Answer
$$\,\frac{{{x^2}}}{{25}} + {y^2} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( { \pm 5,0} \right){\text{ passing through the point }}\left( {4,\frac{3}{5}} \right) \cr
& {\text{The equation of the ellipse is given by: }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,{\text{ with}} \cr
& {\text{Vertices }}\left( { \pm a,0} \right),\,\,\,a = 5 \cr
& \frac{{{x^2}}}{{{{\left( 5 \right)}^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{passing through the point }}\left( {4,\frac{3}{5}} \right) \cr
& \frac{{{{\left( 4 \right)}^2}}}{{25}} + \frac{{{{\left( {3/5} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{16}}{{25}} + \frac{9}{{25{b^2}}} = 1 \cr
& 16 + \frac{9}{{{b^2}}} = 25 \cr
& \frac{9}{{{b^2}}} = 9 \cr
& b = 1 \cr
& \cr
& {\text{Then, the equation is}} \cr
& \,\frac{{{x^2}}}{{25}} + {y^2} = 1 \cr
& \cr
& {\text{Graph}} \cr} $$
