Answer
$$\frac{{9{x^2}}}{{28}} + \frac{{9{y^2}}}{{64}} = 1$$
Work Step by Step
$$\eqalign{
& e = \frac{3}{4};\,\,\,{\text{foci at }}\left( {0, - 2} \right),\,\,\left( {0,2} \right) \cr
& {\text{The }}x{\text{ coordinate is the same, then the ellipse has the equation}} \cr
& \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1\,\,\,\,\,\,\left( {{\text{where }}a > b} \right){\text{ and foci }}\left( {0, \pm c} \right) \cr
& {\text{,then}} \cr
& c = 2 \cr
& e = \frac{c}{a} = \frac{3}{4} \cr
& a = \frac{c}{3} = \frac{2}{{3/4}} = \frac{8}{3} \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = {\left( {\frac{8}{3}} \right)^2} - {2^2} \cr
& {b^2} = \frac{{28}}{9} \cr
& \cr
& \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{Substituting }}a{\text{ and }}b \cr
& \frac{{{x^2}}}{{\frac{{28}}{9}}} + \frac{{{y^2}}}{{\frac{{64}}{9}}} = 1 \cr
& \frac{{9{x^2}}}{{28}} + \frac{{9{y^2}}}{{64}} = 1 \cr} $$
