Answer
$$\frac{{{y^2}}}{{64}} - \frac{{{x^2}}}{{16}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {0, \pm 8} \right) \cr
& {\text{Asymptotes: }}y = \pm 2x \cr
& x{\text{ is constant, so the equation of the hyperbola is }} \cr
& \frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& {\text{With vertices }}\left( {0, \pm a} \right) \cr
& \left( {0, \pm 8} \right) \to a = 8 \cr
& {\text{The asymptores are:}} \cr
& y = \pm \frac{a}{b}x \cr
& \pm \frac{a}{b}x = \pm 2x \cr
& \frac{a}{b} = 2 \to b = \frac{a}{2} = \frac{8}{2} \cr
& b = 4 \cr
& {\text{Substituting the constants }}a{\text{ and }}b{\text{ into }}\left( {\bf{1}} \right) \cr
& \frac{{{y^2}}}{{{{\left( 8 \right)}^2}}} - \frac{{{x^2}}}{{{{\left( 4 \right)}^2}}} = 1 \cr
& \frac{{{y^2}}}{{64}} - \frac{{{x^2}}}{{16}} = 1 \cr} $$
