Answer
$$\frac{{{x^2}}}{{40}} + \frac{{{y^2}}}{{121}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Foci: }}\left( {0, \pm 9} \right) \cr
& {\text{Major axis length: 22}} \cr
& {\text{22 = 2}}a \cr
& a = 11 \cr
& {\text{Foci}}\left( {\underbrace 0_x,\underbrace { - 9}_y} \right){\text{ and }}\left( {\underbrace 0_x,\underbrace 9_y} \right) \cr
& {\text{Standard form of the equation}}:\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1 \cr
& {\text{With}} \cr
& {\text{Foci}}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr
& c = 9 \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = {11^2} - {9^2} \cr
& {b^2} = 40 \cr
& \underbrace {\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1}_ \Downarrow \cr
& \frac{{{x^2}}}{{40}} + \frac{{{y^2}}}{{121}} = 1 \cr} $$
