8. Let $P$ be any point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $F_{1} 、 F_{2}$ be the foci of the ellipse, and $P F_{1} 、 P F_{2}$ intersect the ellipse at points $A 、 B$ respectively. Then
$$
\frac{\left|P F_{1}\right|}{\left|F_{1} A\right|}+\frac{\left|P F_{2}\right|}{\left|F_{2} B\right|}=
$$
$\qquad$
8. $\frac{4 a^{2}}{b^{2}}-2$.

As shown in Figure 1.
By the definition of an ellipse,
$$
\left|P P_{1}\right|=\frac{\left|P F_{1}\right|}{e},\left|F_{1} M\right|=p,\left|A A_{1}\right|=\frac{\left|F A_{1}\right|}{e} \text {, }
$$

where $e$ is the eccentricity of the ellipse, and $p$ is the focal parameter of the ellipse.
By the properties of similar figures and the ratio of sums and differences, we get
$\frac{|A P|}{\left|A F_{1}\right|}=\frac{\left|A F_{1}\right|+\left|F_{1} P\right|}{\left|A F_{1}\right|}$
$=\frac{\frac{\left|P F_{1}\right|}{e}-\frac{\left|A F_{1}\right|}{e}}{p-\frac{\left|A F_{1}\right|}{e}}=\frac{\left|P F_{1}\right|-\left|A F_{1}\right|}{e p-\left|A F_{1}\right|}$
$=\frac{2\left|P F_{1}\right|}{e p}$.
Thus, $\frac{\left|P F_{1}\right|}{\left|A F_{1}\right|}=\frac{2\left|P F_{1}\right|}{e p}-1$.
Similarly, $\frac{\left|P F_{2}\right|}{\left|B F_{2}\right|}=\frac{2\left|P F_{2}\right|}{e p}-1$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\left|P F_{1}\right|}{\left|F_{1} A\right|}+\frac{\left|P F_{2}\right|}{\left|F_{2} B\right|} \\
=\frac{2}{e p}\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)-2 \\
=\frac{4 a}{e p}-2=\frac{4 a^{2}}{b^{2}}-2 .
\end{array}
$$