【答案】
分析:(1)將A、B、C的坐標(biāo)代入拋物線的解析式中,即可求得待定系數(shù)的值;
(2)根據(jù)(1)得到的拋物線的解析式,可求出其對(duì)稱軸方程聯(lián)立直線OD的解析式即可求出D點(diǎn)的坐標(biāo);由于⊙D與x軸相切,那么D點(diǎn)縱坐標(biāo)即為⊙D的半徑;欲求劣弧EF的長(zhǎng),關(guān)鍵是求出圓心角∠EDF的度數(shù),連接DE、DF,過(guò)D作y軸的垂線DM,則DM即為D點(diǎn)的橫坐標(biāo),通過(guò)解直角三角形易求得∠EDM和∠FDM的度數(shù),即可得到∠EDF的度數(shù),進(jìn)而可根據(jù)弧長(zhǎng)計(jì)算公式求出劣弧EF的長(zhǎng);
(3)易求得直線AC的解析式,設(shè)直線AC與PG的交點(diǎn)為N,設(shè)出P點(diǎn)的橫坐標(biāo),根據(jù)拋物線與直線AC的解析式即可得到P、N的縱坐標(biāo),進(jìn)而可求出PN,NG的長(zhǎng);Rt△PGA中,△PNA與△NGA同高不等底,那么它們的面積比等于底邊PN、NG的比,因此本題可分兩種情況討論:
①△PNA的面積是△NGA的2倍,則PN:NG=2:1;②△PNA的面積是△NGA的
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,則NG=2PN;
可根據(jù)上述兩種情況所得的不同等量關(guān)系求出P點(diǎn)的橫坐標(biāo),進(jìn)而由拋物線的解析式確定出P點(diǎn)的坐標(biāo).
解答:解:(1)∵拋物線y=ax
2+bx+c經(jīng)過(guò)點(diǎn)A(2,0),B(6,0),
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;
∴
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,
解得
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;
∴拋物線的解析式為:
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;(3分)
(2)易知拋物線的對(duì)稱軸是x=4,
把x=4代入y=2x,得y=8,
∴點(diǎn)D的坐標(biāo)為(4,8);
∵⊙D與x軸相切,∴⊙D的半徑為8;(1分)
連接DE、DF,作DM⊥y軸,垂足為點(diǎn)M;
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在Rt△MFD中,F(xiàn)D=8,MD=4,
∴cos∠MDF=
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;
∴∠MDF=60°,
∴∠EDF=120°;(2分)
∴劣弧EF的長(zhǎng)為:
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;(1分)
(3)設(shè)直線AC的解析式為y=kx+b;
∵直線AC經(jīng)過(guò)點(diǎn)
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,
∴
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,
解得
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;
∴直線AC的解析式為:
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;(1分)
設(shè)點(diǎn)
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,PG交直線AC于N,
則點(diǎn)N坐標(biāo)為
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,
∵S
△PNA:S
△GNA=PN:GN;
∴①若PN:GN=1:2,則PG:GN=3:2,PG=
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GN;
即
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=
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;
解得:m
1=-3,m
2=2(舍去);
當(dāng)m=-3時(shí),
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=
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;
∴此時(shí)點(diǎn)P的坐標(biāo)為
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;(2分)
②若PN:GN=2:1,則PG:GN=3:1,PG=3GN;
即
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=
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;
解得:m
1=-12,m
2=2(舍去);
當(dāng)m
1=-12時(shí),
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=
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;
∴此時(shí)點(diǎn)P的坐標(biāo)為
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;
綜上所述,當(dāng)點(diǎn)P坐標(biāo)為
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或
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時(shí),△PGA的面積被直線AC分成1:2兩部分.(2分)
點(diǎn)評(píng):此題主要考查了二次函數(shù)解析式的確定、函數(shù)圖象交點(diǎn)、圖形面積的求法等知識(shí),需要特別注意的是(3)題中,△PGA被直線AC所分成的兩部分中,并沒(méi)有明確誰(shuí)大誰(shuí)小,所以要分類討論,以免漏解.