Answer
$$\frac{{5{y^2}}}{{16}} - \frac{{{x^2}}}{4} = 1$$
Work Step by Step
$$\eqalign{
& {\text{hyperbola; }}y{\text{ - intercept }}\left( {0, - 2} \right),{\text{ passing through the point }}\left( {2,3} \right) \cr
& {\text{The parabola intercept the }}y{\text{ - axis then the standard equation is}} \cr
& \frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& {\text{The }}y{\text{ - intercepts are at }}\left( {0, \pm a} \right){\text{, }}\,\,\left( {0, - 2} \right)\, \to a = 2 \cr
& {\text{Passing through the point }}\left( {2,3} \right).{\text{ so,}} \cr
& \frac{{{{\left( 3 \right)}^2}}}{{{{\left( 2 \right)}^2}}} - \frac{{{{\left( 2 \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{Solve for }}b \cr
& \frac{9}{4} - \frac{4}{{{b^2}}} = 1 \cr
& \frac{4}{{{b^2}}} = \frac{5}{4} \cr
& {b^2} = \frac{{16}}{5} \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{5{y^2}}}{{16}} - \frac{{{x^2}}}{4} = 1 \cr} $$
