Example 22 (2004 National High School Competition Question) Given $M=\left\{(x, y) \mid x^{2}+2 y^{2}=3\right\}, N=\{(x, y) \mid y=m x+b\}$. If for all $m \in \mathbf{R}$, there is $M \cap N \neq \varnothing$, then the range of values for $b$ is ( ).
A. $\left[-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right]$
B. $\left(-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right)$
C. $\left(-\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right)$
D. $\left[-\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right]$
Solution: Choose A. Reason: According to the problem, for $y=m x+b$, when $m \in \mathbf{R}$, it is known that it is a family of lines passing through the fixed point $(0, b)$. Therefore, the condition $M \cap N \neq \varnothing$ in the original problem can be changed to (i.e., is equivalent to) the point $(0, b)$ being on the ellipse $\frac{x^{2}}{3}+\frac{2 y^{2}}{3}=1$ or inside it. Thus, substituting the coordinates of the point $(0, b)$ into the left side of the ellipse equation should satisfy $\frac{2 b^{2}}{3} \leqslant 1$, hence solving for $-\frac{\sqrt{6}}{2} \leqslant b \leqslant \frac{\sqrt{6}}{2}$.