Example 2 - (83 College Entrance Examination
Mathematics for Science Students, Question 7)
As shown in the figure, it is known that the length of the major axis of the ellipse

is $\left|A_{1} A_{2}\right|=6$, and the focal distance
$\left|\mathbf{F}_{\mathbf{1}} \mathbf{F}_{\mathbf{2}}\right|=4 \sqrt{\mathbf{2}}$. A line is drawn through
the focus $\mathrm{F}_{1}$ of the ellipse,
intersecting the ellipse at points $\mathrm{M}$ and $\mathrm{N}$. Let $\angle \mathrm{F}_{2} \mathrm{~F}_{1} \mathrm{M}=\alpha(0 \leqslant \alpha \leqslant \pi)$,
for what value of $\alpha$ is $|\mathrm{MN}|$ equal to the length of the minor axis of the ellipse?
Take the focus $\mathrm{F}_{1}$ as the pole, and the ray starting from $\mathrm{F}_{1}$ passing through $F_{3}$ as the polar axis to establish a polar coordinate system.

Given the ellipse's semi-major axis $\mathrm{a}=3$, semi-focal distance $\mathrm{c}=2 \sqrt{2}$, semi-minor axis $\mathrm{b}=1$, eccentricity $\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}=\frac{2 \sqrt{2}}{3}$, and the distance from the focus $\mathrm{F}_{1}$ to the corresponding directrix $\mathrm{x}=-\frac{\mathrm{a}^{2}}{\mathrm{c}}=-\frac{9 \sqrt{2}}{4}$ is $\mathrm{p}=\frac{\sqrt{2}}{4}$.

Since the focus $F_{1}$ is on the side of the corresponding directrix $x-\frac{9 \sqrt{2}}{4}$, the polar equation of the ellipse fits the form (Ｉ), which is
$$
\begin{array}{l}
\rho=\frac{e p}{1-\operatorname{ecos} \theta}, \text { i.e., } \rho=\frac{1}{3-2 \sqrt{2} \cos \theta} \text {. } \\
\therefore\left|F_{1} \mathbf{M}\right|=\rho_{1}=\frac{1}{3-2 \sqrt{2 \cos \alpha}}, \\
\left|F_{1} N\right|=\rho_{2}=\frac{1}{3+2 \sqrt{2} \cos \alpha} . \\
\therefore|M N|=\rho_{1}+\rho_{2}=\frac{1}{3-2 \sqrt{2} \cos \alpha} \\
+\frac{1}{3+2 \sqrt{2} \cos \alpha}=\frac{6}{9-8 \cos ^{2} \alpha} . \\
\text { Let } \\
\therefore-8 \cos \alpha=2
\end{array}
$$

Each step in the solution process is verifiable, so this problem can also be solved as shown in the right figure.

When $\alpha=\frac{\pi}{6}$ or $\alpha=\frac{5 \pi}{6}$,
$|\mathbf{M N}|$ equals the length of the semi-minor axis.