【答案】
分析:(1)由于四邊形ABCO是平行四邊形,那么對(duì)邊AB和OC相等,由此可求出AB的長(zhǎng),由于A、B關(guān)于拋物線的對(duì)稱軸(即y軸)對(duì)稱,由此可得到A、B的橫坐標(biāo),將它們代入拋物線的解析式中即可求出A、B的坐標(biāo),也就得到了M點(diǎn)的坐標(biāo);
(2)①根據(jù)C、M的坐標(biāo),易求得OM、OC的長(zhǎng);過Q作QH⊥x軸于H,易證得△HQP∽△OMC,根據(jù)相似三角形得到的比例線段,即可求出t、x的函數(shù)關(guān)系式;
在求自變量的取值范圍時(shí),可參考兩個(gè)方面:一、P、C重合時(shí),不能構(gòu)成四邊形PCMQ;二、Q與B或A重合時(shí),四邊形PCMQ是平行四邊形;只要x不取上述兩種情況所得的值即可;
②由于CM、PQ的長(zhǎng)不確定,因此要分類討論:
一、CM>PQ,則CM:PQ=2:1,由(2)的相似三角形知OM=2QH,即M點(diǎn)縱坐標(biāo)為Q點(diǎn)縱坐標(biāo)的2倍,由此可求得t的值;
二、CM<PQ,則CM:PQ=1:2,后同一.
解答:解:(1)∵OABC是平行四邊形,∴AB∥OC,且AB=OC=4,
∵A,B在拋物線上,y軸是拋物線的對(duì)稱軸,
∴A,B的橫坐標(biāo)分別是2和-2,
代入y=
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+1得,A(2,2),B(-2,2),
∴M(0,2),(2分)
(2)①過點(diǎn)Q作QH⊥x軸,設(shè)垂足為H,則HQ=y,HP=x-t,
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由△HQP∽△OMC,得:
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=
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,即:t=x-2y,
∵Q(x,y)在y=
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+1上,
∴t=-
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+x-2.(2分)
當(dāng)點(diǎn)P與點(diǎn)C重合時(shí),梯形不存在,此時(shí),t=-4,解得x=1±
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,
當(dāng)Q與B或A重合時(shí),四邊形為平行四邊形,此時(shí),x=±2
∴x的取值范圍是x≠1±
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,且x≠±2的所有實(shí)數(shù);(2分)
②連接MC.分兩種情況討論:
∵CM∥PQ,
∴∠QPC=∠MCO,
∵∠COM=∠PHQ=90°,
∴△HQP∽△OMC,
(1)當(dāng)CM>PQ時(shí),則點(diǎn)P在線段OC上,
∵CM∥PQ,CM=2PQ,
∴點(diǎn)M縱坐標(biāo)為點(diǎn)Q縱坐標(biāo)的2倍,即2=2(
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+1),解得x=0,
∴t=-
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+0-2=-2;(2分)
(2)當(dāng)CM<PQ時(shí),則點(diǎn)P在OC的延長(zhǎng)線上,
∵CM∥PQ,CM=
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PQ,
∴點(diǎn)Q縱坐標(biāo)為點(diǎn)M縱坐標(biāo)的2倍,即
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+1=2×2,
解得:x=±2
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;(2分)
當(dāng)x=-2
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時(shí),得t=-
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-2
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-2=-8-2
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,
當(dāng)x=2
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時(shí),得t=2
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-8.
點(diǎn)評(píng):此題主要考查了平行四邊形的性質(zhì)、拋物線的對(duì)稱性、梯形的判定和性質(zhì)以及相似三角形的性質(zhì)等知識(shí)的綜合應(yīng)用能力.