4. Given that $M$ is any point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $F_{1}, F_{2}$ are the left and right foci of the ellipse, respectively, and the lines $M F_{1}, M F_{2}$ intersect the ellipse at points $A, B$, respectively, then $\frac{b^{2}}{a^{2}}\left(\frac{\left|M F_{1}\right|}{\left|F_{1} A\right|}+\frac{\left|M F_{2}\right|}{\left|F_{2} B\right|}+2\right)=$ $\qquad$
4.4

Analysis: As shown in the figure, let $l$ be the left directrix of the ellipse $x=-\frac{a^{2}}{c}$. Draw perpendiculars from $A, M$ to $l$, intersecting $l$ at $A_{1}, M_{1}$, respectively. Then draw a perpendicular from $A$ to $M M_{1}$, intersecting $M M_{1}$ at $M_{2}$, and intersecting the $x$-axis at $N_{2}$. The projection of $N_{2}$ on $l$ is $N_{1}$.

By the second definition of the ellipse, we have: $M M_{1}=\frac{M F_{1}}{e}, A A_{1}=\frac{A F_{1}}{e}, F_{1} N_{1}=\frac{a^{2}}{c}-c=\frac{b^{2}}{c}$, thus
$$
\frac{\left|M F_{1}\right|+\left|F_{1} A\right|}{\left|F_{1} A\right|}=\frac{|A M|}{\left|F_{1} A\right|}=\frac{\left|M M_{2}\right|}{\left|F_{1} N_{2}\right|}=\frac{\left|M M_{1}\right|-\left|M_{1} M_{2}\right|}{\left|F_{1} N_{1}\right|-\left|N_{1} N_{2}\right|}=\frac{\left|M M_{1}\right|-\left|A A_{1}\right|}{\left|F_{1} N_{1}\right|-\left|A A_{1}\right|}=\frac{\frac{\left|M F_{1}\right|}{e}-\frac{\left|A F_{1}\right|}{e}}{\frac{b^{2}}{c}-\frac{\left|A F_{1}\right|}{e}}=\frac{c\left|M F_{1}\right|-c\left|F_{1} A\right|}{e b^{2}-c\left|F_{1} A\right|}
$$

By the theorem of the ratio of sums, we get $\frac{\left|M F_{1}\right|+\left|F_{1} A\right|}{\left|F_{1} A\right|}=\frac{c\left|M F_{1}\right|-c\left|F_{1} A\right|+\left(\left|M F_{1}\right|+\left|F_{1} A\right|\right)}{e b^{2}-c\left|F_{1} A\right|+\left|F_{1} A\right|}=\frac{2 c\left|M F_{1}\right|}{e b^{2}}$, hence
$\frac{\left|M F_{1}\right|}{\left|F_{1} A\right|}=\frac{2 c\left|M F_{1}\right|}{e b^{2}}-1$, similarly, $\frac{\left|M F_{2}\right|}{\left|F_{2} A\right|}=\frac{2 c\left|M F_{2}\right|}{e b^{2}}-1$, adding the two equations gives
$\frac{\left|M F_{1}\right|}{\left|F_{1} A\right|}+\frac{\left|M F_{2}\right|}{\left|F_{2} A\right|}+2=\frac{2 c\left(\left|M F_{1}\right|+\left|M F_{2}\right|\right)}{e b^{2}}=\frac{4 a c}{\frac{c}{a} \cdot b^{2}}=\frac{4 a^{2}}{b^{2}}$, therefore the value of the original expression is 4.