解：?$(1)$?∵?$AO=1，$??$tan ∠ACO=\frac {1}{5}$?
∴?$OC=5，$?即點(diǎn)?$C$?的坐標(biāo)為?$(0，$??$5)$?
∵二次函數(shù)的圖像與?$x$?軸交于?$A(-1，$??$0)、$??$B(5，$??$0)$?兩點(diǎn)且過點(diǎn)?$C(0，$??$5)$?
設(shè)二次函數(shù)的表達(dá)式為?$y=ax^2+bx+c$?
帶入得?$\begin{cases}{a-b+c=0}\\{25a+5b+c=0}\\{c=5}\end{cases}，$?解得?$\begin{cases}{a=-1}\\{b=4}\\{c=5}\end{cases}$?
∴二次函數(shù)的表達(dá)式為?$y=-x^2+4x+5$?
?$ (2)$?∵?$y=-x^2+4x+5=-(x-2)^2+9$?
∴頂點(diǎn)坐標(biāo)為?$(2，$??$9)$?
如圖①，過點(diǎn)?$D$?作?$DN⊥AB$?于點(diǎn)?$N，$?作?$DM⊥OC$?于點(diǎn)?$M$?
四邊形?$ACDB$?的面積?$=S_{△AOC}+S_{矩形OMDN}-S_{△CDM}+S_{△DNB}$?
?$=\frac {1}{2} ×1×5+2×9- \frac {1}{2} ×2×(9-5)+\frac {1}{2} ×(5-2)×9=30$?
?$ (3)$?如圖②，?$P $?是拋物線上的一點(diǎn)，且在第一象限
當(dāng)?$∠ACO=∠PBC$?時(shí)，連接?$PB，$?過點(diǎn)?$C$?作?$CE⊥BC$?交?$BP $?于點(diǎn)?$E，$?過點(diǎn)?$E$?作?$EF⊥OC$?于點(diǎn)?$F$?
∵?$OC=OB=5，$?則?$BC=5 \sqrt{2}$?
∵?$∠ACO=∠PBC$?
∴?$tan ∠ACO= tan ∠PBC，$?即?$\frac {1}{5} =\frac {CE}{CB} =\frac {CE}{5\sqrt{2}}$?
∴?$CE=\sqrt{2}$?
∵?$CO=BO$?
∴?$∠OCB=45°$?
∵?$∠ECB=90°$?
∴?$∠ECF=45°$?
∴?$△EFC$?是等腰直角三角形
∴?$FC=FE=1$?
∴點(diǎn)?$E$?的坐標(biāo)為?$(1，$??$6)$?
∴易求得過?$B、$??$E$?兩點(diǎn)的直線的表達(dá)式為?$y=- \frac {3}{2} x+ \frac {15}{2}$?
令?$\begin{cases}{ y=-\dfrac {3}{2}x+\dfrac {15}{2}}\\{y=-x^2+4x+5}\end{cases}，$?解得?$\begin{cases}{x=5 }\\{y=0}\end{cases}，$?或?$\begin{cases}{x=\dfrac {1}{2}}\\{y=\dfrac {27}{4}}\end{cases}$?
∴直線?$BE$?與拋物線的兩個(gè)交點(diǎn)為?$B(5，$??$0)、$??$P(\frac {1}{2} ，$??$\frac {27}{4} )$?
∴即所求點(diǎn)?$P $?的坐標(biāo)為?$(\frac {1}{2} ，$??$\frac {27}{4} )$?
