10. (20 points) Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with its upper vertex $B$ and right focus $F$, a line $l$ intersects the ellipse at points $M$ and $N$. Determine whether there exists a line $l$ such that point $F$ is the (1) centroid; (2) orthocenter of $\triangle B M N$. If it exists, find the equation of the corresponding line $l$; if not, explain the reason.
10. (1) Suppose there exists a line $l$ such that point $F$ is the centroid of $\triangle B M N$. Let the coordinates of points $M, N$ be $M(\sqrt{2} \cos \alpha, \sin \alpha), N(\sqrt{2} \cos \beta, \sin \beta)$. According to the problem, we have
$$
\left\{\begin{array}{l}
\sqrt{2} \cos \alpha+\sqrt{2} \cos \beta+0=3 \times 1 \\
\sin \alpha+\sin \beta+1=3 \times 0
\end{array}\right.
$$

which simplifies to
$$
\left\{\begin{array}{l}
\cos \alpha+\cos \beta=\frac{3}{\sqrt{2}}, \\
\sin \alpha+\sin \beta=-1,
\end{array}\right. \text {. }
$$

Squaring and adding, we get $2+2 \cos (\alpha-\beta)=\frac{11}{2}$, which simplifies to $\cos (\alpha-\beta)=\frac{7}{4}>1$, a contradiction.
Therefore, point $F$ cannot be the centroid of $\triangle B M N$.
(2) Suppose there exists a line $l$ such that point $F$ is the orthocenter of $\triangle B M N$. Given $B(0,1), F(1,0)$, the slope of line $B F$ is -1, so by $B F \perp M N$, the slope of line $l$ is 1. Let the equation of line $l$ be $y=x+b$. From
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2}+y^{2}=1 \\
y=x+b
\end{array}\right.
$$

eliminating $y$, we get
$$
3 x^{2}+4 b x+2\left(b^{2}-1\right)=0 \text {. }
$$

Let the coordinates of points $M, N$ be $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. By Vieta's formulas, we have
$$
x_{1}+x_{2}=-\frac{4 b}{3}, \quad x_{1} x_{2}=\frac{2 b^{2}-2}{3} .
$$

Since $N F \perp B M$, we have $\overrightarrow{N F} \cdot \overrightarrow{B M}=0$, and $\overrightarrow{N F}=\left(1-x_{2},-y_{2}\right), \overrightarrow{B M}=\left(x_{1}, y_{1}-1\right)$, so
$$
\begin{aligned}
\overrightarrow{N F} \cdot \overrightarrow{B M} & =\left(1-x_{2}\right) x_{1}-y_{2}\left(y_{1}-1\right)=x_{1}+y_{2}-x_{1} x_{2}-y_{1} y_{2} \\
& =x_{1}+x_{2}+b-x_{1} x_{2}-\left(x_{1}+b\right)\left(x_{2}+b\right)=-2 x_{1} x_{2}+(1-b)\left(x_{1}+x_{2}\right)+b-b^{2} \\
& =-2 \cdot \frac{2 b^{2}-2}{3}+(1-b) \cdot\left(-\frac{4 b}{3}\right)+b-b^{2}=0 .
\end{aligned}
$$

Solving, we get $b=1$ or $b=\frac{4}{3}$. When $b=1$, point $B$ is the intersection point of line $l$ and the ellipse, which is not valid. Therefore, the equation of the required line $l$ is $y=x-\frac{4}{3}$.