4. Given the parabola $y=a x^{2}(a>0)$ and the line $y=b x+c(b \neq 0)$ have two common points, their x-coordinates are $x_{1} 、 x_{2}$, and the line $y=b x+c$ intersects the x-axis at the point $\left(x_{3}, 0\right)$. Then the relationship satisfied by $x_{1} 、 x_{2} 、 x_{3}$ is ( ).
(A) $x_{1}+x_{2}=x_{3}$
(B) $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}}$
(C) $x_{3}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}$
(D) $x_{1} x_{2}+x_{2} x_{3}=x_{3} x_{1}$
4. B.

Solving the system of equations $\left\{\begin{array}{l}y=a x^{2}, \\ y=b x+c\end{array}\right.$ yields $a x^{2}-b x-c=0$.
Therefore, $x_{1}+x_{2}=\frac{b}{a}, x_{1} x_{2}=-\frac{c}{a}$.
Substituting $y=0$ into $y=b x+c$, we get $x_{3}=-\frac{c}{b}$.
Thus, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=-\frac{b}{c}$.
Hence, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}}$.