Answer
$$\frac{{{{\left( {x - 8} \right)}^2}}}{9} - \frac{{{{\left( {y - 7} \right)}^2}}}{{16}} = 1$$
Work Step by Step
$$\eqalign{
& {\bf{e}} = \frac{5}{3};{\text{ center at }}\left( {8,7} \right);{\text{ focus at }}\left( {3,7} \right) \cr
& {\text{The }}y{\text{ - coordinate in the focus and center are the same, then }} \cr
& {\text{the hyperbola has the equation }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{center at }}\left( {h,k} \right),\,\,\,h = 8,\,\,\,k = 7 \cr
& {\text{foci at }}\left( {h \pm c,k} \right){\text{,}}\,{\text{ focus at }}\left( {3,7} \right) \cr
& h - c = 3 \cr
& 8 - c = 3 \cr
& c = 5 \cr
& \cr
& {\text{excentricity }}{\bf{e}} = \frac{c}{a} \cr
& \frac{5}{3} = \frac{5}{a} \cr
& a = 3 \cr
& {b^2} = {c^2} - {a^2} \cr
& {b^2} = {5^2} - {3^2} \cr
& {b^2} = 16 \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{{\left( {x - 8} \right)}^2}}}{9} - \frac{{{{\left( {y - 7} \right)}^2}}}{{16}} = 1 \cr} $$
