解???$：(1)$???設(shè)???$OD=x，$???則???$AD=CD=8-x$???
在???$Rt\triangle OCD$???中，由勾股定理得
???$C{D}^2=O{D}^2+O{C}^2，$???即???${(8-x)}^2={x}^2+{4}^2$???
解得???$x=3$???
???$∴OD=3$???
???$∴D(3，0)$???
???$∵OA=8，$??????$OC=4$???
???$∴B(8，-4)，$??????$C(0，-4)$???且拋物線(xiàn)過(guò)???$B，$??????$C$???點(diǎn)
∴拋物線(xiàn)的對(duì)稱(chēng)軸為???$x=4$???
???$∵D(3，0)$???
∴另一個(gè)交點(diǎn)???$E(5，0).$???
???$(2)$???不存在這樣的點(diǎn)???$P，$???理由如下：
???${S}_{矩形OABC}=OA·OB=32$???
若存在這樣的點(diǎn)???$P，$???設(shè)點(diǎn)???$P$???到???$BC$???的距離為???$h$???
則???${S}_{△PBC}=\frac {1}{2}BC·h=32$???
???$∴h=8$???
設(shè)拋物線(xiàn)的解析式為???$y=a{x}^2+bx+c$???
∵拋物線(xiàn)過(guò)???$B(8，-4)，$??????$C(0，-4)，$??????$D(3，0)$???
???$∴\{\begin{array}{l}-4=64a+8b+c\\-4=c\\0=9a+3b+c\end{array}.$???
解得???$\{\begin{array}{l}a=-\frac {4}{15}\\b=\frac {32}{15}\\c=-4\end{array}.$???
∴拋物線(xiàn)解析式為???$y=-\frac {4}{15}{x}^2+\frac {32}{15}x-4=-\frac {4}{15}{(x-4)}^2+\frac {4}{15}$???
???$∴p$???拋物線(xiàn)的頂點(diǎn)為???$(4，\frac {4}{15})$???
∴頂點(diǎn)到???$BC$???的距離為???$4+\frac {4}{15}=\frac {64}{15}<8$???
∴不存在這樣的點(diǎn)???$P，$???使???$\triangle PBC$???的面積等于矩形???$OABC$???的面積.
