【答案】
分析:(1)因?yàn)橐訫(4,0),N(8,0)為斜邊端點(diǎn)作的等腰直角三角形PMN,點(diǎn)P在第一象限,所以可作PK⊥MN于K,則PK=KM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/0.png)
NM=2,進(jìn)而可求KO=6,所以P(6,2);
(2)需分情況討論:當(dāng)0<b≤2時(shí),S=0;當(dāng)2<b≤3時(shí),重合部分是一個(gè)等腰直角三角形,可設(shè)AC交PM于H,AM=HA=2b-4,所以S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/1.png)
(2b-4)
2;當(dāng)3<b<4時(shí),重合部分是一個(gè)四邊形,因此可設(shè)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?yàn)镹A=HA=8-2b,所以S=-2(4-b)
2+4,當(dāng)b≥4時(shí),重合部分就是直角三角形PMN,所以S=4.
(3)因?yàn)橹本€y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/2.png)
x+b(b>0)上存在點(diǎn)Q,使∠OQM等于90°,利用90°的圓周角對的弦是直徑,所以以O(shè)M為直徑作圓,當(dāng)直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/3.png)
x+b(b>0)與此圓相切時(shí),求得的就是b的最大值,而此時(shí)b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/4.png)
+1;
(4)因?yàn)椤鱌CD為等腰三角形,所以需分情況討論,當(dāng)PC=PD時(shí),b=4.當(dāng)PC=CD時(shí),b
1=2(舍),b
2=5.當(dāng)PD=CD時(shí),b=8±2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/5.png)
.
解答:解:(1)作PK⊥MN于K,則PK=KM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/6.png)
NM=2,
∴KO=6,
∴P(6,2);
(2)①當(dāng)點(diǎn)A落在線段OM上(可與點(diǎn)M重合)時(shí),如圖(一),此時(shí)0<b≤2,S=0;
②當(dāng)點(diǎn)A落在線段AK上(可與點(diǎn)K重合)時(shí),如圖(二),此時(shí)2<b≤3,設(shè)AC交PM于H,MA=AH=2b-4,
∴S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/7.png)
(2b-4)
2=2b
2-8b+8,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/images8.png)
③當(dāng)點(diǎn)A落在線段KN上(可與點(diǎn)N重合)時(shí),如圖(三),此時(shí)3<b≤4,設(shè)AC交PN于H,AN=AH=8-2b,
∴S=S
△PMN-S
△ANH=4-2(4-b)
2=-2b
2+16b-28,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/images9.png)
④當(dāng)點(diǎn)A落在線段MN的延長線上時(shí),b>4,如圖(四),S=4;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/images10.png)
(3)以O(shè)M為直徑作圓,當(dāng)直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/8.png)
x+b(b>0)與圓相切時(shí),b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/9.png)
+1,如圖(五);
當(dāng)b≥4時(shí),重合部分是△PMN,S=4
設(shè)Q(x,b-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/10.png)
x),因?yàn)椤螼QM=90°,O(0,0),M(4,0)所以O(shè)Q
2+QM
2=OM
2,
即[x
2+(b-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/11.png)
x)
2]+[(x-4)
2+(b-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/12.png)
x)
2]=4
2,
整理得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/13.png)
x
2-(2b+8)x+2b
2=0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/14.png)
x
2-(b+4)x+b
2=0,
根據(jù)題意這個(gè)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-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/15.png)
≤b≤1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/16.png)
,由于b>0,所以0<b≤1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/17.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/images21.png)
故0<b≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/18.png)
+1;
(4)b的值為4,5,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/19.png)
.
∵點(diǎn)C、D的坐標(biāo)分別為(2b,b),(b,b)
當(dāng)PC=PD時(shí),b=4;
當(dāng)PC=CD時(shí),b
1=2(P、C、D三點(diǎn)共線,舍去),b
2=5;
當(dāng)PD=CD時(shí),b=8±2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160109132297623/SYS201310221601091322976024_DA/20.png)
.
點(diǎn)評:本題是一道綜合性極強(qiáng)的題目,解決這類問題常用到分類討論、數(shù)形結(jié)合、方程和轉(zhuǎn)化等數(shù)學(xué)思想方法.