【答案】
分析:(1)可連接OA,通過(guò)證∠AOE=60°,即與旋轉(zhuǎn)角相同來(lái)得出OE在y軸上的結(jié)論.
(2)已知了AB,OB的長(zhǎng)即可求出A的坐標(biāo),在直角三角形OEF中,可用勾股定理求出OE的長(zhǎng),也就能求得E點(diǎn)的坐標(biāo),要想得出拋物線的解析式還少D點(diǎn)的坐標(biāo),可過(guò)D作x軸的垂線,通過(guò)構(gòu)建直角三角形,根據(jù)OD的長(zhǎng)和∠DOx的正弦和余弦值來(lái)求出D的坐標(biāo).
求出A、E、D三點(diǎn)坐標(biāo)后即可用待定系數(shù)法求出拋物線的解析式.
(3)可先求出矩形的面積,進(jìn)而可得出平行四邊形OBPQ的面積.由于平行四邊形中OB邊的長(zhǎng)是定值,因此可根據(jù)平行四邊形的面積求出P點(diǎn)的縱坐標(biāo)(由于P點(diǎn)在x軸上方,因此P的縱坐標(biāo)為正數(shù)),然后將P點(diǎn)的縱坐標(biāo)代入拋物線中可求出P點(diǎn)的坐標(biāo).求出P點(diǎn)的坐標(biāo)后,將P點(diǎn)分別向左、向右平移OB個(gè)單位即可得出Q點(diǎn)的坐標(biāo),由此可得出符合條件的兩個(gè)P點(diǎn)坐標(biāo)和四個(gè)Q點(diǎn)坐標(biāo).
解答:解:(1)點(diǎn)E在y軸上
理由如下:
連接AO,如圖所示,在Rt△ABO中,∵AB=1,BO=

,
∴AO=2∴sin∠AOB=

,∴∠AOB=30°
由題意可知:∠AOE=60°∴∠BOE=∠AOB+∠AOE=30°+60°=90°
∵點(diǎn)B在x軸上,∴點(diǎn)E在y軸上.
(2)過(guò)點(diǎn)D作DM⊥x軸于點(diǎn)M,
∵OD=1,∠DOM=30°

∴在Rt△DOM中,DM=

,OM=

∵點(diǎn)D在第一象限,
∴點(diǎn)D的坐標(biāo)為

由(1)知EO=AO=2,點(diǎn)E在y軸的正半軸上
∴點(diǎn)E的坐標(biāo)為(0,2)
∴點(diǎn)A的坐標(biāo)為(-

,1)
∵拋物線y=ax
2+bx+c經(jīng)過(guò)點(diǎn)E,
∴c=2
由題意,將A(-

,1),D(

,

)代入y=ax
2+bx+2中,
得

解得

∴所求拋物線表達(dá)式為:y=-

x
2-

x+2
(3)存在符合條件的點(diǎn)P,點(diǎn)Q.
理由如下:∵矩形ABOC的面積=AB•BO=

∴以O(shè),B,P,Q為頂點(diǎn)的平行四邊形面積為

.
由題意可知OB為此平行四邊形一邊,
又∵OB=

∴OB邊上的高為2
依題意設(shè)點(diǎn)P的坐標(biāo)為(m,2)
∵點(diǎn)P在拋物線y=-

x
2-

x+2上
∴-

m
2-

m+2=2
解得,m
1=0,m
2=-

∴P
1(0,2),P
2(-

,2)
∵以O(shè),B,P,Q為頂點(diǎn)的四邊形是平行四邊形,
∴PQ∥OB,PQ=OB=

,
∴當(dāng)點(diǎn)P
1的坐標(biāo)為(0,2)時(shí),點(diǎn)Q的坐標(biāo)分別為Q
1(-

,2),Q
2(

,2);
當(dāng)點(diǎn)P
2的坐標(biāo)為(-

,2)時(shí),點(diǎn)Q的坐標(biāo)分別為Q
3(-

,2),Q
4(

,2).
點(diǎn)評(píng):本題著重考查了待定系數(shù)法求二次函數(shù)解析式、圖形旋轉(zhuǎn)變換、平行四邊形的性質(zhì)等知識(shí)點(diǎn),綜合性強(qiáng),能力要求較高.考查學(xué)生數(shù)形結(jié)合的數(shù)學(xué)思想方法.