【答案】
分析:(1)將A、B的坐標(biāo)代入拋物線的解析式中,即可求出待定系數(shù)的值;
(2)先求出直線AC的解析式,由于BD∥AC,那么直線BD的斜率與直線AC的相同,可據(jù)此求出直線BD的解析式,聯(lián)立拋物線的解析式即可求出D點(diǎn)的坐標(biāo);由圖知四邊形ACBD的面積是△ABC和△ABD的面積和,由此可求得其面積;
(3)易知OA=OB=OC=1,那么△ACB是等腰直角三角形,由于AC∥BD,則∠CBD=90°;根據(jù)B、C的坐標(biāo)可求出BC、BD的長,進(jìn)而可求出它們的比例關(guān)系;若以A、M、N為頂點(diǎn)的三角形與△BCD相似,那么兩個(gè)直角三角形的對應(yīng)直角邊應(yīng)該成立,可據(jù)此求出△AMN兩條直角邊的比例關(guān)系,連接拋物線的解析式即可求出M點(diǎn)的坐標(biāo).
解答:解:(1)依題意,得:
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,解得
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;
∴拋物線的解析式為:y=-x
2+1;
(2)易知A(-1,0),C(0,1),則直線AC的解析式為:y=x+1;
由于AC∥BD,可設(shè)直線BD的解析式為y=x+h,則有:1+h=0,h=-1;
∴直線BD的解析式為y=x-1;聯(lián)立拋物線的解析式得:
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,解得
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,
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;
∴D(-2,-3);
∴S
四邊形ACBD=S
△ABC+S
△ABD=
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×2×1+
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×2×3=4;
(3)∵OA=OB=OC=1,
∴△ABC是等腰Rt△;
∵AC∥BD,
∴∠CBD=90°;
易求得BC=
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,BD=3
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;
∴BC:BD=1:3;
由于∠CBD=∠MNA=90°,若以A、M、N為頂點(diǎn)的三角形與△BCD相似,則有:
△MNA∽△CBD或△MNA∽△DBC,得:
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=
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或
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=3;
即MN=
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AN或MN=3AN;
設(shè)M點(diǎn)的坐標(biāo)為(x,-x
2+1),
①當(dāng)x>1時(shí),AN=x-(-1)=x+1,MN=x
2-1;
∴x
2-1=
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(x+1)或x
2-1=3(x+1)
解得x=
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,x=-1(舍去)或x=4,x=-1(舍去);
∴M點(diǎn)的坐標(biāo)為:M(
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,-
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)或(4,-15);
②當(dāng)x<-1時(shí),AN=-1-x,MN=x
2-1;
∴x
2-1=
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(-x-1)或x
2-1=3(-x-1)
解得x=
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,x=-1(兩個(gè)都不合題意,舍去)或x=-2,x=-1(舍去);
∴M(-2,-3);
故存在符合條件的M點(diǎn),且坐標(biāo)為:M(
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,-
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)或(4,-15)或(-2,-3).
點(diǎn)評:此題主要考查了二次函數(shù)解析式的確定、圖形面積的求法以及相似三角形的判定和性質(zhì)等重要知識點(diǎn),同時(shí)還考查了分類討論的數(shù)學(xué)思想.