【答案】
分析:(1)根據(jù)B點(diǎn)的坐標(biāo)以及矩形的面積可以求出矩形的四個頂點(diǎn)的坐標(biāo),根據(jù)待定系數(shù)法就可以求出拋物線的解析式;
(2)①過點(diǎn)B作BN⊥PS,垂足為N,可以設(shè)P的坐標(biāo)是(a,
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a
2+1),根據(jù)勾股定理就可以用a表示出PB=PS的長,由此可以證明;
②判斷△SBR的形狀,根據(jù)①同理可知BQ=QR,根據(jù)等邊對等角就可以證明∠SBR=90度,則△SBR為直角三角形;
③若以P、S、M為頂點(diǎn)的三角形與以Q、M、R為頂點(diǎn)的三角形相似,有△PSM∽△MRQ和△PSM∽△QRM兩種情況,根據(jù)相似三角形的對應(yīng)邊的比相等就可以求出.
解答:解:(1)方法一:
∵B點(diǎn)坐標(biāo)為(0.2),
∴OB=2,
∵矩形CDEF面積為8,
∴CF=4.
∴C點(diǎn)坐標(biāo)為(-2,2).F點(diǎn)坐標(biāo)為(2,2).
設(shè)拋物線的解析式為y=ax
2+bx+c.
其過三點(diǎn)A(0,1),C(-2.2),F(xiàn)(2,2).
得
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,
解這個方程組,得a=
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,b=0,c=1,
∴此拋物線的解析式為y=
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x
2+1.(3分)
方法二:
∵B點(diǎn)坐標(biāo)為(0.2),
∴OB=2,
∵矩形CDEF面積為8,
∴CF=4.
∴C點(diǎn)坐標(biāo)為(-2,2),
根據(jù)題意可設(shè)拋物線解析式為y=ax
2+c.
其過點(diǎn)A(0,1)和C(-2.2)
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解這個方程組,得a=
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,c=1
此拋物線解析式為y=
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x
2+1.
(2)①證明:如圖(2)過點(diǎn)B作BN⊥PS,垂足為N.
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∵P點(diǎn)在拋物線y=
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x
2+1上.可設(shè)P點(diǎn)坐標(biāo)為(a,
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a
2+1).
∴PS=
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a
2+1,OB=NS=2,BN=-a.
∴PN=PS-NS=
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,
在Rt△PNB中.
PB
2=PN
2+BN
2=(
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a
2-1)
2+a
2=(
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a
2+1)
2∴PB=PS=
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.(6分)
②根據(jù)①同理可知BQ=QR.
∴∠1=∠2,
又∵∠1=∠3,
∴∠2=∠3,
同理∠SBP=∠5(7分)
∴2∠5+2∠3=180°
∴∠5+∠3=90°
∴∠SBR=90度.
∴△SBR為直角三角形.(8分)
③方法一:如圖(3)作QN⊥PS,
設(shè)PS=b,QR=c,
∵由①知PS=PB=b.QR=QB=c,PQ=b+c.PN=b-c.
∴QN
2=SR
2=(b+c)
2-(b-c)
2∴
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.(9分)
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假設(shè)存在點(diǎn)M.且MS=x,則MR=
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.
若使△PSM∽△MRQ,
則有
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.
即x
2-2
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x+bc=0
∴
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.
∴SR=2
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∴M為SR的中點(diǎn).(11分)
若使△PSM∽△QRM,
則有
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.
∴
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.
∴
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.
∴M點(diǎn)即為原點(diǎn)O.
綜上所述,當(dāng)點(diǎn)M為SR的中點(diǎn)時.△PSM∽△MRQ;
當(dāng)點(diǎn)M為原點(diǎn)時,△PSM∽△MRQ.(13分)
方法二:
若以P、S、M為頂點(diǎn)的三角形與以Q、M、R為頂點(diǎn)的三角形相似,
∵∠PSM=∠MRQ=90°,
∴有△PSM∽△MRQ和△PSM∽△QRM兩種情況.
當(dāng)△PSM∽△MRQ時.∠SPM=∠RMQ,∠SMP=∠RQM.
由直角三角形兩銳角互余性質(zhì).知∠PMS+∠QMR=90度.
∴∠PMQ=90度.(9分)
取PQ中點(diǎn)為T.連接MT.則MT=
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PQ=
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(QR+PS).(10分)
∴MN為直角梯形SRQP的中位線,
∴點(diǎn)M為SR的中點(diǎn)(11分)
∴
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=1
當(dāng)△PSM∽△QRM時,
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∴QB=BP
∵PS∥OB∥QR
∴點(diǎn)M為原點(diǎn)O.
綜上所述,當(dāng)點(diǎn)M為SR的中點(diǎn)時,△PSM∽△MRQ;
當(dāng)點(diǎn)M為原點(diǎn)時,△PSM∽△QRM.(13分)
點(diǎn)評:本題主要考查了待定系數(shù)法求函數(shù)解析式,以及相似三角形的對應(yīng)邊的比相等.