解:(1)作PK⊥MN于K,則PK=KM=
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NM=2,
∴KO=6,
∴P(6,2);
(2)①當點A落在線段OM上(可與點M重合)時,如圖(一),此時0<b≤2,S=0;
②當點A落在線段AK上(可與點K重合)時,如圖(二),此時2<b≤3,設AC交PM于H,MA=AH=2b-4,
∴S=
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(2b-4)
2=2b
2-8b+8,
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③當點A落在線段KN上(可與點N重合)時,如圖(三),此時3<b≤4,設AC交PN于H,AN=AH=8-2b,
∴S=S
△PMN-S
△ANH=4-2(4-b)
2=-2b
2+16b-28,
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④當點A落在線段MN的延長線上時,b>4,如圖(四),S=4;
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(3)以OM為直徑作圓,當直線y=-
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x+b(b>0)與圓相切時,b=
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+1,如圖(五);
當b≥4時,重合部分是△PMN,S=4
設Q(x,b-
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x),因為∠OQM=90°,O(0,0),M(4,0)所以OQ
2+QM
2=OM
2,
即[x
2+(b-
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x)
2]+[(x-4)
2+(b-
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x)
2]=4
2,
整理得
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x
2-(2b+8)x+2b
2=0,
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x
2-(b+4)x+b
2=0,
根據(jù)題意這個方程必須有解,也就是判別式△≥0,即(b+4)
2-5b
2≥0,-b
2+2b+4≥0,b
2-2b-4≤0,可以解得 1-
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≤b≤1+
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,由于b>0,所以0<b≤1+
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.
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故0<b≤
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+1;
(4)b的值為4,5,
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.
∵點C、D的坐標分別為(2b,b),(b,b)
當PC=PD時,b=4;
當PC=CD時,b
1=2(P、C、D三點共線,舍去),b
2=5;
當PD=CD時,b=8±2
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.
分析:(1)因為以M(4,0),N(8,0)為斜邊端點作的等腰直角三角形PMN,點P在第一象限,所以可作PK⊥MN于K,則PK=KM=
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NM=2,進而可求KO=6,所以P(6,2);
(2)需分情況討論:當0<b≤2時,S=0;當2<b≤3時,重合部分是一個等腰直角三角形,可設AC交PM于H,AM=HA=2b-4,所以S=
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(2b-4)
2;當3<b<4時,重合部分是一個四邊形,因此可設AC交PN于H,四邊形的面積=三角形PMN的面積-三角形HAN的面積,因為NA=HA=8-2b,所以S=-2(4-b)
2+4,當b≥4時,重合部分就是直角三角形PMN,所以S=4.
(3)因為直線y=-
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x+b(b>0)上存在點Q,使∠OQM等于90°,利用90°的圓周角對的弦是直徑,所以以OM為直徑作圓,當直線y=-
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x+b(b>0)與此圓相切時,求得的就是b的最大值,而此時b=
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+1;
(4)因為△PCD為等腰三角形,所以需分情況討論,當PC=PD時,b=4.當PC=CD時,b
1=2(舍),b
2=5.當PD=CD時,b=8±2
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.
點評:本題是一道綜合性極強的題目,解決這類問題常用到分類討論、數(shù)形結合、方程和轉化等數(shù)學思想方法.