Answer
$$\eqalign{
& {\text{Vertices: }}\left( { \pm 3,0} \right);\,\,{\text{Foci:}}\,\,\left( { \pm \sqrt 5 ,0} \right) \cr
& {\text{major axis has length 6}} \cr
& {\text{minor axis has length 4}} \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1 \cr
& {\text{The equation has the standard form}} \cr
& \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1 \Rightarrow a = 3,b = 2 \cr
& c = \sqrt {{a^2} - {b^2}} = \sqrt {{3^2} - {2^2}} = \sqrt 5 \cr
& {\text{With}} \cr
& {\text{Orientation of Major Axis Horizontal along the }}x{\text{ - axis, then}} \cr
& {\text{Vertex }}\left( { - a,0} \right){\text{ and }}\left( {a,0} \right) \cr
& {\text{Vertex}}\left( { - 3,0} \right){\text{ and }}\left( {3,0} \right) \cr
& \cr
& {\text{Foci }}\left( { - c,0} \right){\text{ and }}\left( {c,0} \right) \cr
& {\text{Foci}}\left( { - \sqrt 5 ,0} \right){\text{ and }}\left( {\sqrt 5 ,0} \right) \cr
& \cr
& {\text{Length of the minor axis}} \cr
& 2b = 2\left( 2 \right) = 4 \cr
& {\text{Length of the major axis}} \cr
& 2a = 2\left( 3 \right) = 6 \cr
& \cr
& {\text{Graph }} \cr} $$
