(13) The line $x+y=1$ intersects the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at points $A$ and $B$, and $C$ is the midpoint of segment $AB$. If $|AB|=2 \sqrt{2}$, and the slope of line $OC$ is $k_C=\frac{1}{\sqrt{2}}$, find the equation of the ellipse.
(13) Let $\frac{1}{a^{2}}=\alpha, \frac{1}{b^{2}}=\beta$, substituting $y=1-x$ into the ellipse equation $\alpha x^{2}+\beta y^{2}=1$ yields
$$
(\alpha+\beta) x^{2}-2 \beta x+(\beta-1)=0 .
$$

Let the coordinates of points $A$ and $B$ be $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, respectively, then
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{2 \beta}{\alpha+\beta}, \\
x_{1} x_{2}=\frac{\beta-1}{\alpha+\beta},
\end{array}
$$

Therefore,
$$
\begin{aligned}
|A B|^{2} & =\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2} \\
& =2\left(x_{1}-x_{2}\right)^{2} \\
& =2\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right] \\
& =\frac{8(\alpha+\beta-\alpha \beta)}{(\alpha+\beta)^{2}},
\end{aligned}
$$

Given $|A B|=2 \sqrt{2}$, we have
$$
\frac{\alpha+\beta-\alpha \beta}{(\alpha+\beta)^{2}}=1,
$$

which simplifies to
$$
\alpha^{2}+3 \alpha \beta+\beta^{2}-\alpha-\beta=0 .
$$

Let point $C\left(x_{0}, y_{0}\right)$, then
$$
\begin{array}{l}
x_{0}=\frac{x_{1}+x_{2}}{2}=\frac{\beta}{\alpha+\beta}, \\
y_{0}=1-x_{0}=\frac{\alpha}{\alpha+\beta},
\end{array}
$$

Given $k_{O C}=\frac{1}{\sqrt{2}}$, we have $\frac{x_{0}}{y_{0}}=\sqrt{2}$, so
$$
\beta=\sqrt{2} \alpha .
$$

From equations (1) and (2), we get $\alpha=\frac{1}{3}, \beta=\frac{\sqrt{2}}{3}$, thus the equation of the ellipse is $\frac{x^{2}}{3}+\frac{\sqrt{2} y^{2}}{3}=1$.