解:(1)∵C(0,5),D(2,5),
∴拋物線的對稱軸為直線x=
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=1,
∵A(-1,0),
∴2×1-(-1)=3,
∴點(diǎn)B的坐標(biāo)為(3,0);
(2)如圖,連接CD,則∠DCF=90°,
∵四邊形DFBG為矩形,
∴∠DFC+∠OFB=180°-90°=90°,
∵∠OFB+∠OBF=90°,
∴∠DFC=∠OBF,
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又∵∠DCF=∠FOB=90°,
∴△CDF∽△OFB,
∴
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=
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,
∵B(3,0),C(0,5),D(2,5),
∴CD=2,OB=3,OC=5,
∴CF=5-OF,
∴
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=
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,
整理得,OF
2-5OF+6=0,
解得OF=2或OF=3,
∴點(diǎn)F的坐標(biāo)為(0,2)或(0,3);
(3)連接BD,設(shè)FG、BD相交于點(diǎn)H,
∵四邊形DFBG是平行四邊形,
∴FG、BD互相平分,
∴FG=2FH,
又∵B(3,0),D(2,5),
∴點(diǎn)H的坐標(biāo)為(2.5,2.5),
根據(jù)垂線段最短,F(xiàn)H⊥y軸時,F(xiàn)H最短,
此時,F(xiàn)H=2.5,
FG=2FH=2×2.5=5;
(4)設(shè)拋物線解析式為y=a(x-1)
2+k(a≠0),
把點(diǎn)A、C的坐標(biāo)代入得,
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,
解得
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,
∴拋物線解析式為y=-
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(x-1)
2+
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,
∵E為AB中點(diǎn),
∴點(diǎn)E的坐標(biāo)為(1,0),
∴以E為圓心,以2為半徑的圓為(x-1)
2+y
2=4,
與拋物線解析式聯(lián)立消掉(x-1)
2得,-
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(4-y
2)+
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=y,
整理得,5y
2-3y=0,
解得y
1=0,y
2=
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,
y=
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時,-
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(x-1)
2+
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=
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,
整理得,(x-1)
2=
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,
解得x
1=
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,x
2=
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,
∴-1<x<
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或
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<x<3時,拋物線上的點(diǎn)到E點(diǎn)的距離小于2.
故答案為:(1)(3,0);(4)-1<x<
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或
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<x<3.
分析:(1)根據(jù)點(diǎn)C、D的縱坐標(biāo)相等求出拋物線的對稱軸,然后根據(jù)二次函數(shù)的對稱性求出點(diǎn)B的坐標(biāo)即可;
(2)連接CD,然后求出△CDF和△OFB相似,根據(jù)相似三角形對應(yīng)邊成比例列式求出OF,然后寫出點(diǎn)F的坐標(biāo)即可;
(3)連接BD,設(shè)FG、BD相交于點(diǎn)H,根據(jù)平行四邊形的對角線互相平分可得FG=2FH,再求出點(diǎn)H的坐標(biāo),再根據(jù)垂線段最短可得FH⊥y軸時,F(xiàn)H最短,從而求出FH,再求出FG即可;
(4)利用待定系數(shù)法求出函數(shù)解析式,再寫出以點(diǎn)E為圓心,以2為半徑的圓的解析式,然后消掉x得到關(guān)于y的一元二次方程,求解得到y(tǒng)的值,再代入拋物線解析式求出到點(diǎn)E的距離等于2的橫坐標(biāo)x的值,然后根據(jù)函數(shù)圖象解答.
點(diǎn)評:本題是二次函數(shù)綜合題型,主要利用了二次函數(shù)的對稱性,相似三角形的判定與性質(zhì),平行四邊形的對角線互相平分的性質(zhì),待定系數(shù)法求二次函數(shù)解析式,利用圓的解析式求出拋物線到點(diǎn)E的距離等于2的點(diǎn)的縱坐標(biāo)是解題的關(guān)鍵,也是本題的難點(diǎn).