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