【答案】
分析:(1)根據(jù)拋物線過C(0,4)點(diǎn),可確定c=4,然后可將A的坐標(biāo)代入拋物線的解析式中,即可得出二次函數(shù)的解析式.
(2)可先設(shè)Q的坐標(biāo)為(m,0);通過求△CEQ的面積與m之間的函數(shù)關(guān)系式,來得出△CQE的面積最大時點(diǎn)Q的坐標(biāo).
△CEQ的面積=△CBQ的面積-△BQE的面積.
可用m表示出BQ的長,然后通過相似△BEQ和△BCA得出△BEQ中BQ邊上的高,進(jìn)而可根據(jù)△CEQ的面積計(jì)算方法得出△CEQ的面積與m的函數(shù)關(guān)系式,可根據(jù)函數(shù)的性質(zhì)求出△CEQ的面積最大時,m的取值,也就求出了Q的坐標(biāo).
(3)本題要分三種情況進(jìn)行求解:
①當(dāng)OD=OF時,OD=DF=AD=2,又有∠OAF=45°,那么△OFA是個等腰直角三角形,于是可得出F的坐標(biāo)應(yīng)該是(2,2).由于P,F(xiàn)兩點(diǎn)的縱坐標(biāo)相同,因此可將F的縱坐標(biāo)代入拋物線的解析式中即可求出P的坐標(biāo).
②當(dāng)OF=DF時,如果過F作FM⊥OD于M,那么FM垂直平分OD,因此OM=1,在直角三角形FMA中,由于∠OAF=45°,因此FM=AM=3,也就得出了F的縱坐標(biāo),然后根據(jù)①的方法求出P的坐標(biāo).
③當(dāng)OD=OF時,OF=2,由于O到AC的最短距離為2
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,因此此種情況是不成立的.
綜合上面的情況即可得出符合條件的P的坐標(biāo).
解答:解:(1)由題意,得
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解得
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(2分)
∴所求拋物線的解析式為:y=-
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x
2+x+4.
(2)設(shè)點(diǎn)Q的坐標(biāo)為(m,0),過點(diǎn)E作EG⊥x軸于點(diǎn)G.
由-
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x
2+x+4=0,
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得x
1=-2,x
2=4
∴點(diǎn)B的坐標(biāo)為(-2,0)
∴AB=6,BQ=m+2
∵QE∥AC
∴△BQE∽△BAC
∴
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即
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∴
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∴S
△CQE=S
△CBQ-S
△EBQ=
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BQ•CO-
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BQ•EG
=
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(m+2)(4-
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)
=
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=-
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(m-1)
2+3
又∵-2≤m≤4
∴當(dāng)m=1時,S
△CQE有最大值3,此時Q(1,0).
(3)存在.在△ODF中.
(ⅰ)若DO=DF
∵A(4,0),D(2,0)
∴AD=OD=DF=2
又在Rt△AOC中,OA=OC=4
∴∠OAC=45度
∴∠DFA=∠OAC=45度
∴∠ADF=90度.此時,點(diǎn)F的坐標(biāo)為(2,2)
由-
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x
2+x+4=2,
得x
1=1+
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,x
2=1-
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此時,點(diǎn)P的坐標(biāo)為:P(1+
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,2)或P(1-
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,2).
(ⅱ)若FO=FD,過點(diǎn)F作FM⊥x軸于點(diǎn)M
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由等腰三角形的性質(zhì)得:OM=
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OD=1
∴AM=3
∴在等腰直角△AMF中,MF=AM=3
∴F(1,3)
由-
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x
2+x+4=3,
得x
1=1+
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,x
2=1-
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此時,點(diǎn)P的坐標(biāo)為:P(1+
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,3)或P(1-
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,3).
(ⅲ)若OD=OF
∵OA=OC=4,且∠AOC=90°
∴AC=
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∴點(diǎn)O到AC的距離為
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,而OF=OD=2
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,與OF≥2
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矛盾,所以AC上不存在點(diǎn)使得OF=OD=2,
此時,不存在這樣的直線l,使得△ODF是等腰三角形
綜上所述,存在這樣的直線l,使得△ODF是等腰三角形
所求點(diǎn)P的坐標(biāo)為:P(1+
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,2)或P(1-
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,2)或P(1+
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,3)或P(1-
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,3).
點(diǎn)評:本題著重考查了圖形平移變換、三角形相似、以及二次函數(shù)的綜合應(yīng)用等重要知識點(diǎn),要注意的是(3)中不確定等腰三角形的腰是哪些線段時,要分類進(jìn)行討論.