【答案】
分析:(1)作CH⊥x軸,垂足為H,CH必經(jīng)過圓心D,易得CH=6,則點(diǎn)C的坐標(biāo)可以得到.
(2)連接OA,OC則陰影部分的面積S=S
扇形DAC-S
△DAC;
(3)設(shè)OC的中點(diǎn)是E,E點(diǎn)的坐標(biāo)就可以求出,利用待定系數(shù)法就可以求出直線DE的解析式,直線與拋物線的交點(diǎn)就是所求的點(diǎn)P.
解答:
解:(1)如圖,作CH⊥x軸,垂足為H,
∵直線CH為拋物線對稱軸,
∴CH垂直平分AB,
∴CH必經(jīng)過圓心D(-2,-2).
∵DC=4,
∴CH=6
∴C點(diǎn)的坐標(biāo)為(-2,-6).(3分)
(2)連接AD.
在Rt△ADH中,AD=4,DH=2,
∴∠HAD=30°,AH=

(4分)
∴∠ADC=120°
∴S
扇形DAC=

π(5分)
S
△DAC=

AH•CD=

×2

×4=4

.(6分)
∴陰影部分的面積S=S
扇形DAC-S
△DAC=

π-4

.(7分)
(3)又∵AH=2

,H點(diǎn)坐標(biāo)為(-2,0),H為AB的中點(diǎn),
∴A點(diǎn)坐標(biāo)為(-2-2

,0),B點(diǎn)坐標(biāo)為(

,0).(8分)
又∵拋物線頂點(diǎn)C的坐標(biāo)為(-2,-6),
設(shè)拋物線解析式為y=a(x+2)
2-6.
∵B(

,0)在拋物線上,
∴a(2

-2+2)
2-6=0,
解得

.
∴拋物線的解析式為y=

(x+2)
2-6(9分).
設(shè)OC的中點(diǎn)為E,過E作EF⊥x軸,垂足為F,連接DE,

∵CH⊥x軸,EF⊥x軸,
∴CH∥EF
∵E為OC的中點(diǎn),
∴EF=

CH=3,OF=

OH=1.
即點(diǎn)E的坐標(biāo)為(-1,-3).
設(shè)直線DE的解析式為y=kx+b(k≠0),
∴

,
解得k=-1,b=-4,
∴直線DE的解析式為y=-x-4.(10分)
若存在P點(diǎn)滿足已知條件,則P點(diǎn)必在直線DE和拋物線上.
設(shè)點(diǎn)P的坐標(biāo)為(m,n),
∴n=-m-4,即點(diǎn)P坐標(biāo)為(m,-m-4),
∴-m-4=

(m+2)
2-6,
解這個(gè)方程,得m
1=0,m
2=-6
∴點(diǎn)P的坐標(biāo)為(0,-4)和(-6,2).
故在拋物線上存在點(diǎn)P,使DP所在直線平分線段OC.(12分)
點(diǎn)評:本題主要考查了待定系數(shù)法求函數(shù)的解析式,以及弓形面積的求法,轉(zhuǎn)化為扇形的面積與三角形的面積的差的問題.