2} = 16 \cr
& {\text{Divide both sides by 16}} \cr
& \frac{{8{y^2}}}{{16}} - \frac{{2{x^2}}}{{16}} = 1 \cr
& \frac{{{y^2}}}{2} - \frac{{{x^2}}}{8} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& \frac{{{y^2}}}{2} - \frac{{{x^2}}}{8} = 1,{\text{ then }}{a^2} = 2{\text{ and }}{b^2} = 8 \cr
& {\text{The excentricity of a hyperbola is giving by}} \cr
& {\bf{e}} = \frac{{\sqrt {{a^2} + {b^2}} }}{a} \cr
& {\bf{e}} = \frac{{\sqrt {2 + 8} }}{{\sqrt 2 }} \cr
& {\bf{e}} = \sqrt 5 \cr
& {\bf{e}} \approx 2.2 \cr} $$
