2. Let $a^{2}+b^{2}=1(b \neq 0)$. If the line $a x+b y$ $=2$ and the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$ have a common point, then the range of $\frac{a}{b}$ is ( ).
(A) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(B) $[-1,1]$
(C) $(-\infty,-1] \cup[1,+\infty)$
(D) $[-2,2]$
2. C.

Substitute $y=\frac{2-a x}{b}$ into the ellipse equation and rearrange to get $\left(3 a^{2}+b^{2}\right) x^{2}-12 a x+12-6 b^{2}=0$. Since the line and the ellipse have a common point, the discriminant $\Delta=(12 a)^{2}-4\left(3 a^{2}+b^{2}\right)\left(12-6 b^{2}\right) \geqslant 0$. Using $a^{2}+b^{2}=1$, simplify to get $a^{2} \geqslant b^{2}$.

Therefore, $\left|\frac{a}{b}\right| \geqslant 1$.
Hence $\frac{a}{b} \in(-\infty,-1] \cup[1,+\infty)$.