Answer
$$\eqalign{
& {\bf{e}} = \sqrt 3 \cr
& {\bf{e}} \approx 1.7 \cr} $$
Work Step by Step
$$\eqalign{
& 16{y^2} - 8{x^2} = 16 \cr
& {\text{Divide both sides by 16}} \cr
& \frac{{16{y^2}}}{{16}} - \frac{{8{x^2}}}{{16}} = 1 \cr
& {y^2} - \frac{{{x^2}}}{2} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& {y^2} - \frac{{{x^2}}}{2} = 1,{\text{ then }}{a^2} = 1{\text{ and }}{b^2} = 2 \cr
& {\text{The excentricity of a hyperbola is giving by}} \cr
& {\bf{e}} = \frac{{\sqrt {{a^2} + {b^2}} }}{a} \cr
& {\bf{e}} = \frac{{\sqrt {1 + 2} }}{{\sqrt 1 }} \cr
& {\bf{e}} = \sqrt 3 \cr
& {\bf{e}} \approx 1.7 \cr} $$
