3.(2024·宿迁宿城一模)如图,抛物线$y = x^{2}+bx + c$经过$A(0,3)$,$B(4,3)$两点,过点$B$作$BC\perp x$轴于点$C$．
(1)求该抛物线对应的函数表达式．
(2)连接$OB$．若$D$是抛物线上一点,满足$\angle BOC=\angle OBD$,求点$D$的坐标．
(3)若$P$为抛物线上一点,且在第四象限内．已知直线$PA$,$PB$与$x$轴分别交于$E$,$F$两点．当点$P$运动时,$\frac{1}{OE}+\frac{1}{CF}$是否为定值？若是,求出该定值；若不是,请说明理由．

[跟踪训练]3.(1)$\because$抛物线$y=x^{2}+bx + c$经过$A(0,3)$，$B(4,3)$两点，$\therefore\begin{cases}c = 3,\\4^{2}+4b + c = 3,\end{cases}$解得$\begin{cases}b=-4,\\c = 3.\end{cases}$$\therefore$该抛物线对应的函数表达式为$y=x^{2}-4x + 3$.
(2)①当点$D$在直线$OB$的上方时，如图①所示.$\because\angle BOC=\angle OBD$，$\therefore BD// x$轴.$\because$点$A$与点$B$的纵坐标相等，$\therefore AB// x$轴，此时点$A$与点$D$重合，即点$D$的坐标为$(0,3)$.
②当点$D$在直线$OB$的下方时，设$BD$与$x$轴交于点$M$，如图②所示.$\because\angle BOC=\angle OBD$，$\therefore OM = BM$.$\because BC\perp x$轴于点$C$，点$B$的坐标为$(4,3)$，$\therefore$点$C$的坐标为$(4,0)$，$OC = 4$，$BC = 3$.设$OM = BM = m$，则$CM=OC - CM=4 - m$.在$Rt\triangle BCM$中，$CM^{2}+BC^{2}=BM^{2}$，即$(4 - m)^{2}+3^{2}=m^{2}$，解得$m=\frac{25}{8}$.$\therefore$点$M$的坐标为$(\frac{25}{8},0)$.设直线$BD$对应的函数表达式为$y=k_{1}x + b_{1}$.$\because$点$B(4,3)$，$M(\frac{25}{8},0)$在直线$BD$上，$\therefore\begin{cases}4k_{1}+b_{1}=3,\\\frac{25}{8}k_{1}+b_{1}=0,\end{cases}$解得$\begin{cases}k_{1}=\frac{24}{7},\\b_{1}=-\frac{75}{7}.\end{cases}$$\therefore$直线$BD$对应的函数表达式为$y=\frac{24}{7}x-\frac{75}{7}$.联立$\begin{cases}y=\frac{24}{7}x-\frac{75}{7},\\y=x^{2}-4x + 3,\end{cases}$解得$\begin{cases}x=\frac{24}{7},\\y=\frac{51}{49},\end{cases}$或$\begin{cases}x = 4,\\y = 3\end{cases}$（舍去）.$\therefore$点$D$的坐标为$(\frac{24}{7},\frac{51}{49})$.综上所述，点$D$的坐标为$(0,3)$或$(\frac{24}{7},\frac{51}{49})$.
(3)$\frac{1}{OE}+\frac{1}{CF}$是定值.在$y=x^{2}-4x + 3$中，令$y = 0$，得$x_{1}=1$，$x_{2}=3$.$\therefore$抛物线与$x$轴的交点坐标分别是$(1,0)$，$(3,0)$.设点$P$的坐标为$(t,t^{2}-4t + 3)$，则$1＜t＜3$.设直线$AP$对应的函数表达式为$y=k_{2}x + b_{2}$.$\because$点$A(0,3)$，$P(t,t^{2}-4t + 3)$在直线$AP$上，$\therefore\begin{cases}b_{2}=3,\\k_{2}t + b_{2}=t^{2}-4t + 3,\end{cases}$解得$\begin{cases}k_{2}=t - 4,\\b_{2}=3.\end{cases}$$\therefore$直线$AP$对应的函数表达式为$y=(t - 4)x + 3$.在$y=(t - 4)x + 3$中，令$y = 0$，得$x=\frac{3}{4 - t}$.$\therefore$点$E$的坐标为$(\frac{3}{4 - t},0)$.$\therefore OE=\frac{3}{4 - t}$.设直线$BP$对应的函数表达式为$y=k_{3}x + b_{3}$.$\because$点$B(4,3)$，$P(t,t^{2}-4t + 3)$在直线$BP$上，$\therefore\begin{cases}4k_{3}+b_{3}=3,\\k_{3}t + b_{3}=t^{2}-4t + 3,\end{cases}$解得$\begin{cases}k_{3}=t,\\b_{3}=-4t + 3.\end{cases}$$\therefore$直线$BP$对应的函数表达式为$y=tx - 4t + 3$.在$y=tx - 4t + 3$中，令$y = 0$，得$x=\frac{4t - 3}{t}$.$\therefore$点$F$的坐标为$(\frac{4t - 3}{t},0)$.$\therefore OF=\frac{4t - 3}{t}$.$\therefore CF=OC - OF=4-\frac{4t - 3}{t}=\frac{3}{t}$.$\therefore\frac{1}{OE}+\frac{1}{CF}=\frac{4 - t}{3}+\frac{t}{3}=\frac{4}{3}$.$\therefore\frac{1}{OE}+\frac{1}{CF}$是定值，该定值为$\frac{4}{3}$.