Answer
$$\frac{{25{{\left( {x + 1} \right)}^2}}}{{324}} - \frac{{25{{\left( {y + 1} \right)}^2}}}{{2176}} = 1$$
Work Step by Step
$$\eqalign{
& {\bf{e}} = \frac{{25}}{9};{\text{ foci at }}\left( {9, - 1} \right){\text{, }}\left( { - 11, - 1} \right) \cr
& {\text{The }}y{\text{ - coordinate in the focus and center are the same, then }} \cr
& {\text{the hyperbola has the equation }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{foci at }}\left( {h \pm c,k} \right){\text{,}}\,{\text{ foci at }}\left( {9, - 1} \right){\text{, }}\left( { - 11, - 1} \right),\,\,\,\,\,\,\,\,\,k = - 1 \cr
& h + c = 9{\text{ and }}h - c = - 11 \cr
& {\text{Solving both equations, we obtain}} \cr
& h = - 1,\,\,\,\,c = 10 \cr
& \cr
& {\text{excentricity }}{\bf{e}} = \frac{c}{a} \cr
& \frac{{25}}{9} = \frac{{10}}{a} \cr
& a = \frac{{18}}{5} \cr
& \cr
& {b^2} = {c^2} - {a^2} \cr
& {b^2} = {10^2} - {\left( {\frac{{18}}{5}} \right)^2} \cr
& {b^2} = \frac{{2176}}{{25}} \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{25{{\left( {x + 1} \right)}^2}}}{{324}} - \frac{{25{{\left( {y + 1} \right)}^2}}}{{2176}} = 1 \cr} $$
