解:(1)A(-3,0),B(0,4).
當(dāng)y=2時,
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,
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.
所以直線AB與CD交點的坐標(biāo)為
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.
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(2)①當(dāng)0<t<
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時,△MPH與矩形AOCD重合部分的面積即△MPH的面積.
過點M作MN⊥OA,垂足為N.
由△AMN∽△ABO,得
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.
∵AO=3,BO=4,
∴AB=
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=5,
∴
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.
∴AN=t.
∴△MPH的面積為
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.
當(dāng)3-2t=1時,t=1.
當(dāng)
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<t≤3時,設(shè)MH與CD相交于點E,
△MPH與矩形AOCD重合部分的面積即△PEH的面積.
過點M作MG⊥AO于G,MF⊥HP交HP的延長線于點F.
FM=AG-AH=AM×cos∠BAO-(AO-HO)=
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.
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.
由△HPE∽△HFM,得
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.
∴
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.
∴
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.
∴△PEH的面積為
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.
當(dāng)
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時,
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.
經(jīng)檢驗,t=
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是原方程的解,
綜上所述,若△MPH與矩形AOCD重合部分的面積為1,t為1或
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.
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②BP+PH+HQ有最小值.
連接PB,CH,則四邊形PHCB是平行四邊形.
∴BP=CH.
∴BP+PH+HQ=CH+HQ+2.
當(dāng)點C,H,Q在同一直線上時,CH+HQ的值最�。�
∵點C,Q的坐標(biāo)分別為(0,2),(-6,-4),
∴直線CQ的解析式為y=x+2,
∴點H的坐標(biāo)為(-2,0).因此點P的坐標(biāo)為(-2,2).
分析:(1)讓y=0求得x的值可得A的坐標(biāo),(0,b)為B的坐標(biāo),讓y=
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可得交點的縱坐標(biāo),代入直線解析式可得交點的橫坐標(biāo);
(2)由△AMN∽△ABO,得出△MPH的面積,再利用由△HPE∽△HFM,表示出△PEH的面積,即可得出答案.
(3)當(dāng)點C,H,Q在同一直線上時,CH+HQ的值最小,利用平行四邊形的性質(zhì)得出即可.
點評:此題主要考查了相似三角形的應(yīng)用以及平行四邊形的性質(zhì),利用數(shù)形結(jié)合進(jìn)行分類討論是解決問題的關(guān)鍵,分析時注意不要漏解.