Answer
$$\frac{{{y^2}}}{9} - \frac{{{{\left( {x - 2} \right)}^2}}}{{9/4}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {2, \pm 3} \right) \cr
& {\text{Vertices }}\left( {\underbrace 2_x,\underbrace 3_y} \right){\text{ and }}\left( {\underbrace 2_x,\underbrace { - 3}_y} \right) \cr
& x = x{\text{ so the hyperbola has the standard form}} \cr
& \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{Characteristics of the hyperbola}} \cr
& {\text{Vertices }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right) \cr
& {\text{Vertices }}\left( {2,\underbrace { - 3}_{k - a},} \right){\text{ and }}\left( {0,\underbrace 3_{k + a}} \right) \cr
& k - a = - 3 \cr
& k + a = 3 \cr
& k = 0,{\text{ }}a = 3,{\text{ }}h = 2 \cr
& \cr
& \frac{{{{\left( {y - 0} \right)}^2}}}{{{3^2}}} - \frac{{{{\left( {x - 2} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{{y^2}}}{{{3^2}}} - \frac{{{{\left( {x - 2} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{We know the point on the graph }}\left( {0,5} \right) \cr
& \frac{{{{\left( 5 \right)}^2}}}{{{3^2}}} - \frac{{{{\left( {0 - 2} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{25}}{9} - \frac{4}{{{b^2}}} \to {b^2} = \frac{9}{4} \cr
& \cr
& \underbrace {\frac{{{y^2}}}{{{3^2}}} - \frac{{{{\left( {x - 2} \right)}^2}}}{{{b^2}}} = 1}_ \Downarrow \cr
& \frac{{{y^2}}}{9} - \frac{{{{\left( {x - 2} \right)}^2}}}{{9/4}} = 1 \cr} $$
