4. Given a line $l$ passing through the point $F(a, 0)$ (not perpendicular to the $x$-axis) intersects the curve $C: y^{2}=4 a x$ at points $A$ and $B$. Let point $K(-a, 0)$, and the angle between $\boldsymbol{K A}$ and $\boldsymbol{K B}$ be $\theta$. Then the range of $\theta$ is $\qquad$.
4. $\left(0, \frac{\pi}{2}\right)$.
Let the equation of $l$ be $y=k(x-a)$.
From $\left\{\begin{array}{l}y^{2}=4 a x, \\ y=k(x-a),\end{array}\right.$ we get $y^{2}-\frac{4 a}{k} y-4 a^{2}=0$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
y_{1} y_{2}=-4 a^{2}, \boldsymbol{K A}=\left(x_{1}+a, y_{1}\right), \boldsymbol{K B}=\left(x_{2}+a, y_{2}\right) . \\
\text { Hence } \boldsymbol{K A} \cdot \boldsymbol{K B}=x_{1} x_{2}+a\left(x_{1}+x_{2}\right)+a^{2}+y_{1} y_{2} \\
=\frac{y_{1}^{2} y_{2}^{2}}{(4 a)^{2}}+a\left(\frac{y_{1}^{2}}{4 a}+\frac{y_{2}^{2}}{4 a}\right)+a^{2}-4 a^{2} \\
=\frac{1}{4}\left(y_{1}^{2}+y_{2}^{2}\right)-2 a^{2} \\
>\frac{1}{4} \times 2\left|y_{1} y_{2}\right|-2 a^{2}=\frac{1}{2} \times 4 a^{2}-2 a^{2}=0 .
\end{array}
$$

Therefore, $\boldsymbol{K A} \cdot \boldsymbol{K B}>0$, which means $\cos \theta=\frac{\boldsymbol{K A} \cdot \boldsymbol{K B}}{|\boldsymbol{K B} \cdot \boldsymbol{K B}|}>0$.
Thus, $0<\theta<\frac{\pi}{2}$.