3. As shown in Figure 1, given the parabola $C: y=x^{2}-2 x+4$ and the line $l: y=-2 x + 8$, the line $y=k x$ $(k>0)$ intersects the parabola $C$ at two different points $A$ and $B$, and intersects the line $l$ at point $P$. Perpendiculars are drawn from $A, B, P$ to the $x$-axis, with the feet of the perpendiculars being $A_{1}, B_{1}, P_{1}$, respectively. If $\frac{1}{O A_{1}}+\frac{1}{O B_{1}}=\frac{u}{O P_{1}}$, then $u=(\quad$.
(A) 1
(B) 2
(C) 3
(D) 4
3. B.
$$
\begin{array}{l}
\text { Let } A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) . \\
\text { From }\left\{\begin{array}{l}
y=k x, \\
y=x^{2}-2 x+4
\end{array}\right. \\
\Rightarrow x^{2}-(2+k) x+4=0 \\
\Rightarrow x_{1}+x_{2}=2+k, x_{1} x_{2}=4 \\
\Rightarrow \frac{1}{O A_{1}}+\frac{1}{O B_{1}}=\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1} x_{2}}{x_{1}+x_{2}}=\frac{2+k}{4} . \\
\text { From }\left\{\begin{array}{l}
y=k x, \\
y=-2 x+8
\end{array}\right. \\
\Rightarrow(k+2 .) x=8 \\
\Rightarrow x=O P_{1}=\frac{8}{k+2} . \\
\Rightarrow u=2 .
\end{array}
$$