10. (20 points) Given the curve $C_{1}: \frac{x^{2}}{4}+y^{2}=1$, the curve $C_{2}: y=x^{2}-1$, and the point $M(0,-1)$. A line $l$ passing through the origin $O$ intersects $C_{2}$ at points $A, B$. The lines $M A, M B$ intersect $C_{1}$ at points $D, E$ $(D, E \neq M)$.
(1) Prove: $M D \perp M E$;
(2) Let the areas of $\triangle M A B$ and $\triangle M D E$ be $S_{1}, S_{2}$, respectively. Is there a line $l$ such that $\frac{S_{1}}{S_{2}}=\frac{17}{32}$? If it exists, find the equation of the line $l$; if not, explain the reason.
10. (1) From the problem, we know that the slope of line $l$ exists, let it be $k$, then the equation of line $l$ is $y=k x$. From
$$
\left\{\begin{array}{l}
y=k x \\
y=x^{2}-1
\end{array}\right.
$$

we get
$$
x^{2}-k x-1=0 .
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $x_{1}, x_{2}$ are the two real roots of the above equation, so $x_{1}+x_{2}=k, x_{1} x_{2}=-1$. Also, $M(0,-1)$, so
$$
\begin{aligned}
k_{M A} \cdot k_{M B} & =\frac{y_{1}+1}{x_{1}} \cdot \frac{y_{2}+1}{x_{2}}=\frac{\left(k x_{1}+1\right)\left(k x_{2}+1\right)}{x_{1} x_{2}}=\frac{k^{2} x_{1} x_{2}+k\left(x_{1}+x_{2}\right)+1}{x_{1} x_{2}} \\
& =\frac{-k^{2}+k^{2}+1}{-1}=-1 .
\end{aligned}
$$

Thus, $M D \perp M E$.
(2) Let the slope of line $M A$ be $k_{1}$, then the equation of line $M A$ is $y=k_{1} x-1$. From
$$
\left\{\begin{array}{l}
y=k_{1} x-1 \\
y=x^{2}-1
\end{array}\right.
$$

we solve to get $(x, y)=(0,-1)$ or $\left(k_{1}, k_{1}^{2}-1\right)$. Therefore, the coordinates of point $A$ are $\left(k_{1}, k_{1}^{2}-1\right)$. Also, the slope of line $M B$ is $-\frac{1}{k_{1}}$, similarly, we get the coordinates of point $B$ as $\left(-\frac{1}{k_{1}}, \frac{1}{k_{1}^{2}}-1\right)$. Thus,
$$
S_{1}=\frac{1}{2}|M A| \cdot|M B|=\frac{1}{2} \sqrt{1+k_{1}^{2}} \cdot\left|k_{1}\right| \cdot \sqrt{1+\frac{1}{k_{1}^{2}}} \cdot\left|-\frac{1}{k_{1}}\right|=\frac{1+k_{1}^{2}}{2\left|k_{1}\right|}
$$

From
$$
\left\{\begin{array}{l}
y=k_{1} x-1 \\
x^{2}+4 y^{2}-4=0
\end{array}\right.
$$

we get
$$
\left(1+4 k_{1}^{2}\right) x^{2}-8 k_{1} x=0 .
$$

Solving, we get $(x, y)=(0,-1)$ or $\left(\frac{8 k_{1}}{1+4 k_{1}^{2}}, \frac{4 k_{1}^{2}-1}{1+4 k_{1}^{2}}\right)$. Therefore, the coordinates of point $D$ are $\left(\frac{8 k_{1}^{2}}{1+4 k_{1}^{2}}, \frac{4 k_{1}^{2}-1}{1+4 k_{1}^{2}}\right)$. Also, the slope of line $M E$ is $-\frac{1}{k_{1}}$, similarly, we get the coordinates of point $E$ as $\left(\frac{-8 k_{1}}{4+k_{1}^{2}}, \frac{4-k_{1}^{2}}{4+k_{1}^{2}}\right)$, thus,
$$
S_{2}=\frac{1}{2}|M D| \cdot|M E|=\frac{32\left(1+k_{1}^{2}\right) \cdot\left|k_{1}\right|}{\left(1+4 k_{1}^{2}\right)\left(k_{1}^{2}+4\right)} .
$$

Therefore,
$$
\frac{S_{1}}{S_{2}}=\frac{1}{64}\left(4 k_{1}^{2}+\frac{4}{k_{1}^{2}}+17\right)=\frac{17}{32} .
$$

Solving, we get $k_{1}^{2}=4$ or $\frac{1}{4}$. Also,
$$
k=\frac{y_{A}-y_{B}}{x_{A}-x_{B}}=\frac{k_{1}^{2}-\frac{1}{k_{1}^{2}}}{k_{1}+\frac{1}{k_{1}}}=k_{1}-\frac{1}{k_{1}}= \pm \frac{3}{2} .
$$

Therefore, the line $l$ that satisfies the condition exists, and the equation of the line $l$ that satisfies the condition is $y= \pm \frac{3}{2} x$.