【答案】
分析:(1)由旋轉(zhuǎn)的性質(zhì)可知:△AOC≌△ABC,由此可得出四邊形AOCB的兩組對邊分別對應(yīng)相等.由此可得證.
(2)由于拋物線過A點(diǎn),因此可將A點(diǎn)的坐標(biāo)代入拋物線的解析式中即可得出a的值和拋物線的解析式.
要判斷B是否在拋物線的解析式上,首先要求出B點(diǎn)的坐標(biāo),由于四邊形APCB是平行四邊形,OA=2,因此將C點(diǎn)向右平移2個單位即可得出B點(diǎn)的坐標(biāo),然后將B的坐標(biāo)代入拋物線的解析式中即可判斷出B是否在拋物線上.
(3)先根據(jù)(2)的拋物線的解析式求出頂點(diǎn)D的坐標(biāo),然后求出OB、AD的長,當(dāng)∠APD=∠OAB時,可得出△APD∽△OAB,進(jìn)而可得出關(guān)于AP,AD、OA、OB的比例關(guān)系式.設(shè)出P點(diǎn)的坐標(biāo),然后用P的橫坐標(biāo)表示出AP的長,即可根據(jù)上面的比例關(guān)系式求出P點(diǎn)的坐標(biāo).
(4)
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分三種情況進(jìn)行討論:
①如第一個圖:此時QD=AP=1,因此OP=OA-1=1,P點(diǎn)的坐標(biāo)為(1,0);
②如第二個圖:此時OP=OA+AP=3,P點(diǎn)的坐標(biāo)為(3,0);
③如第三個圖:此時D,Q兩點(diǎn)的縱坐標(biāo)互為相反數(shù),因此Q點(diǎn)的坐標(biāo)為(0,
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),根據(jù)A,D的坐標(biāo)可求出直線AD的解析式為y=
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x-2
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,由于QP∥AD,因此直線PQ的解析式為y=
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x+
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,可求得P點(diǎn)的坐標(biāo)為(-1,0).
因此共有3個符合條件的P點(diǎn)的坐標(biāo).
解答:
(1)證明:∵△AOC繞AC的中點(diǎn)旋轉(zhuǎn)180°,
點(diǎn)O落到點(diǎn)B的位置,
∴△ACO≌△CAB.
∴AO=CB,CO=AB,
∴四邊形ABCO是平行四邊形.
(2)解:∵拋物線y=ax
2-2
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x經(jīng)過點(diǎn)A,
點(diǎn)A的坐標(biāo)為(2,0),
∴
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,
解得:
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.
∴y=
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x
2-2
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x.
∵四邊形ABCO是平行四邊形,
∴OA∥CB.
∵點(diǎn)C的坐標(biāo)為(1,
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),
∴點(diǎn)B的坐標(biāo)為(3,3
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).
把x=3代入此函數(shù)解析式,得:y=
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×3
2-2
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×3=3
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.
∴點(diǎn)B的坐標(biāo)滿足此函數(shù)解析式,點(diǎn)B在此拋物線上.
∴頂點(diǎn)D的坐標(biāo)為(1,-
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).
(3)連接BO,
過點(diǎn)B作BE⊥x軸于點(diǎn)E,
過點(diǎn)D作DF⊥x軸于點(diǎn)F.
tan∠BOE=
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,tan∠DAF=
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,
∴tan∠BOE=tan∠DAF.
∴∠BOE=∠DAF.
∵∠APD=∠OAB,
∴△APD∽△OAB.
設(shè)點(diǎn)P的坐標(biāo)為(x,0),
∴
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,
∴
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,
解得:x=
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.
∴點(diǎn)P的坐標(biāo)為(
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,0).
(4)P
1(1,0),P
2(-1,0),P
3(3,0).
點(diǎn)評:本題著重考查了待定系數(shù)法求二次函數(shù)解析式、圖形旋轉(zhuǎn)變換、三角形相似、平行四邊形的判定等知識點(diǎn),綜合性強(qiáng),考查學(xué)生分類討論,數(shù)形結(jié)合的數(shù)學(xué)思想方法.