10. (20 points) Let the line $y=x+\sqrt{2}$ intersect the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ at points $M, N$, and $O M \perp O N$ (where $O$ is the origin). If $M N=\sqrt{6}$, find the equation of the ellipse.
10. Let \( M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right) \). Substituting \( y=x+\sqrt{2} \) into the ellipse equation and eliminating \( y \), we get
\[
\frac{x^{2}}{a^{2}}+\frac{(x+\sqrt{2})^{2}}{b^{2}}=1,
\]

which simplifies to
\[
\left(a^{2}+b^{2}\right) x^{2}+2 \sqrt{2} a^{2} x+a^{2}\left(2-b^{2}\right)=0 .
\]

Since the intersection points \( M, N \) lie on the ellipse, \( x_{1}, x_{2} \) are the roots of the equation, so
\[
x_{1}+x_{2}=-\frac{2 \sqrt{2} a^{2}}{a^{2}+b^{2}}, \quad x_{1} x_{2}=\frac{a^{2}\left(2-b^{2}\right)}{a^{2}+b^{2}} .
\]

Since \( O M \perp O N \), then \( x_{1} x_{2}+y_{1} y_{2}=0 \), which means
\[
0=x_{1} x_{2}+\left(x_{1}+\sqrt{2}\right)\left(x_{2}+\sqrt{2}\right)=2 x_{1} x_{2}+\sqrt{2}\left(x_{1}+x_{2}\right)+2 \text{. }
\]

Substituting equation (1), we get
\[
\frac{2 a^{2}\left(2-b^{2}\right)}{a^{2}+b^{2}}-\frac{4 a^{2}}{a^{2}+b^{2}}+2=0,
\]

which simplifies to
\[
a^{2}+b^{2}=a^{2} b^{2} \text{. }
\]

Thus, equation (1) becomes
\[
x_{1}+x_{2}=-\frac{2 \sqrt{2}}{b^{2}}, \quad x_{1} x_{2}=\frac{2}{b^{2}}-1 .
\]

Given \( M N=\sqrt{6} \), we have
\[
\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=6 \text{. }
\]

Since \( M, N \) lie on the line \( y=x+\sqrt{2} \), it follows that \( x_{1}-x_{2}=y_{1}-y_{2} \), thus \( \left(x_{1}-x_{2}\right)^{2}=3 \), i.e.,
\[
\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=3 \text{. }
\]

From equation (3), we get
\[
\frac{8}{b^{4}}-4\left(\frac{2}{b^{2}}-1\right)=3,
\]

which simplifies to
\[
b^{4}-8 b^{2}+8=0 .
\]

From equations (2), (4), and \( a^{2}>b^{2} \), we solve to get \( a^{2}=4+2 \sqrt{2}, b^{2}=4-2 \sqrt{2} \). Therefore, the equation of the ellipse is
\[
\frac{x^{2}}{4+2 \sqrt{2}}+\frac{y^{2}}{4-2 \sqrt{2}}=1
\]