Answer
$\frac{{{x^2}}}{{49}} - \frac{{{{\left( {y + 1} \right)}^2}}}{{32}} = 1{\text{ }}$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( { \pm 7, - 1} \right) \cr
& {\text{Foci: }}\left( { \pm 9, - 1} \right) \cr
& y{\text{ is constant}}{\text{, then the equation of the hyperbola is }} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Foci: }}\left( {h - c,k} \right){\text{ and }}\left( {h + c,k} \right) \cr
& h - c = - 9,{\text{ }}h + c = 9 \to h = 0,{\text{ }}c = 9 \cr
& {\text{Vertices: }}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right) \cr
& h - a = - 7,{\text{ }}h + a = 7 \to h = 0,{\text{ }}a = 7 \cr
& \cr
& {c^2} = {a^2} + {b^2},{\text{ then}} \cr
& {b^2} = {c^2} - {a^2} \cr
& {b^2} = {9^2} - {7^2} \cr
& {b^2} = 32 \cr
& \cr
& {\text{Substituting into }}\left( {\bf{1}} \right) \cr
& \frac{{{{\left( {x - 0} \right)}^2}}}{{49}} - \frac{{{{\left( {y - \left( { - 1} \right)} \right)}^2}}}{{32}} = 1{\text{ }} \cr
& \frac{{{x^2}}}{{49}} - \frac{{{{\left( {y + 1} \right)}^2}}}{{32}} = 1{\text{ }} \cr} $$
