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Solution :

Since the foci on the x-axis, the equation of the hyperbola is of the form `(x^(2))/(a^(2))-(y^(2))/(b^(2)) =1`. <br> Since, vertices are `(pm3, 0)`, therefore a = 3 i.e., `a^(2) = 9` <br> Also, since foci are `(pm5, 0)`, therefore c = 5 <br> Now, `b^(2) = c^(2) - a^(2)` <br> `rArr b^(2) = 25 - 9` <br> `rArr b^(2) =- 16` <br> Therefore, the equation of the hyperbola is `(x^(2))/(9) - (y^(2))/(16) = 1`.