Answer
$$\frac{{2{y^2}}}{{25}} - 2{x^2} = 1$$
Work Step by Step
$$\eqalign{
& {\text{foci at }}\left( {0,\sqrt {13} } \right),\left( {0, - \sqrt {13} } \right);{\text{ asymptotes }}y = \pm 5x \cr
& {\text{The }}x{\text{ - coordinate in the foci are the same, then the hyperbola}} \cr
& {\text{has the equation }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& {\text{With foci at }}\left( {0, \pm c} \right){\text{ and asymptotes }}y = \pm \frac{a}{b}x \cr
& \frac{a}{b} = 5 \cr
& a = 5b \cr
& \cr
& {\text{foci at }}\left( {0, \pm \sqrt {13} } \right) \to c = \sqrt {13} {\text{ }} \cr
& {c^2} = {a^2} + {b^2} \cr
& 13 = {a^2} + {b^2} \cr
& {\text{Substituting }}5b{\text{ for }}a \cr
& 13 = {\left( {5b} \right)^2} + {b^2} \cr
& 13 = 25{b^2} + {b^2} \cr
& 26{b^2} = 13 \cr
& {b^2} = \frac{1}{2} \cr
& \cr
& {a^2} = 25{b^2} \cr
& {a^2} = \frac{{25}}{2} \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& \frac{{{y^2}}}{{25/2}} - \frac{{{x^2}}}{{1/2}} = 1 \cr
& \frac{{2{y^2}}}{{25}} - 2{x^2} = 1 \cr} $$
