Answer
$${\text{The equation describes a hyperbola}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& {y^2} - 4{x^2} = 16 \cr
& \frac{{{y^2}}}{{16}} - \frac{{4{x^2}}}{{16}} = \frac{{16}}{{16}} \cr
& \frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{4} = 1 \cr
& \left( a \right) \cr
& {\text{The equation is written in the form }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& {\text{then, the equation describes a hyperbola}}{\text{.}} \cr
& \underbrace {\frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{4} = 1}_{\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1} \to a = 4,{\text{ }}b = 2 \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {16 + 4} = \sqrt {20} = 2\sqrt 5 \cr
& \cr
& \left( b \right) \cr
& {\text{Foci: }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr
& {\text{Foci: }}\left( {0, - 2\sqrt 5 } \right){\text{ and }}\left( {0,2\sqrt 5 } \right) \cr
& {\text{Vertices: }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr
& {\text{Vertices: }}\left( {0, - 4} \right){\text{ and }}\left( {0,4} \right) \cr
& \cr
& \left( c \right) \cr
& {\text{Eccentricity }} \cr
& e = \frac{c}{a} \cr
& e = \frac{{2\sqrt 5 }}{4} \cr
& e = \frac{{\sqrt 5 }}{2} \cr
& \cr
& \left( d \right) \cr
& {\text{Graph}} \cr} $$
