Example 3 Given the parabola $C: y^{2}=4 x, F$ is the focus of the parabola $C$, a line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$, and let $\overrightarrow{F B}=\lambda \overrightarrow{A F}$. If $\lambda \in[4,9]$, find the range of the y-intercept of line $l$.
Explanation: The line $l$ is not fixed, and its position relationship satisfies $\overrightarrow{F B}=\lambda \overrightarrow{A F}, \lambda$ varies within the range $[4,9]$. Since $\overrightarrow{F B}$ and $\overrightarrow{A F}$ are in the same direction, $\lambda$ represents the ratio of the lengths of $\overrightarrow{F B}$ and $\overrightarrow{A F}$. Therefore, we can consider using the polar equation of a parabola to solve the problem.

As shown in Figure 2, with $F$ as the pole and $F x$ as the polar axis, we establish a polar coordinate system. Thus, we only need to find the range of the inclination angle $\angle B F x$.

The polar equation of the parabola is
$$
\rho=\frac{2}{1-\cos \theta} .
$$

According to the problem, point $B$ may be on the parabola in the first quadrant or the fourth quadrant.
(1) When $B$ is a point on the parabola in the first quadrant, let $B\left(\rho_{1}, \theta\right)\left(\angle B F x=\theta, 0<\theta<\frac{\pi}{2}\right)$. Then point $A\left(\rho_{2}, \theta+\pi\right)$.
Thus, $|\overrightarrow{F B}|=\rho_{1}=\frac{2}{1-\cos \theta}$,
$|\overrightarrow{A F}|=\rho_{2}=\frac{2}{1+\cos \theta}$.
From $\overrightarrow{F B}=\lambda \overrightarrow{A F}$, we get $|\overrightarrow{F B}|=\lambda|\overrightarrow{A F}|$.
Therefore, $\lambda=\frac{|\overrightarrow{F B}|}{|\overrightarrow{A F}|}=\frac{1+\cos \theta}{1-\cos \theta}$.
Since $\lambda \in[4,9]$, we have
$4 \leqslant \frac{1+\cos \theta}{1-\cos \theta} \leqslant 9\left(0<\theta<\frac{\pi}{2}\right)$.
Solving this, we get $\cos \theta \in\left[\frac{3}{5}, \frac{4}{5}\right]$.
Noting that the slope $k$ of line $l$ satisfies
$k^{2}=\tan ^{2} \theta=\frac{1}{\cos ^{2} \theta}-1 \in\left[\frac{9}{16}, \frac{16}{9}\right]$.
Since $0<\theta<\frac{\pi}{2}$, we get $\frac{3}{4} \leqslant k \leqslant \frac{4}{3}$.
(2) When $B$ is a point on the parabola in the fourth quadrant, by symmetry, we easily find $-\frac{4}{3} \leqslant k \leqslant-\frac{3}{4}$.
In the original Cartesian coordinate system, the equation of $l$ is $y=k(x-1)$.
Thus, the y-intercept of $l$ is
$$
b=-k \in\left[-\frac{4}{3},-\frac{3}{4}\right] \cup\left[\frac{3}{4}, \frac{4}{3}\right] .
$$