Answer
$$\eqalign{
& {\text{Vertices: }}\left( {0, \pm 4} \right);\,\,{\text{Foci:}}\,\,\left( {0, \pm \sqrt 5 } \right) \cr
& {\text{major axis has length 8}} \cr
& {\text{minor axis has length 4}} \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{16}} = 1 \cr
& {\text{The equation has the standard form}} \cr
& \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1,\,\,\,\,\,a > b > 0 \cr
& \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{16}} = 1 \Rightarrow a = 4,b = 2 \cr
& c = \sqrt {{a^2} - {b^2}} = \sqrt {{4^2} - {2^2}} = 2\sqrt 3 \cr
& {\text{With}} \cr
& {\text{Orientation of Major Axis Horizontal along the }}x{\text{ - axis, then}} \cr
& {\text{Vertices }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr
& {\text{Vertices }}\left( {0, - 4} \right){\text{ and }}\left( {0,4} \right) \cr
& \cr
& {\text{Foci }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr
& {\text{Foci}}\left( {0, - \sqrt 5 } \right){\text{ and }}\left( {0,\sqrt 5 } \right) \cr
& \cr
& {\text{Length of the minor axis}} \cr
& 2b = 2\left( 2 \right) = 4 \cr
& {\text{Length of the major axis}} \cr
& 2a = 2\left( 4 \right) = 8 \cr
& \cr
& {\text{Graph }} \cr} $$
