Answer
$$\frac{{{{\left( {x - 2} \right)}^2}}}{{41}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{16}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{foci at }}\left( { - 3, - 3} \right){\text{ and }}\left( {7, - 3} \right) \cr
& {\text{The }}y{\text{ coordinate is the same, then the ellipse has the equation}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{With foci }}\left( {h \pm c,k} \right),\,\,\,k = - 3 \cr
& h + c = 7{\text{ and }}h - c = - 3 \cr
& {\text{Solving we obtain}} \cr
& h = 2,\,\,\,\,c = 5 \cr
& \cr
& {\text{We know the point }}\left( {2, - 7} \right){\text{ lies on the ellipse, then}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{{{\left( {2 - 2} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( { - 7 + 3} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{16}}{{{b^2}}} = 1\,\,\, \cr
& {b^2} = 16 \cr
& \cr
& {a^2} = {b^2} + {c^2} \cr
& {a^2} = 16 + {\left( 5 \right)^2} \cr
& {a^2} = 41 \cr
& \cr
& {\text{The equation of the ellipse is}} \cr
& \frac{{{{\left( {x - 2} \right)}^2}}}{{41}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{16}} = 1 \cr} $$
