17. Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$, and $M$ is any point on the circle $x^{2}+y^{2}=3$, $M A, M B$ are tangent to the ellipse $C$ at points $A, B$, find the range of the area of $\triangle O A B$ (where $O$ is the origin).
Given $M(a, b), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then
$M A: \frac{x_{1} x}{2}+y_{1} y=1, M B: \frac{x_{2} x}{2}+y_{2} y=1$
$\Rightarrow A B: \frac{a x}{2}+b y=1$. Since $M$ is on the circle $O$,
then the line $A B: \cos \theta \cdot x+2 \sin \theta \cdot y=\frac{2}{\sqrt{3}}$.
Substituting into the ellipse equation $x^{2}+2 y^{2}=2$ gives
$$
\begin{array}{l}
2\left(\frac{1}{\sqrt{3}}-\sin \theta \cdot y\right)^{2}+\cos ^{2} \theta \cdot y^{2}=\cos ^{2} \theta \\
\Rightarrow\left(2 \sin ^{2} \theta+\cos ^{2} \theta\right) y^{2}-\frac{4 \sin \theta}{\sqrt{3}} y+\frac{2}{3}-\cos ^{2} \theta=0 .
\end{array}
$$

Thus, $|A B|=\sqrt{1+\left(\frac{2 \sin \theta}{\cos \theta}\right)^{2}} \cdot \frac{\sqrt{\frac{16 \sin ^{2} \theta}{3}-4\left(\frac{2}{3}-\cos ^{2} \theta\right)\left(2 \sin ^{2} \theta+\cos ^{2} \theta\right)}}{2 \sin ^{2} \theta+\cos ^{2} \theta}$
$$
=\sqrt{4 \sin ^{2} \theta+\cos ^{2} \theta} \cdot \frac{2 \sqrt{2 \sin ^{2} \theta+\cos ^{2} \theta-\frac{2}{3}}}{2 \sin ^{2} \theta+\cos ^{2} \theta} \text {, }
$$

The distance from the origin $O$ to the line $A B$ is $d=\frac{\frac{2}{\sqrt{3}}}{\sqrt{4 \sin ^{2} \theta+\cos ^{2} \theta}}$, thus $S_{\triangle O A B}=\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{2 \sin ^{2} \theta+\cos ^{2} \theta-\frac{2}{3}}}{2 \sin ^{2} \theta+\cos ^{2} \theta}$, let $t=\sqrt{2 \sin ^{2} \theta+\cos ^{2} \theta-\frac{2}{3}}=\sqrt{\sin ^{2} \theta+\frac{1}{3}} \in\left[\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right]$,
so $S_{\triangle O A B}=\frac{2}{\sqrt{3}} \cdot \frac{t}{t^{2}+\frac{2}{3}}=\frac{2}{\sqrt{3}} \cdot \frac{1}{t+\frac{2}{3 t}} \in\left[\frac{2}{3}, \frac{\sqrt{2}}{2}\right]$.