【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(duì)邊分別對(duì)應(yīng)相等.由此可得證.
(2)由于拋物線過(guò)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(gè)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(zhǎng),當(dāng)∠APD=∠OAB時(shí),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(zhǎng),即可根據(jù)上面的比例關(guān)系式求出P點(diǎn)的坐標(biāo).
(4)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/images0.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/images1.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/images2.png)
分三種情況進(jìn)行討論:
①如第一個(gè)圖:此時(shí)QD=AP=1,因此OP=OA-1=1,P點(diǎn)的坐標(biāo)為(1,0);
②如第二個(gè)圖:此時(shí)OP=OA+AP=3,P點(diǎn)的坐標(biāo)為(3,0);
③如第三個(gè)圖:此時(shí)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,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/0.png)
),根據(jù)A,D的坐標(biāo)可求出直線AD的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/1.png)
x-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/2.png)
,由于QP∥AD,因此直線PQ的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/3.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/4.png)
,可求得P點(diǎn)的坐標(biāo)為(-1,0).
因此共有3個(gè)符合條件的P點(diǎn)的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/images8.png)
(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
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/5.png)
x經(jīng)過(guò)點(diǎn)A,
點(diǎn)A的坐標(biāo)為(2,0),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/6.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/7.png)
.
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/8.png)
x
2-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/9.png)
x.
∵四邊形ABCO是平行四邊形,
∴OA∥CB.
∵點(diǎn)C的坐標(biāo)為(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/10.png)
),
∴點(diǎn)B的坐標(biāo)為(3,3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/11.png)
).
把x=3代入此函數(shù)解析式,得:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/12.png)
×3
2-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/13.png)
×3=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/14.png)
.
∴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,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/15.png)
).
(3)連接BO,
過(guò)點(diǎn)B作BE⊥x軸于點(diǎn)E,
過(guò)點(diǎn)D作DF⊥x軸于點(diǎn)F.
tan∠BOE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/16.png)
,tan∠DAF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/17.png)
,
∴tan∠BOE=tan∠DAF.
∴∠BOE=∠DAF.
∵∠APD=∠OAB,
∴△APD∽△OAB.
設(shè)點(diǎn)P的坐標(biāo)為(x,0),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/18.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/19.png)
,
解得:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/20.png)
.
∴點(diǎn)P的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101558329693896/SYS201311031015583296938025_DA/21.png)
,0).
(4)P
1(1,0),P
2(-1,0),P
3(3,0).
點(diǎn)評(píng):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(shí)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.