4. Let $A B$ be the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the major axis $A B$ is divided into 100 equal parts, and perpendicular lines are drawn through each division point to $A B$, intersecting the upper half of the ellipse at $P_{1}, P_{2}$, $P_{3}, \cdots, P_{99}$, and $F_{1}$ is the left focus of the ellipse. Then the value of $\left|F_{1} A\right|$ $+\left|F_{1} P_{1}\right|+\left|F_{1} P_{2}\right|+\cdots+\left|F_{1} P_{99}\right|+$ $\left|F_{1} B\right|$ is $(\quad)$.
(A) $101 a$
(B) $100 a$
(C) $99 a$
(D) $98 a$
4. (A).

From the definition of an ellipse, we know
$$
\begin{array}{l}
\left|F_{1} P_{i}\right|+\left|F_{2} P_{i}\right|=2 a(i=1,2, \cdots, 99) . \\
\therefore \sum_{i=1}^{99}\left(\left|F_{1} P_{i}\right|+\left|F_{2} P_{i}\right|\right)=99 \times 2 a=198 a .
\end{array}
$$

According to the problem, $P_{1}, P_{2}, \cdots, P_{99}$ are symmetrically distributed about the $y$-axis,
$$
\begin{array}{l}
\therefore \sum_{i=1}^{99}\left|F_{1} P_{i}\right|=\frac{1}{2} \sum_{i=1}^{99}\left(\left|F_{1} P_{i}\right|+\left|F_{2} P_{i}\right|\right) \\
\quad=99 a . \\
\text { Also, } \because\left|F_{1} A\right|+\left|F_{1} B\right|=2 a, \\
\therefore\left|F_{1} A_{1}\right|+\left|F_{1} P_{1}\right|+\left|F_{1} P_{2}\right|+\cdots+\left|F_{1} P_{99}\right| \\
+\left|F_{1} B\right|=99 a+2 a=101 a .
\end{array}
$$