Answer
$$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$$
Work Step by Step
$$\eqalign{
& e = \frac{4}{5};\,\,\,{\text{vertices at }}\left( { - 5,0} \right),\,\,\left( {5,0} \right) \cr
& {\text{The excentricity of the ellipse is }}e = \frac{c}{a} \cr
& e = \frac{4}{5}\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,c = 4{\text{ }}\,\,\,\,\,{\text{and }}\,\,\,\,\,a = 5 \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = 25 - 16 \cr
& {b^2} = 9 \cr
& \cr
& {\text{vertices at }}\left( { - 5,0} \right),\,\,\left( {5,0} \right) \cr
& {\text{The }}y{\text{ coordinate is the same, then the ellipse has the equation}} \cr
& \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{Substituting }}a{\text{ and }}b \cr
& \frac{{{x^2}}}{{{{\left( 5 \right)}^2}}} + \frac{{{y^2}}}{9} = 1 \cr
& \frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1 \cr} $$
