1. Given the length of the major axis of an ellipse $\left|A_{1} A_{2}\right|=6$, and the focal distance $\left|F_{1} F_{2}\right|$ $=4 \sqrt{2}$, a line is drawn through the left focus $F_{1}$ intersecting the ellipse at points $M$ and $N$. Let $\angle F_{2} F_{1} M=\alpha(0 \leqslant \alpha<\pi)$. When $|M N|$ equals the length of the minor axis of the ellipse, $\alpha=$ $\qquad$ .
(Tip: The polar equation of the ellipse is
$$
\rho=\frac{1}{3-2 \sqrt{2} \cos \theta} \text {. }
$$

Let $M\left(\rho_{1}, \alpha\right) 、 N\left(\rho_{2}, \alpha+\pi\right)$. Then
$$
\begin{array}{l}
|M N|=\frac{1}{3-2 \sqrt{2} \cos \alpha}+\frac{1}{3-2 \sqrt{2} \cos (\alpha+\pi)} \\
=\frac{6}{9-8 \cos ^{2} \alpha} .
\end{array}
$$

Therefore, $\frac{6}{9-8 \cos ^{2} \theta}=2$. Solving this, we get $\cos \theta= \pm \frac{\sqrt{3}}{2}$. Since $0 \leqslant \alpha<\pi$, hence $\alpha=\frac{\pi}{6}$ or $\frac{5 \pi}{6}$. )