Answer
$${\text{The equation describes an ellipse}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 1 \cr
& \left( a \right) \cr
& {\text{The equation is written in the form }}\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1,{\text{ }}a > b > 0 \cr
& {\text{then, the equation describes an ellipse}}{\text{.}} \cr
& \underbrace {\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 1}_{\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1} \to a = 5,{\text{ }}b = 2 \cr
& c = \sqrt {{a^2} - {b^2}} = \sqrt {25 - 4} = \sqrt {21} \cr
& \cr
& \left( b \right) \cr
& {\text{Foci: }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr
& {\text{Foci: }}\left( {0, - \sqrt {21} } \right){\text{ and }}\left( {0,\sqrt {21} } \right) \cr
& {\text{Vertices: }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr
& {\text{Vertices: }}\left( {0, - 5} \right){\text{ and }}\left( {0,5} \right) \cr
& \cr
& \left( c \right) \cr
& {\text{Eccentricity }} \cr
& e = \frac{c}{a} \cr
& e = \frac{{\sqrt {21} }}{5} \cr
& \cr
& \left( d \right) \cr
& {\text{Graph}} \cr} $$
