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