Answer
$\frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{4} = 1$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {0, \pm 4} \right) \cr
& {\text{Asymptotes }}y = \pm 2x \cr
& {\text{Vertices }}\left( {\underbrace 0_x,\underbrace 4_y} \right){\text{ and }}\left( {\underbrace 0_x,\underbrace { - 4}_y} \right) \cr
& {\text{the x coordinates of the vertices are equal, }} \cr
& {\text{ so the hyperbola has the standard form}} \cr
& \frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& {\text{Characteristics of the hyperbola}} \cr
& {\text{Vertices }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr
& {\text{Vertices }}\left( {0,\underbrace { - 4}_{ - a},} \right){\text{ and }}\left( {0,\underbrace 4_a} \right) \Rightarrow a = 4 \cr
& \cr
& {\text{Asymptotes:}} \pm \frac{a}{b}x \cr
& \pm \frac{a}{b}x = \pm 2x \cr
& \frac{a}{b} = 2 \to \frac{4}{b} = 2,{\text{ }}b = 2 \cr
& \cr
& \underbrace {\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1}_ \Downarrow \cr
& \frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{4} = 1 \cr} $$
