Answer
$$\frac{{{x^2}}}{{40}} + \frac{{{y^2}}}{{49}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{ellipse with foci at }}\left( {0, \pm 3} \right){\text{and vertex }}\left( {0, - 7} \right) \cr
& {\text{The coordinates of the foci are }}\left( {0, \pm c} \right),{\text{ then}} \cr
& {\text{The equation of the ellipse is of the form }}\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1 \cr
& {\text{Foci at }}\left( {0, \pm 3} \right),\,\,\left( {0, \pm c} \right) \to c = 3 \cr
& {\text{Vertices }}\left( {0, \pm a} \right),\,\left( {0, - 7} \right) \to a = 7 \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = {7^2} - {3^2} = 40 \cr
& \cr
& {\text{The equation of the ellipse is}} \cr
& \frac{{{x^2}}}{{40}} + \frac{{{y^2}}}{{49}} = 1 \cr} $$
