解:(1)∵S
△BOC=3S
△AOC,
∴OB=3OA,
∵B(3,0),
∴A點坐標(biāo)為(-1,0),
∴1+(m+2)+3(m-1)=0,
解得m=0,
故拋物線的解析式為y=x
2-2x-3;
(2)假設(shè)存在點P,
則在△PCB中,∠PCB=∠APC-∠ABC(三角形的一個外角等于與它不相鄰的兩個內(nèi)角的和),
∵∠PCB=∠CAB-∠ABC,
∴∠CAB=∠APC,
∴AC=PC,
又CO⊥AP,
∴AO=PO(等腰三角形三線合一),
∴點P的坐標(biāo)為(1,0);
故存在點P(1,0),使∠PCB=∠CAB-∠ABC;
(3)當(dāng)E點運動時,AQ•O
1E的值不變.
∵A(-1,0),B(3,0),
∴
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=1,
∴圓心坐標(biāo)為O
1(1,0),
∴OM=ON=
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=
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,
∴點MN的坐標(biāo)為M(0,
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),N(0,-
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),
設(shè)點E坐標(biāo)為(2a,0),則點F坐標(biāo)為(a,
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),
設(shè)直線O
1F的解析式為y=kx+b,
則
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,
解得
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,
∴直線O
1F的解析式為:y=
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x-
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①,
又點A、N的坐標(biāo)為A(-1,0),N(-
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,0),
∴直線AN的解析式為y=-
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x-
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②,
①②聯(lián)立得
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,
解得
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,
∴點Q坐標(biāo)為(
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,-
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),
∴AQ=
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=|
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|,
又∵O
1E=1-2a,
∴AQ•O
1E=|
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|•(1-2a)=4的值不變.
即當(dāng)E點運動時,AQ•O
1E的值不變,不變值為4.
分析:(1)根據(jù)等高的三角形的面積的比等于底邊長的比,由S
△BOC=3S
△AOC可知,OB=3OA,根據(jù)B(3,0)可得A(-1,0),然后把點A坐標(biāo)代入函數(shù)表達式即可求出m的值,拋物線關(guān)系式即可求出;
(2)根據(jù)三角形的外角性質(zhì),三角形的一個外角等于與它不相鄰的兩個內(nèi)角的和,只要∠APC=∠CAB即可,即PC=AC,然后求出點P的坐標(biāo);
(3)先分別求出點M、N的坐標(biāo),設(shè)點E的坐標(biāo)為(2b,0),根據(jù)點F為EM的中點表示出點F的坐標(biāo),然后利用待定系數(shù)法分別求出直線O
1F與NQ的解析式,從而點Q的坐標(biāo)可得,再利用兩點之間距離公式求出AQ的長度,而O
1E=1-2b,從而便可確定正確的結(jié)論并求出其值.
點評:本題主要考查了待定系數(shù)法求二次函數(shù)解析式,三角形的外角性質(zhì),等腰三角形三線合一的性質(zhì)以及兩點間距離的求法,并且運算量較大,需要小心計算,以避免出錯,此題難度較大.