(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 common points, 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) Substituting $y=\frac{2-a x}{b}$ into the ellipse equation and simplifying, we 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 common points, the discriminant is
$$
(12 a)^{2}-4\left(3 a^{2}+b^{2}\right)\left(12-6 b^{2}\right) \geqslant 0,
$$

Using $a^{2}+b^{2}=1$, simplifying yields $a^{2} \geqslant b^{2}$, so $\left|\frac{a}{b}\right| \geqslant 1$. Therefore,
$$
\frac{a}{b} \in(-\infty,-1] \cup[1,+\infty) \text {. }
$$

Hence, the answer is C.