Answer
$$\frac{{{{\left( {y - 3} \right)}^2}}}{4} - \frac{{49{{\left( {x - 4} \right)}^2}}}{4} = 1$$
Work Step by Step
$$\eqalign{
& {\text{ertices at }}\left( {4,5} \right),\left( {4,1} \right);{\text{ asymptotes }}y = \pm 7\left( {x - 4} \right) + 3 \cr
& {\text{The }}x{\text{ - coordinate in the vertices are the same, then the hyperbola}} \cr
& {\text{has the equation }}\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{With vertices at }}\left( {h,k \pm a} \right){\text{ and asymptotes }}y = \pm \frac{a}{b}\left( {x - h} \right) + k \cr
& {\text{,then}} \cr
& {\text{vertices }}\left( {4,5} \right),\left( {4,1} \right) \to \,\,\,h = 4 \cr
& {\text{asymptotes }}y = \pm 7\left( {x - 4} \right) + 3 \to k = 3 \cr
& k + a = 5 \cr
& 3 + a = 5 \cr
& a = 2 \cr
& \cr
& \frac{a}{b} = 7 \cr
& b = \frac{2}{7} \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{{\left( {y - 3} \right)}^2}}}{{{2^2}}} - \frac{{{{\left( {x - 4} \right)}^2}}}{{{{\left( {2/7} \right)}^2}}} = 1 \cr
& \frac{{{{\left( {y - 3} \right)}^2}}}{4} - \frac{{49{{\left( {x - 4} \right)}^2}}}{4} = 1 \cr} $$
