【答案】
分析:(1)由已知得OA=2,將線段OA繞原點(diǎn)O順時(shí)針旋轉(zhuǎn)120°,則OB與x軸的正方向夾角為60°,過點(diǎn)B作BD⊥x軸于點(diǎn)D,解直角三角形可得OD、BD的長(zhǎng),可表示B點(diǎn)的坐標(biāo);
(2)直接將A、O、B三點(diǎn)坐標(biāo)代入拋物線解析式的一般式,可求解析式;
(3)因?yàn)辄c(diǎn)A,O關(guān)于對(duì)稱軸對(duì)稱,連接AB交對(duì)稱軸于C點(diǎn),C點(diǎn)即為所求,求直線AB的解析式,再根據(jù)C點(diǎn)的橫坐標(biāo)值,求縱坐標(biāo);
(4)設(shè)P(x,y)(-2<x<0,y<0),用割補(bǔ)法可表示△PAB的面積,根據(jù)面積表達(dá)式再求取最大值時(shí),x的值.
解答:
解:(1)過點(diǎn)B作BD⊥x軸于點(diǎn)D,由已知可得:OB=OA=2,∠BOD=60°,
在Rt△OBD中,∠ODB=90°,∠OBD=30°
∴OD=1,DB=

∴點(diǎn)B的坐標(biāo)是(1,

).(2分)
(2)設(shè)所求拋物線的解析式為y=ax
2+bx+c(a≠0),
由已知可得:
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,
解得:a=
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,b=

,c=0,
∴所求拋物線解析式為y=

x
2+

x.(4分)
(3)存在,
由y=

x
2+

x配方后得:y=
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(x+1)
2-

∴拋物線的對(duì)稱軸為x=-1(6分)
(也可用頂點(diǎn)坐標(biāo)公式求出)
∵點(diǎn)C在對(duì)稱軸x=-1上,△BOC的周長(zhǎng)=OB+BC+CO;
∵OB=2,要使△BOC的周長(zhǎng)最小,必須BC+CO最小,
∵點(diǎn)O與點(diǎn)A關(guān)于直線x=-1對(duì)稱,有CO=CA
△BOC的周長(zhǎng)=OB+BC+CO=OB+BC+CA
∴當(dāng)A、C、B三點(diǎn)共線,即點(diǎn)C為直線AB與拋物線對(duì)稱軸的交點(diǎn)時(shí),BC+CA最小,此時(shí)△BOC的周長(zhǎng)最�。�
設(shè)直線AB的解析式為y=kx+b,則有:
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,
解得:k=
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,b=

,
∴直線AB的解析式為y=

x+

,(7分)
當(dāng)x=-1時(shí),y=

,
∴所求點(diǎn)C的坐標(biāo)為(-1,

),(8分)
(4)設(shè)P(x,y)(-2<x<0,y<0),

則y=

x
2+

x①
過點(diǎn)P作PQ⊥y軸于點(diǎn)Q,PG⊥x軸于點(diǎn)G,過點(diǎn)A作AF⊥PQ軸于點(diǎn)F,過點(diǎn)B作BE⊥PQ軸于點(diǎn)E,
則PQ=-x,PG=-y,
由題意可得:S
△PAB=S
梯形AFEB-S
△AFP-S
△BEP(9分)
=

(AF+BE)•FE-
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AF•FP-

PE•BE
=

(-y+

-y)(1+2)-

(-y)(x+2)-

(1-x)(

-y)
=

②
將①代入②,
化簡(jiǎn)得:S
△PAB=-

x
2-

x+

(10分)
=
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(x+

)
2+

∴當(dāng)

時(shí),△PAB得面積有最大值,最大面積為

.(11分)
此時(shí)

∴點(diǎn)P的坐標(biāo)為

.(12分)
點(diǎn)評(píng):本題考查了坐標(biāo)系中點(diǎn)的坐標(biāo)求法,拋物線解析式的求法,根據(jù)對(duì)稱性求線段和最小的問題,也考查了在坐標(biāo)系里表示面積及求面積最大值等問題;
解答本題(4)也可以將直線AB向下平移至與拋物線相切的位置,聯(lián)立此時(shí)的直線解析式與拋物線解析式,可求唯一交點(diǎn)P的坐標(biāo).