The $y$-coordinate of foci is zero $\therefore$ Major axis is on $\mathrm{X}$-axis
$$
\therefore \quad \text { ae }=4
$$
Let, equation of ellipse be
$$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\
& \therefore \quad \frac{(4 \sqrt{2})^2}{a^2}+\frac{(2 \sqrt{6})^2}{a^2-16} \\
& {\left[\because b^2=a^2\left(1-e^2\right)=a^2-16\right.} \\
& \Rightarrow \quad \frac{32}{a^2}+\frac{24}{a^2-16}=1 \\
& \Rightarrow \quad 32 a^2-512+24 a^2=a^2\left(a^2-16\right) \\
& \Rightarrow \quad 56 a^2-512=a^4-16 a^2 \\
& \Rightarrow \quad a^4-72 a^2+512=0 \\
& \Rightarrow \quad a^2-64 a^2-8 a^2+512=0 \\
& \Rightarrow \quad a^2\left(a^2-64\right)-8\left(a^2-64\right)=0 \\
& \Rightarrow \quad\left(a^2-8\right)\left(a^2-64\right)=0 \\
& \Rightarrow \quad a^2=64 \Rightarrow a=8 \\
& \left(\because a^2=8\right. \text { is not possible) } \\
& \because \quad \text { ae }=4 \Rightarrow 8 \times e=4 \\
&
\end{aligned}
$$
$\Rightarrow \quad e=\frac{1}{2}$