2. Given $M=\left\{(x, y) \mid x^{2}+2 y^{2}=3\right\}, N=\{(x, y) \mid y=m x+b\}$. If for all $m \in \mathbf{R}$, we have $M \cap N$ $\neq \varnothing$, then the range of values for $b$ is
(A) $\left[-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right]$
(B) $\left(-\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{2}\right)$
(C) $\left(-\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3)}\right]$
(D) $\left[-\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right]$
(A)
2.【Analysis and Solution】 $M \cap N \neq \varnothing$ is equivalent to the point $(0, b)$ being on or inside the ellipse $x^{2}+2 y^{2}=3$ $\therefore \frac{2 b^{2}}{3} \leqslant 1$, $\therefore-\frac{\sqrt{6}}{2} \leqslant b \leqslant \frac{\sqrt{6}}{2}$. Therefore, the answer is A.