Let the length of the latus rectum of an ellipse with its

Let the ellipse be $\frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}}=1$

Then, $\frac{2 b^{2}}{a}=8,2 a e=b^{2}$ and $b^{2}=a^{2}\left(1-e^{2}\right)$\

$\Rightarrow \quad a=8, b^{2}=32$

Then, the equation of the ellipse

$\frac{x^{2}}{64}+\frac{y^{2}}{32}=1$

Hence, the point $(4 \sqrt{3}, 2 \sqrt{2})$ lies on the ellipse.