Answer
$$\eqalign{
& {\text{Vertices: }}\left( {0, \pm 3} \right);\,\,{\text{Foci:}}\,\,\left( {0, \pm 2\sqrt 2 } \right) \cr
& {\text{major axis has length 6}} \cr
& {\text{minor axis has length 2}} \cr} $$
Work Step by Step
$$\eqalign{
& {x^2} + \frac{{{y^2}}}{9} = 1 \cr
& {\text{The equation has the standard form}} \cr
& \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1,\,\,\,\,\,a > b > 0 \cr
& {x^2} + \frac{{{y^2}}}{9} = 1 \Rightarrow a = 3,b = 1 \cr
& c = \sqrt {{a^2} - {b^2}} = \sqrt {{3^2} - {1^2}} = 2\sqrt 2 \cr
& {\text{With}} \cr
& {\text{Orientation of Major Axis Horizontal along the }}x{\text{ - axis, then}} \cr
& {\text{Vertices }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr
& {\text{Vertices }}\left( {0, - 3} \right){\text{ and }}\left( {0,3} \right) \cr
& \cr
& {\text{Foci }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr
& {\text{Foci}}\left( {0, - 2\sqrt 2 } \right){\text{ and }}\left( {0,2\sqrt 2 } \right) \cr
& \cr
& {\text{Length of the minor axis}} \cr
& 2b = 2\left( 1 \right) = 2 \cr
& {\text{Length of the major axis}} \cr
& 2a = 2\left( 3 \right) = 6 \cr
& \cr
& {\text{Graph }} \cr} $$
