10. (20 points) Given points $M(-1,0), N(1,0)$, the perimeter of $\triangle M N Q$ is 6, and the trajectory of the moving point $Q$ is the curve $C$. $P$ is any point on the circle $x^{2}+y^{2}=4$ (not on the x-axis), and $P A, P B$ are tangent to the curve $C$ at points $A, B$ respectively. Find the maximum value of the area of $\triangle O A B$ (where $O$ is the origin).
10. According to the problem, the trajectory of the moving point $Q$ is an ellipse with foci at $M$ and $N$, and a major axis length of $2a = 4$ (excluding the intersection points with the $x$-axis).
Thus, the equation of the curve $C$ is
$$
\frac{x^{2}}{4}+\frac{y^{2}}{3}=1(y \neq 0).
$$

Let the points be $P\left(x_{0}, y_{0}\right)$, $A\left(x_{1}, y_{1}\right)$, and $B\left(x_{2}, y_{2}\right)$, and the slope of the line $AB$ be $k$. Then the equation of the chord of contact $AB$ is
$$
\frac{x_{0} x}{4}+\frac{y_{0} y}{3}=1.
$$

By combining this with the equation of the ellipse, we get
$$
\left(3+\frac{9 x_{0}^{2}}{4 y_{0}^{2}}\right) x^{2}-\frac{18 x_{0}}{y_{0}^{2}} x+\frac{36}{y_{0}^{2}}-12=0.
$$

Given that $x_{0}^{2}+y_{0}^{2}=4$, we have
$$
\left\{\begin{array}{c}
x_{1}+x_{2}=\frac{\frac{18 x_{0}}{y_{0}^{2}}}{3+\frac{9 x_{0}^{2}}{4 y_{0}^{2}}}=\frac{24 x_{0}}{4 y_{0}^{2}+3 x_{0}^{2}}=\frac{24 x_{0}}{y_{0}^{2}+12}, \\
x_{1} x_{2}=\frac{\frac{36}{y_{0}^{2}}-12}{3+\frac{9 x_{0}^{2}}{4 y_{0}^{2}}}=\frac{16\left(3-y_{0}^{2}\right)}{4 y_{0}^{2}+3 x_{0}^{2}}=\frac{16\left(3-y_{0}^{2}\right)}{y_{0}^{2}+12}.
\end{array}\right.
$$

The distance from point $O$ to the line $AB$ is
$$
d=\frac{\frac{3}{\left|y_{0}\right|}}{\sqrt{1+k^{2}}},
$$

and
$|AB|=\sqrt{1+k^{2}}\left|x_{1}-x_{2}\right|$.
Thus, $S_{\triangle OAB}=\frac{1}{2}|AB| d=\frac{3}{2\left|y_{0}\right|}\left|x_{1}-x_{2}\right|$
$$
\begin{array}{l}
=3 \sqrt{\frac{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}{4 y_{0}^{2}}} \\
=3 \sqrt{\frac{\left(\frac{24 x_{0}}{y_{0}^{2}+12}\right)^{2}-\frac{64\left(3-y_{0}^{2}\right)}{y_{0}^{2}+12}}{4 y_{0}^{2}}} \\
=12 \sqrt{\frac{9 x_{0}^{2}+y_{0}^{4}+9 y_{0}^{2}-36}{y_{0}^{2}\left(y_{0}^{2}+12\right)^{2}}} \\
=\frac{12\left|y_{0}\right|}{y_{0}^{2}+12}=\frac{12}{\left|y_{0}\right|+\frac{12}{\left|y_{0}\right|}}.
\end{array}
$$

It is easy to see that this function is monotonically increasing on $(0,2]$.
Therefore, when $\left|y_{0}\right|=2$, $S_{\triangle OCD}$ reaches its maximum value $\frac{3}{2}$.