Answer
$$\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{24}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Center: }}\left( {0,0} \right) \cr
& {\text{Focus: }}\left( {5,0} \right) \cr
& {\text{Vertex: }}\left( {7,0} \right) \cr
& {\text{The }}y{\text{ - coordinate is the same, so the equation of the ellipse}} \cr
& {\text{is of the form }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{\left( {y - k} \right)}}{{{b^2}}} = 1 \cr
& {\text{With:}} \cr
& {\text{Center: }}\left( {h,k} \right):\left( {0,0} \right) \to h = 0,{\text{ }}k = 0 \cr
& {\text{Foci: }}\left( {h \pm c,k} \right) \to \left( {5,0} \right) \to h + c = 5 \to c = 5 \cr
& {\text{Vertices: }}\left( {h \pm a,k} \right) \to h + a = 7 \to a = 7 \cr
& {b^2} = {a^2} - {c^2} = {7^2} - {5^2} = 24 \cr
& \underbrace {\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{\left( {y - k} \right)}}{{{b^2}}} = 1}_ \downarrow \cr
& \frac{{{{\left( {x - 0} \right)}^2}}}{{{7^2}}} + \frac{{{{\left( {y - 0} \right)}^2}}}{{24}} = 1 \cr
& \frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{24}} = 1 \cr} $$
