2.(2024·宿迁三模)如图①,抛物线$y = ax^{2}+bx + c(a\neq0)$与$x$轴交于$A$,$B$两点,与$y$轴交于点$C$,$OB = OC = 2OA$,点$C$的坐标为$(0,-2)$,连接$BC$．
(1)求抛物线对应的函数表达式．
(2)如图①,$D(x,y)$是线段$BC$下方抛物线上的一动点,连接$DO$交线段$BC$于点$E$,连接$AC$,$CD$．设$\frac{OE}{DE}=k$,当$\angle ACD = 90^{\circ}$时,求$k$的值．
(3)如图②,在线段$BC$上方有一条动直线$KL$始终与线段$BC$平行,且与抛物线$y = ax^{2}+bx + c(a\neq0)$交于$K$,$L$两点,直线$CK$与$BL$交于点$P$．$\triangle BCP$的面积能否为$4$？若能,直接写出点$P$的坐标；若不能,请说明理由．

[跟踪训练]2.(1)$\because$点$C$的坐标为$(0,-2)$，$\therefore OC = 2$.$\because OB = OC = 2OA$，$\therefore OA = 1$，$OB = 2$.$\therefore$点$A$，$B$的坐标分别为$(-1,0)$，$(2,0)$.$\because$点$A(-1,0)$，$B(2,0)$，$C(0,-2)$在抛物线$y=ax^{2}+bx + c$上，$\therefore\begin{cases}0=a - b + c,\\0=4a + 2b + c,\\-2=c,\end{cases}$解得$\begin{cases}a = 1,\\b=-1,\\c=-2.\end{cases}$$\therefore$抛物线对应的函数表达式为$y=x^{2}-x - 2$.
(2)由点$A$，$C$的坐标，得直线$AC$对应的函数表达式为$y=-2x - 2$.当$\angle ACD = 90^{\circ}$时，易得直线$CD$对应的函数表达式为$y=\frac{1}{2}x - 2$.联立$\begin{cases}y=\frac{1}{2}x - 2,\\y=x^{2}-x - 2,\end{cases}$消去$y$、整理，得$2x^{2}-3x = 0$，解得$x = 0$（不合题意，舍去）或$x=\frac{3}{2}$.$\therefore$点$D$的坐标为$(\frac{3}{2},-\frac{5}{4})$.由点$B$，$C$的坐标，得直线$BC$对应的函数表达式为$y=x - 2$.如图①，过点$D$作$DH// y$轴，交$BC$于点$H$，则点$H$的坐标为$(\frac{3}{2},-\frac{1}{2})$.$\therefore DH=-\frac{1}{2}-(-\frac{5}{4})=\frac{3}{4}$.易得$\triangle COE\sim\triangle HDE$，则$k=\frac{OE}{DE}=\frac{OC}{DH}=\frac{2}{\frac{3}{4}}=\frac{8}{3}$.
(3)设点$K$，$L$的坐标分别为$(m,m^{2}-m - 2)$，$(n,n^{2}-n - 2)$.由点$K$，$L$的坐标，得直线$KL$对应的函数表达式为$y=(m + n - 1)(x - n)+n^{2}-n - 2$.$\because KL// BC$，$\therefore m + n - 1 = 1$，即$m = 2 - n$.由点$K$，$C$的坐标，得直线$KC$对应的函数表达式为$y=(m - 1)x - 2$.同理，可得直线$BL$对应的函数表达式为$y=(n + 1)(x - 2)$.联立$\begin{cases}y=(m - 1)x - 2,\\y=(n + 1)(x - 2),\end{cases}$消去$y$，得$(n + 1)(x - 2)=(m - 1)x - 2$，解得$x=\frac{-2n}{m - n - 2}=\frac{-2n}{-2n}=1$.$\therefore$点$P$的坐标为$(1,-n - 1)$.如图②，过点$P$作直线$PN// BC$，交$y$轴于点$N$，过点$C$作$CT\perp PN$于点$T$，则易得$CT=\frac{\sqrt{2}}{2}CN$，直线$PN$对应的函数表达式为$y=x - 1 - n - 1=x - n - 2$.令$x = 0$，得$y=-n - 2$.$\therefore$点$N$的坐标为$(0,-n - 2)$.$\therefore CN=-2 + n + 2 = n$.易得$BC = 2\sqrt{2}$，$\therefore S_{\triangle BCP}=\frac{1}{2}BC\cdot CT=\frac{1}{2}BC\cdot\frac{\sqrt{2}}{2}CN=\frac{1}{2}\times2\sqrt{2}\times\frac{\sqrt{2}}{2}n=n$.当$S_{\triangle BCP}=4$时，$n = 4$，此时点$P$的坐标为$(1,-5)$.