Answer
$$\frac{{{{\left( {x - 4} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 3} \right)}^2}}}{7} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {0,3} \right),\left( {8,3} \right),{\text{ Eccentricity :}}\frac{3}{4} \cr
& {\text{Vertices }}\left( {\underbrace 0_x,\underbrace 3_y} \right){\text{and}}\left( {\underbrace 8_x,\underbrace 3_y} \right) \cr
& y = y{\text{, so the equation has the standard form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{With}} \cr
& {\text{Vertex }}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right) \cr
& {\text{Vertices }}\left( {\underbrace 0_{h - a},\underbrace 3_k} \right){\text{and}}\left( {\underbrace 8_{h + a},\underbrace 3_k} \right) \Rightarrow k = 3 \cr
& {\text{Find }}a{\text{ and }}h \cr
& h - a = 0 \cr
& h + a = 8 \cr
& 2h = 8 \cr
& h = 4 \cr
& a = 8 - h \Rightarrow a = 4 \cr
& {\text{Eccentricity: }}e = \frac{c}{a} \cr
& \frac{3}{4} = \frac{c}{4} \to c = 3 \cr
& {b^2} = {a^2} - {c^2} = 7 \cr
& \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{{{\left( {x - 4} \right)}^2}}}{{{4^2}}} + \frac{{{{\left( {y - 3} \right)}^2}}}{7} = 1 \cr
& \frac{{{{\left( {x - 4} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 3} \right)}^2}}}{7} = 1 \cr} $$
