5. In an isosceles right triangle $\triangle ABC$, the hypotenuse $BC=4\sqrt{2}$. An ellipse has $C$ as one of its foci, the other focus lies on the line segment $AB$, and the ellipse passes through points $A$ and $B$. The standard equation of the ellipse (with foci on the $x$-axis) is ( ).
(A) $\frac{x^{2}}{6+4 \sqrt{2}}+\frac{y^{2}}{4 \sqrt{2}}=1$
(B) $\frac{x^{2}}{6+4 \sqrt{2}}+\frac{y^{2}}{3+4 \sqrt{2}}=1$
(C) $\frac{x^{2}}{4 \sqrt{2}}+\frac{y^{2}}{6+4 \sqrt{2}}=1$
(D) $\frac{x^{2}}{3+4 \sqrt{2}}+\frac{y^{2}}{6+4 \sqrt{2}}=1$
5.A.

Since $B C=4 \sqrt{2}$, let the other focus of the ellipse be $D$. Establish a rectangular coordinate system with $D C$ as the $x$-axis and the midpoint of $D C$ as the origin. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. Therefore,
$$
|A D|+|B D|+|A C|+|B C|=4 a,
$$

which means $8+4 \sqrt{2}=4 a$.
Solving for $a$ gives $a=2+\sqrt{2}$.
Thus, $|A D|=2 a-|A C|=2 \sqrt{2}$.
In the right triangle $\triangle A D C$, since
$|C D|^{2}=8+16=24, c^{2}=6, b^{2}=a^{2}-c^{2}=4 \sqrt{2}$, the equation $\frac{x^{2}}{6+4 \sqrt{2}}+\frac{y^{2}}{4 \sqrt{2}}=1$ is the desired equation.