Answer
$$\frac{{{x^2}}}{{51}} + \frac{{{y^2}}}{{100}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Foci }}\left( {0, \pm 7} \right) \cr
& {\text{Foci }}\left( {0, \pm 7} \right) \to {\text{The orientation of the major axis is vertical}} \cr
& \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1,{\text{ }}a > b > 0{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Foci }}\underbrace {\left( {0, \pm 7} \right)}_{\left( {0, \pm c} \right)} \to c = 7 \cr
& {\text{Major axis length: 20}} \cr
& 2a = 20 \to a = 10 \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = {10^2} - {7^2} \cr
& {b^2} = 51 \cr
& \cr
& {\text{Substituting the constants }}{a^2}{\text{ and }}{b^2}{\text{ into }}\left( {\bf{1}} \right) \cr
& \frac{{{x^2}}}{{51}} + \frac{{{y^2}}}{{100}} = 1 \cr} $$
