【答案】
分析:(1)欲求拋物線的解析式,就必須先求出D點的坐標(biāo),也就要求出CD的長;根據(jù)折疊的性質(zhì)知:AB=A′B=OC=4,易證得△OCD≌△BA′D,那么CD=A′D,BD=BC-CD=8-CD,在Rt△A′BD中,利用勾股定理即可求出CD的長,從而得到點D的坐標(biāo),進而可由待定系數(shù)法求得拋物線的解析式.
(2)△PAA′中,AA′的長是定值,若此三角形的周長最小,那么PA+PA′的長最小,由于O、A關(guān)于拋物線的對稱軸對稱,那么P點必為直線OA′與拋物線對稱軸的交點;過A′作x軸的垂線,交BC于M,交OA于N,在Rt△A′BD中,利用射影定理即可求得MD的長,利用直角三角形面積的不同表示方法即可求出A′N的長,由此求得點A′的坐標(biāo),進而得到直線OA′的解析式,聯(lián)立拋物線對稱軸方程,即可得到點P的坐標(biāo).
(3)此題應(yīng)分三種情況考慮:
①點D為直角頂點,那么QD⊥AD,易得直線AD的解析式,由于QD⊥AD,那么直線QD和直線AD的斜率的乘積為-1,結(jié)合D點坐標(biāo)即可求得直線DQ的解析式,聯(lián)立拋物線的對稱軸方程,即可求得點Q的坐標(biāo);
②點A為直角頂點,方法同①;
③點Q為直角頂點,設(shè)出點Q的坐標(biāo),由于DQ⊥AQ,那么兩條直線的斜率乘積為-1,可據(jù)此列出關(guān)于Q點縱坐標(biāo)的方程,從而求得點Q的坐標(biāo).
解答:解:(1)根據(jù)折疊的性質(zhì)知:∠DA′B=∠OAB=90°,A′B=AB=4;
∵OC=A′B,∠DA′B=∠DCO=90°,∠ODC=∠BDA′,
∴△OCD≌△BA′D,
∴CD=A′D;
設(shè)CD=A′D=x,則BD=8-x;
Rt△A′BD中,由勾股定理得:x
2+4
2=(8-x)
2,
解得x=3;
故D(3,4);
設(shè)拋物線的解析式為:y=ax(x-8)
2,
則有:3a(3-8)=4,
a=-
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;
∴y=-
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x(x-8)
2=-
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x
2+
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x.
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(2)過A′作x軸的垂線,交BC于M,交OA于N;
在Rt△A′BD中,A′M⊥BD,則:
A′M=A′D•A′B÷BD=
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,
DM=A′D
2÷BD=
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;
故CM=
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,A′N=
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,A′(
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,
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);
△A′AP中,AA′的長為定值,若周長最小,那么PA+PA′最��;
由于O、A關(guān)于拋物線的對稱軸對稱,則點P必為直線OA′與拋物線對稱軸的交點;
易求得直線OA′:y=
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x,
拋物線對稱軸:x=4;
當(dāng)x=4時,y=
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,即P(4,
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).
(3)假設(shè)存在符合條件的Q點,則有:
①D為△ADQ的直角頂點;
易求得直線AD的斜率:k=
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=-
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,
所以設(shè)直線DQ:y=
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x+h,
則有:
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×3+h=4,
解得h=
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,
即y=
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x+
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,
當(dāng)x=4時,y=
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;
故Q(4,
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);
②A為△ADQ的直角頂點,同①可求得Q(4,-5);
③Q為△ADQ的直角頂點,設(shè)Q(4,m),
則有:
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=-1,
即m
2-4m-4=0;
解得m=2±2
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;
即Q(4,2+2
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)或(4,2-2
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);
綜上可知:存在符合條件的Q點,且坐標(biāo)為:
Q(4,-5)或(4,
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)或(4,2+2
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)或(4,2-2
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).
點評:此題考查了矩形的性質(zhì)、圖形的翻折變換、二次函數(shù)解析式的確定、平面展開-最短路徑問題、直角三角形的判定、互相垂直的兩直線的斜率關(guān)系等重要知識,(3)題中,一定要根據(jù)不同直角頂點來分類討論,以免漏解.