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解:(1)∵拋物線的頂點D的坐標(biāo)為(1,
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),
∴設(shè)拋物線的解析式為
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,
將(0,0)代入,得
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,
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,
∴拋物線的解析式為
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,
即
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,
設(shè)y=0,則x=0或2,
∴點C的坐標(biāo)為(0,2),
∵點P為CD的中點,
∴
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;
(2)在y軸右側(cè)的拋物線上存在點Q,使以Q為圓心的圓同時與y軸、直線OP相切,
理由如下:
①若⊙Q在直線OP上方,則Q與D點重合,此時Q
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;
②若⊙Q在直線OP下方,與y軸、直線OP切于E、F,
則QE=QF,QE⊥y軸,QF⊥OP,
∴OQ平分∠EOF,
∵∠EOF=120°,
∴∠FOQ=60°,
∵∠POC=30°,則∠QOC=30°,
設(shè)Q
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,則
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,
解得m
1=0(舍去),
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,
∴
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;
(3)證明:∵在過點O、M、D的圓中,有∠MOD=∠NOD,
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∴
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,
∴MD=ND,
易得OD平分∠AOP,DA⊥y軸,DP⊥OP,
∴DA=DP,
可證得△NAD≌△MPD(HL),
∴MP=AN,
∴OM+ON=OP-MP+OA+AN=OP+OA=2OA=
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,
則OM+ON=
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,即OM+ON為定值;
(4)作過P、D兩點且與y軸相切于點H的圓S,
則由圓周角大于圓外角可知,∠PHD最大.
設(shè)S(x,y),則由HS=SD=SP,
可得,y=2
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±6,
∵0<y<
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,
∴H(0,2
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-6).
分析:(1)因為拋物線的頂點D的坐標(biāo)為(1,
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),所以可設(shè)設(shè)拋物線的解析式為
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,又因為函數(shù)圖象過原點,所以把(0,0)代入求出a的值即可求出拋物線的解析式,設(shè)y=0,則可求出拋物線和x軸的交點坐標(biāo),進(jìn)而求出點P的坐標(biāo);
(2)在y軸右側(cè)的拋物線上存在點Q,使以Q為圓心的圓同時與y軸、直線OP相切,此題要分兩種情況討論:①若⊙Q在直線OP上方②若⊙Q在直線OP下方,再分別求出符合題意的Q點的坐標(biāo)即可;
(3)由圓周角定理可證明MD=ND,進(jìn)而證明△NAD≌△MPD(HL),由全等三角形的性質(zhì)可得MP=AN,所以O(shè)M+ON=OP-MP+OA+AN=OP+OA=2OA=

,則OM+ON=
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,即OM+ON為定值;
(4)作過P、D兩點且與y軸相切于點H的圓S,則由圓周角大于圓外角可知,∠PHD最大,S(x,y),則由HS=SD=SP,繼而求出點H坐標(biāo).
點評:本題著重考查了待定系數(shù)法求二次函數(shù)解析式、中點坐標(biāo)公式、全等三角形的判定和性質(zhì)、圓周角定理的運用、角平分線的性質(zhì)以及考查學(xué)生分類討論,數(shù)形結(jié)合的數(shù)學(xué)思想方法,題目的綜合性強(qiáng),難度大,對學(xué)生的綜合解題能力要求很高.