Four, (20 points) Through the left focus $F$ of the ellipse $x^{2}+2 y^{2}=2$, draw a line $l$ with an inclination angle of $\alpha$ intersecting the ellipse at points $P, Q$, and the two directrices at points $A, B$. If $P Q$, $||A F|-| B F||$, and $|A B|$ form a geometric sequence, find the value of $|\cos \alpha|$.

Draw the line $l$ through the left focus $F$ of the ellipse $x^{2}+2 y^{2}=2$ with an inclination angle of $\alpha$, intersecting the ellipse at points $P, Q$, and the two directrices at points $A, B$. If $P Q$, $||A F|-| B F||$, and $|A B|$ form a geometric sequence, find the value of $|\cos \alpha|$.
Four, Solution: From $\frac{x^{2}}{2}+y^{2}=1$, we know
$$
a^{2}=2, b^{2}=1, c^{2}=1 .
$$

Thus, $F:(-1,0)$.
Let the parametric equation of the line be
$$
\left\{\begin{array}{l}
x=-1+t \cos \alpha, \\
y=t \sin \alpha .
\end{array} \text { ( } t\right. \text { is the parameter) }
$$

Substituting into $x^{2}+$
$2 y^{2}=2$, we get
$$
\begin{array}{l}
(1+ \\
\left.\sin ^{2} \alpha\right) t^{2}- \\
2(\cos \alpha) t- \\
1=0, \\
\therefore|P Q|=\left|t_{1}-t_{2}\right|=\frac{2 \sqrt{2}}{1+\sin ^{2} \alpha} . \\
|A B|=\frac{2 \cdot \frac{a^{2}}{c}}{|\cos \alpha|}=\frac{4}{|\cos \alpha|}, \\
\quad|| A F|-| B F||=\frac{2 c}{|\cos \alpha|}=\frac{2}{|\cos \alpha|}, \\
\therefore\left(\frac{2}{|\cos \alpha|}\right)^{2}=\frac{2 \sqrt{2}}{1+\sin ^{2} \alpha} \cdot \frac{4}{|\cos \alpha|},
\end{array}
$$

which simplifies to $\cos ^{2} \alpha+2 \sqrt{2}|\cos \alpha|-2=0$.
Solving this, we get $|\cos \alpha|=2-\sqrt{2}$.