【答案】
分析:(1)設(shè)直線BC的解析式為y=mx+n,將B(5,0),C(0,5)兩點(diǎn)的坐標(biāo)代入,運(yùn)用待定系數(shù)法即可求出直線BC的解析式;同理,將B(5,0),C(0,5)兩點(diǎn)∑的坐標(biāo)代入y=x
2+bx+c,運(yùn)用待定系數(shù)法即可求出拋物線的解析式;
(2)MN的長是直線BC的函數(shù)值與拋物線的函數(shù)值的差,據(jù)此可得出一個關(guān)于MN的長和M點(diǎn)橫坐標(biāo)的函數(shù)關(guān)系式,根據(jù)函數(shù)的性質(zhì)即可求出MN的最大值;
(3)先求出△ABN的面積S
2=5,則S
1=6S
2=30.再設(shè)平行四邊形CBPQ的邊BC上的高為BD,根據(jù)平行四邊形的面積公式得出BD=3
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,過點(diǎn)D作直線BC的平行線,交拋物線與點(diǎn)P,交x軸于點(diǎn)E,在直線DE上截取PQ=BC,則四邊形CBPQ為平行四邊形.證明△EBD為等腰直角三角形,則BE=
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BD=6,求出E的坐標(biāo)為(-1,0),運(yùn)用待定系數(shù)法求出直線PQ的解析式為y=-x-1,然后解方程組
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,即可求出點(diǎn)P的坐標(biāo).
解答:解:(1)設(shè)直線BC的解析式為y=mx+n,
將B(5,0),C(0,5)兩點(diǎn)的坐標(biāo)代入,
得
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,解得
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,
所以直線BC的解析式為y=-x+5;
將B(5,0),C(0,5)兩點(diǎn)的坐標(biāo)代入y=x
2+bx+c,
得
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,解得
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,
所以拋物線的解析式為y=x
2-6x+5;
(2)設(shè)M(x,x
2-6x+5)(1<x<5),則N(x,-x+5),
∵M(jìn)N=(-x+5)-(x
2-6x+5)=-x
2+5x=-(x-
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)
2+
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,
∴當(dāng)x=
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時,MN有最大值
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;
(3)∵M(jìn)N取得最大值時,x=2.5,
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∴-x+5=-2.5+5=2.5,即N(2.5,2.5).
解方程x
2-6x+5=0,得x=1或5,
∴A(1,0),B(5,0),
∴AB=5-1=4,
∴△ABN的面積S
2=
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×4×2.5=5,
∴平行四邊形CBPQ的面積S
1=6S
2=30.
設(shè)平行四邊形CBPQ的邊BC上的高為BD,則BC⊥BD.
∵BC=5
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,∴BC•BD=30,∴BD=3
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.
過點(diǎn)D作直線BC的平行線,交拋物線與點(diǎn)P,交x軸于點(diǎn)E,在直線DE上截取PQ=BC,則四邊形CBPQ為平行四邊形.
∵BC⊥BD,∠OBC=45°,
∴∠EBD=45°,
∴△EBD為等腰直角三角形,BE=
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BD=6,
∵B(5,0),
∴E(-1,0),
設(shè)直線PQ的解析式為y=-x+t,
將E(-1,0)代入,得1+t=0,解得t=-1
∴直線PQ的解析式為y=-x-1.
解方程組
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,得
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,
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,
∴點(diǎn)P的坐標(biāo)為P
1(2,-3)(與點(diǎn)D重合)或P
2(3,-4).
點(diǎn)評:本題是二次函數(shù)的綜合題,其中涉及到運(yùn)用待定系數(shù)法求一次函數(shù)、二次函數(shù)的解析式,二次函數(shù)的性質(zhì),三角形的面積,平行四邊形的判定和性質(zhì)等知識點(diǎn),綜合性較強(qiáng),考查學(xué)生運(yùn)用方程組、數(shù)形結(jié)合的思想方法.(2)中弄清線段MN長度的函數(shù)意義是關(guān)鍵,(3)中確定P與Q的位置是關(guān)鍵.