7. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with eccentricity $\frac{\sqrt{17}}{3}, F$ is the right focus, points $A, B$ are on the right branch, and $D$ is the point symmetric to $A$ with respect to the origin $O$, and $D F \perp A B$. If $\overrightarrow{A F}=\lambda \overrightarrow{F B}$, then $\lambda=$ $\qquad$ .
Given the left focus of the hyperbola $F_{1}$, and $|A F|=x$,
then $A F D F_{1}$ is a rectangle $\Rightarrow x^{2}+(2 a+x)^{2}=4 c^{2}$
$$
\Rightarrow x^{2}+2 a x-\frac{16}{9} a^{2}=0 \Rightarrow x=\frac{2}{3} a \text {, }
$$

Thus, $\cos \angle O F A=\frac{x}{2 c}=\frac{1}{3} \cdot \frac{a}{c}=\frac{1}{\sqrt{17}}$.
Using the definition of the hyperbola, we have
$$
\cos \theta=\cos \angle O F A=\frac{p-\frac{|A F|}{e}}{|A F|},
$$

Therefore, $e \cos \theta=\frac{e p-|A F|}{|A F|}=\frac{e p}{|A F|}-1 \Rightarrow|A F|=\frac{e p}{1+e \cos \theta}$, similarly, $|B F|=\frac{e p}{1-e \cos \theta}$. Thus, $\lambda=\frac{|A F|}{|B F|}=\frac{1-e \cos \theta}{1+e \cos \theta}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{1}{2}$.