Answer
$$\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{36}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{foci at }}\left( { - 3\sqrt 5 ,0} \right),\left( {3\sqrt 5 ,0} \right);{\text{ asymptotes }}y = \pm 2x \cr
& {\text{The }}y{\text{ - coordinate in the foci are the same, then the hyperbola}} \cr
& {\text{has the equation }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{With foci at }}\left( { \pm c,0} \right){\text{ and asymptotes }}y = \pm \frac{b}{a}x \cr
& \frac{b}{a} = 2 \cr
& b = 2a \cr
& \cr
& {\text{foci at }}\left( { \pm 3\sqrt 5 ,0} \right) \to c = 3\sqrt 5 \cr
& {c^2} = {a^2} + {b^2} \cr
& 45 = {a^2} + {b^2} \cr
& {\text{Substituting }}2a{\text{ for }}b \cr
& 45 = {a^2} + 4{a^2} \cr
& 45 = 5{a^2} \cr
& {a^2} = 9 \cr
& \cr
& {b^2} = 45 - {a^2} \cr
& {b^2} = 45 - 9 \cr
& {b^2} = 36 \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{9} - \frac{{{y^2}}}{{36}} = 1 \cr} $$
