3. 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 . $\qquad$
3. $(-\infty,-1] \cup[1,+\infty)$ Detailed Explanation: 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 $(12 a)^{2}-4\left(3 a^{2}+b^{2}\right)\left(12-6 b^{2}\right) \geqslant 0$, using $a^{2}+b^{2}=1$, simplifying yields $a^{2} \geqslant b^{2}$, so $\left|\frac{a}{b}\right| \geqslant 1$. Therefore, $\frac{a}{b} \in(-\infty,-1] \cup[1,+\infty)$.