Problem 980

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
$$\frac{|D F|}{|D E|}=\frac{y_{1}}{y_{1}-y_{2}}=\frac{1}{1-\frac{y_{2}}{y_{1}}}=\frac{3}{2} .$$