【答案】
分析:(1)根據(jù)折疊圖形的軸對(duì)稱性,△CED、△CBD全等,首先在Rt△CEO中求出OE的長(zhǎng),進(jìn)而可得到AE的長(zhǎng);在Rt△AED中,AD=AB-BD、ED=BD,利用勾股定理可求出AD的長(zhǎng).進(jìn)一步能確定D點(diǎn)坐標(biāo),利用待定系數(shù)法即可求出拋物線的解析式.
(2)由于∠DEC=90°,首先能確定的是∠AED=∠OCE,若以P、Q、C為頂點(diǎn)的三角形與△ADE相似,那么∠QPC=90°或∠PQC=90°,然后在這兩種情況下,分別利用相似三角形的對(duì)應(yīng)邊成比例求出對(duì)應(yīng)的t的值.
(3)由于以M,N,C,E為頂點(diǎn)的四邊形,邊和對(duì)角線都沒(méi)明確指出,所以要分情況進(jìn)行討論:
①EC做平行四邊形的對(duì)角線,那么EC、MN必互相平分,由于EC的中點(diǎn)正好在拋物線對(duì)稱軸上,所以M點(diǎn)一定是拋物線的頂點(diǎn);
②EC做平行四邊形的邊,那么EC、MN平行且相等,首先設(shè)出點(diǎn)N的坐標(biāo),然后結(jié)合E、C的橫、縱坐標(biāo)差表示出M點(diǎn)坐標(biāo),再將點(diǎn)M代入拋物線的解析式中,即可確定M、N的坐標(biāo).
解答:解:(1)∵四邊形ABCO為矩形,
∴∠OAB=∠AOC=∠B=90°,AB=CO=8,AO=BC=10.
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由題意,△BDC≌△EDC.
∴∠B=∠DEC=90°,EC=BC=10,ED=BD.
由勾股定理易得EO=6.
∴AE=10-6=4,
設(shè)AD=x,則BD=ED=8-x,由勾股定理,得x
2+4
2=(8-x)
2,
解得,x=3,∴AD=3.
∵拋物線y=ax
2+bx+c過(guò)點(diǎn)D(3,10),C(8,0),O(0,0)
∴
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,
解得
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∴拋物線的解析式為:y=-
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x
2+
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x.
(2)∵∠DEA+∠OEC=90°,∠OCE+∠OEC=90°,
∴∠DEA=∠OCE,
由(1)可得AD=3,AE=4,DE=5.
而CQ=t,EP=2t,∴PC=10-2t.
當(dāng)∠PQC=∠DAE=90°,△ADE∽△QPC,
∴
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=
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,即
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=
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,
解得t=
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.
當(dāng)∠QPC=∠DAE=90°,△ADE∽△PQC,
∴
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=
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,即
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=
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,
解得t=
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.
∴當(dāng)t=
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或
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時(shí),以P、Q、C為頂點(diǎn)的三角形與△ADE相似.
(3)假設(shè)存在符合條件的M、N點(diǎn),分兩種情況討論:
①
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EC為平行四邊形的對(duì)角線,由于拋物線的對(duì)稱軸經(jīng)過(guò)EC中點(diǎn),若四邊形MENC是平行四邊形,那么M點(diǎn)必為拋物線頂點(diǎn);
則:M(4,
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);而平行四邊形的對(duì)角線互相平分,那么線段MN必被EC中點(diǎn)(4,3)平分,則N(4,-
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);
②EC為平行四邊形的邊,則EC
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MN,設(shè)N(4,m),則M(4-8,m+6)或M(4+8,m-6);
將M(-4,m+6)代入拋物線的解析式中,得:m=-38,此時(shí) N(4,-38)、M(-4,-32);
將M(12,m-6)代入拋物線的解析式中,得:m=-26,此時(shí) N(4,-26)、M(12,-32);
綜上,存在符合條件的M、N點(diǎn),且它們的坐標(biāo)為:
①M(fèi)
1(-4,-32),N
1(4,-38);②M
2(12,-32),N
2(4,-26);③M
3(4,
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),N
3(4,-
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).
點(diǎn)評(píng):考查了二次函數(shù)綜合題,題目涉及了圖形的折疊變換、相似三角形的判定和性質(zhì)、平行四邊形的判定和性質(zhì)等重點(diǎn)知識(shí).后兩問(wèn)的情況較多,需要進(jìn)行分類討論,以免漏解.