Answer
$$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1$$
Work Step by Step
$$\eqalign{
& e = \frac{1}{2};\,\,\,{\text{vertices at }}\left( { - 4,0} \right),\,\,\left( {4,0} \right) \cr
& {\text{The }}y{\text{ coordinate is the same, then the ellipse has the equation}} \cr
& \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,\,\,\,\left( {{\text{where }}a > b} \right){\text{ and vertices }}\left( { \pm a,0} \right) \cr
& {\text{,then}} \cr
& a = 4 \cr
& e = \frac{c}{a} = \frac{1}{2} \cr
& c = ea = \frac{1}{2}\left( 4 \right) = 2 \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = {4^2} - {2^2} \cr
& {b^2} = 12 \cr
& \cr
& \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{Substituting }}a{\text{ and }}b \cr
& \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1 \cr} $$
