Answer
$$\frac{{{y^2}}}{9} - \frac{{{{\left( {x - 2} \right)}^2}}}{{16}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {2, \pm 3} \right) \cr
& {\text{Vertices }}\left( {\underbrace 2_x,\underbrace 3_y} \right){\text{ and }}\left( {\underbrace 2_x,\underbrace { - 3}_y} \right) \cr
& x = x{\text{ so the hyperbola has the standard form}} \cr
& \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{Characteristics of the hyperbola}} \cr
& {\text{Vertices }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right) \cr
& {\text{Vertices }}\left( {2,\underbrace { - 3}_{ - a},} \right){\text{ and }}\left( {0,\underbrace 3_a} \right) \cr
& k - a = - 3 \cr
& k + a = 3 \cr
& k = 0,{\text{ }}a = 3,{\text{ }}h = 2 \cr
& {\text{Foci }}\left( {h,k - c} \right){\text{ and }}\left( {h,k + c} \right) \cr
& {\text{Foci }}\left( {2,\underbrace { - 5}_{k - c},} \right){\text{ and }}\left( {0,\underbrace 5_{k + c}} \right) \cr
& k - c = - 5 \to c = 5 \cr
& {b^2} = {c^2} - {a^2} = {5^2} - {3^2} = 16 \cr
& \cr
& \underbrace {\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1}_ \downarrow \cr
& \frac{{{{\left( {y - 0} \right)}^2}}}{9} - \frac{{{{\left( {x - 2} \right)}^2}}}{{16}} = 1 \cr
& \frac{{{y^2}}}{9} - \frac{{{{\left( {x - 2} \right)}^2}}}{{16}} = 1 \cr} $$
