An ellipse passing through $(4 \sqrt{2},2 \sqrt{6})$ has foci at (–4,0) and (4,0). Its eccentricity is

Given $\mathrm{ae}=4$

$$\begin{gathered} {\mathrm{a}}^2 {\mathrm{e}}^2=16 \\ {\mathrm{a}}^2-{\mathrm{b}}^2=16 \end{gathered}$$

equation  of ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ passes through $(4\sqrt{2}, 2\sqrt{6})$

$$\begin{gathered} \frac{32}{{\mathrm{a}}^2}+\frac{24}{{\mathrm{a}}^2-16}=1 \\ 32{\mathrm{a}}^2-32\times 16+24{\mathrm{a}}^2={\mathrm{a}}^4-16{\mathrm{a}}^2 \\ {\mathrm{a}}^4-72{\mathrm{a}}^2+5-12=0 \\ ({\mathrm{a}}^2-64) ({\mathrm{a}}^2-8)=0 \\ {\mathrm{a}}^2=64 \Rightarrow \mathrm{a}=8 \\ \mathrm{e}=\frac{4}{\mathrm{a}}=\frac{4}{8}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$