Answer
$$\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 1$$
Work Step by Step
$$\eqalign{
& x{\text{ - intercepts }}\left( { \pm 3,0} \right);{\text{ foci }}\left( { - 5,0} \right),\left( {5,0} \right) \cr
& {\text{The }}y{\text{ coordinate in the foci are the same, then the hyperbola}} \cr
& {\text{has the equation }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{With }}x{\text{ - intercepts at }}\left( { \pm a,0} \right){\text{ and foci }}\left( { \pm c,0} \right) \cr
& {\text{,Then}} \cr
& {\text{foci }}\left( { - 5,0} \right),\left( {5,0} \right) \to \,\,\,c = 5 \cr
& x{\text{ - intercepts at }}\left( { \pm 3,0} \right) \to \,\,a = 3 \cr
& {b^2} = {c^2} - {a^2} \cr
& {b^2} = 25 - 9 \cr
& {b^2} = 16 \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 1 \cr} $$
