10. (20 points) Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, points $B$ and $F$ are the right vertex and right focus of the hyperbola $C$, respectively, and $O$ is the origin. Point $A$ is on the positive $x$-axis and satisfies $|O A|,|O B|,|O F|$ forming a geometric sequence. A line $l$ is drawn from point $F$ perpendicular to the asymptotes of the hyperbola $C$ in the first and third quadrants, with the foot of the perpendicular being $P$.
(1) Prove: $\overrightarrow{P A} \cdot \overrightarrow{O P}=\overrightarrow{P A} \cdot \overrightarrow{F P}$;
(2) Let $a=1, b=2$, and the line $l$ intersects the left and right branches of the hyperbola $C$ at points $D$ and $E$, respectively. Find the value of $\frac{|D F|}{|D E|}$.
10. (1) The equation of line $l$ is $y=-\frac{a}{b}(x-c)$. From
$$
\left\{\begin{array}{l}
y=-\frac{a}{b}(x-c) \\
y=\frac{b}{a} x
\end{array}\right.
$$

we get $P\left(\frac{a^{2}}{c}, \frac{a b}{c}\right)$.
Given that $|O A|,|O B|,|O F|$ form a geometric sequence, we have $A\left(\frac{a^{2}}{c}, 0\right)$, thus
$$
\overrightarrow{P A}=\left(0,-\frac{a b}{c}\right), \quad \overrightarrow{O P}=\left(\frac{a^{2}}{c}, \frac{a b}{c}\right), \quad \overrightarrow{F P}=\left(-\frac{b^{2}}{c}, \frac{a b}{c}\right),
$$

Therefore,
$$
\overrightarrow{P A} \cdot \overrightarrow{O P}=-\frac{a^{2} b^{2}}{c^{2}}, \quad \overrightarrow{P A} \cdot \overrightarrow{F P}=-\frac{a^{2} b^{2}}{c^{2}}
$$

Thus,
$$
\overrightarrow{P A} \cdot \overrightarrow{O P}=\overrightarrow{P A} \cdot \overrightarrow{F P}
$$
(2) Given $a=1, b=2$, we have $c=\sqrt{5}, l: y=-\frac{1}{2}(x-\sqrt{5})$. From
$$
\left\{\begin{array}{l}
y=-\frac{1}{2}(x-\sqrt{5}), \\
x^{2}-\frac{y^{2}}{4}=1
\end{array}\right.
$$

eliminating $x$ and rearranging, we get
$$
15 y^{2}-16 \sqrt{5} y+16=0 .
$$

Let $D\left(x_{1}, y_{1}\right), E\left(x_{2}, y_{2}\right)$, and given that $\left|y_{1}\right|>\left|y_{2}\right|$, and $y_{1}, y_{2}$ are the two roots of the above equation. Thus,
$$
y_{1}+y_{2}=\frac{16 \sqrt{5}}{15}, \quad y_{1} y_{2}=\frac{16}{15} .
$$

Hence,
$$
\frac{y_{1}}{y_{2}}+\frac{y_{2}}{y_{1}}=\frac{\left(y_{1}+y_{2}\right)^{2}-2 y_{1} y_{2}}{y_{1} y_{2}}=\frac{10}{3}
$$

Solving, we get $\frac{y_{2}}{y_{1}}=3$ or $\frac{1}{3}$.
Since $\left|y_{1}\right|>\left|y_{2}\right|$, we have $\frac{y_{2}}{y_{1}}=\frac{1}{3}$, thus
$$
\frac{|D F|}{|D E|}=\frac{y_{1}}{y_{1}-y_{2}}=\frac{1}{1-\frac{y_{2}}{y_{1}}}=\frac{3}{2} .
$$