5. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left and right foci $F_{1}, F_{2}$, and eccentricity $\frac{5}{3}$, if a line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $T$, and $l$ intersects the right branch of the hyperbola $C$ at point $P$, then $\frac{\left|\overrightarrow{F_{1} P}\right|}{\left|\overrightarrow{F_{1} T}\right|}=$ $\qquad$.
Answer
As shown in the figure, $|O T|=a, \left|O F_{1}\right|=c \Rightarrow \left|F_{1} T\right|=b$, $\cos \angle O F_{1} T=\frac{b}{c}$, by the definition of a hyperbola, we get $|P M|=\frac{\left|P F_{1}\right|}{e}$,
$$
\begin{array}{c}
p=|N M|=c-\frac{a^{2}}{c}=\frac{b^{2}}{c}, \cos \angle F_{1} P N=\frac{|P N|}{\left|P F_{1}\right|} \\
=\frac{\frac{\left|P F_{1}\right|}{e}+p}{\left|P F_{1}\right|}=\frac{\left|P F_{1}\right|+e p}{e\left|P F_{1}\right|} \Rightarrow\left|P F_{1}\right|=\frac{e p}{e \cos \alpha-1},
\end{array}
$$

where $\alpha=\angle F_{1} P N=\angle O F_{1} T$. Given $\frac{b}{a}=\sqrt{e^{2}-1}=\frac{4}{3}, e \cos \alpha-1=\frac{b}{a}-1=\frac{1}{3}$, $e p=\frac{c}{a} \cdot \frac{b^{2}}{c}=\frac{b^{2}}{a}=\frac{4}{3} b$, thus $\left|P F_{1}\right|=4 b$, so $\frac{\left|\overrightarrow{F_{1} P}\right|}{\left|\overrightarrow{F_{1} T}\right|}=4$.