Answer
$$\eqalign{
& {\text{Vertices }}\left( {0, \pm 4} \right){\text{ and foci}}\left( {0, \pm 2\sqrt 5 } \right) \cr
& {\text{Asymptotes: }}y = 2x{\text{ and }}y = - 2x \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {0, \pm 4} \right){\text{ }} \cr
& {\text{The equation of the hyperbola is given by: }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1,{\text{ with}} \cr
& {\text{Vertices }}\left( {0, \pm a} \right),\,\,\,a = 4 \cr
& {\text{Then,}} \cr
& {\text{Asymptotes: }}y = \frac{a}{b}x{\text{ and }}y = - \frac{a}{b}x \cr
& \frac{a}{b}x = 2x \cr
& \frac{4}{b}x = 2x \cr
& b = 2 \cr
& {\text{The equation of the hyperbola is }} \cr
& \frac{{{y^2}}}{{{{\left( 4 \right)}^2}}} - \frac{{{x^2}}}{{{{\left( 2 \right)}^2}}} = 1 \cr
& \frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{4} = 1 \cr
& \cr
& {\text{Foci }}\left( {0, \pm c} \right) \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {20} = 2\sqrt 5 \cr
& {\text{Foci }}\left( {0, \pm 2\sqrt 5 } \right) \cr
& \cr
& {\text{Graph}} \cr} $$
