Answer
$$\frac{{{{\left( {y + 7} \right)}^2}}}{{36}} - \frac{{{{\left( {x - 9} \right)}^2}}}{{64}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{center at }}\left( {9, - 7} \right);{\text{ focus at }}\left( {9, - 17} \right);{\text{ vertex at }}\left( {9, - 13} \right) \cr
& {\text{The }}x{\text{ - coordinate in the vertex and focus are the same, then }} \cr
& {\text{the hyperbola has the equation }}\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{With center at }}\left( {h,k} \right) \cr
& {\text{center at }}\left( {9, - 7} \right) \to h = 9,\,\,k = - 7 \cr
& {\text{vertices at }}\left( {h,k \pm a} \right){\text{,}}\,{\text{ vertex at }}\left( {9, - 13} \right) \cr
& k - a = - 13 \cr
& - 7 - a = - 13 \cr
& a = 6 \cr
& \cr
& {\text{foci at }}\left( {h,k \pm c} \right),{\text{focus at }}\left( {9, - 17} \right) \cr
& k - c = - 17 \cr
& - 7 - c = - 17 \cr
& c = 10 \cr
& {b^2} = {c^2} - {a^2} \cr
& {b^2} = 100 - 36 = 64 \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{{\left( {y + 7} \right)}^2}}}{{36}} - \frac{{{{\left( {x - 9} \right)}^2}}}{{64}} = 1 \cr} $$
