3. In an isosceles right triangle $\triangle A B C$, the hypotenuse $B C=4 \sqrt{2}$, an ellipse has $C$ as one of its foci, the other focus lies on the line segment $A B$, and the ellipse passes through points $A, B$, then the standard equation of the ellipse (with foci on the $x$-axis) is $\qquad$
3. $\frac{x^{2}}{6+4 \sqrt{2}}+\frac{y^{2}}{4 \sqrt{2}}=1$ Detailed Explanation: 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 as the origin. Assume the equation of the ellipse is: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, so $|A D|+|B D|+|A C|+|B C|=$ $4 a$. That is, $8+4 \sqrt{2}=4 a, a=2+\sqrt{2} .|A D|=2 a-|A C|=2 \sqrt{2}$. In the right triangle $A D C$, $|C D|^{2}=8+16=24$, $c^{2}=6, b^{2}=a^{2}-c^{2}=4 \sqrt{2}$, hence the equation $\frac{x^{2}}{6+4 \sqrt{2}}+\frac{y^{2}}{4 \sqrt{2}}=1$ is the required equation.