【答案】
分析:(1)設(shè)出切線PH所在直線的解析式,過E點(diǎn)作ET⊥x軸于點(diǎn)T,連接DP、DF,則DF⊥PE,構(gòu)造出直角三角形,利用特殊角的三角函數(shù)值求出E點(diǎn)的坐標(biāo),根據(jù)直線過P、E兩點(diǎn),列出方程組求出未知數(shù)的值,進(jìn)而求出切線的解析式;
(2)分當(dāng)k<0,設(shè)過點(diǎn)A且與半圓相切于P
1點(diǎn)的切線方程為y=k
1x+b
1,P
1點(diǎn)的坐標(biāo)為(x
1,y
1),切線與邊BC交于點(diǎn)S,過點(diǎn)S作ST
1⊥x軸于點(diǎn)T
1.利用三角形相似求出P
1點(diǎn)的坐標(biāo).
k>0時(shí),據(jù)圓的對稱性知P
2點(diǎn)是P
1點(diǎn)關(guān)于直線y=2對稱的點(diǎn),從而可得P
2點(diǎn)的坐標(biāo).
解答:解:(1)設(shè)切線PH所在直線的解析式為y=kx+b.(1分)
解法一:設(shè)E點(diǎn)的坐標(biāo)為(x
E,4),過E點(diǎn)作ET⊥x軸于點(diǎn)T,連接DP、DF,則DF⊥PE,
在Rt△DOP和Rt△DFP中,∵OP=PF,OD=DF,∴△DOP≌△DFP.
在Rt△DOP中,tan∠DPO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/1.png)
.
∴∠DPO=30°,從而知∠OPEe=60度.
在Rt△EPT中,可求得PT=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/2.png)
,
∴E點(diǎn)的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/3.png)
,4).(4分)
∵直線過P、E兩點(diǎn),∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/4.png)
解方程組,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/5.png)
∴切線PF所在直線的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/6.png)
x+6.(6分)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/images7.png)
解法二:∵點(diǎn)P的坐標(biāo)為(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/7.png)
,0),且直線y=kx+b過點(diǎn)P,
∴2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/8.png)
k+b=0,b=-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/9.png)
k.
設(shè)E點(diǎn)的坐標(biāo)為(x
E,4),過E點(diǎn)作ET⊥x軸于點(diǎn)T.
∵切線過E點(diǎn),
∴kx
E+b=4,xE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/10.png)
(4-b).
∵EC=EF,PF=PO,
∴PE=EF+FP.(4分)
在Rt△ETP中,PE
2=ET
2+PT
2,
∴[
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/11.png)
(4-b)+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/12.png)
]
2=4
2+[2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/13.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/14.png)
(4-b)]
2,解方程,得k=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/15.png)
,b=6.
∴切線PF所在直線的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/16.png)
x+6.(6分)
(2)如備用圖,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/images18.png)
(�。┊�(dāng)k<0時(shí),設(shè)過點(diǎn)A且與半圓相切于P
1點(diǎn)的切線方程為y=k
1x+b
1,P
1點(diǎn)的坐標(biāo)為(x
1,y
1),切線與邊BC交于點(diǎn)S,過點(diǎn)S作ST
1⊥x軸于點(diǎn)T
1.
同上理,可得b
1=-6k
1,∴[
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/17.png)
(4-b
1)+6]
2=4
2+[6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/18.png)
(4-b
1)]
2,
解方程,得k
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/19.png)
,b
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/20.png)
.(8分)
∵直線y=k
1x+b
1與邊BC交于點(diǎn)S(x
2,4),
∴4=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/21.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/22.png)
,解方程,得x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/23.png)
.
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/24.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/25.png)
,
∴(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/26.png)
+6)y
1=6×4,解得y
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/27.png)
,代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/28.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/29.png)
,解得x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/30.png)
.
∴所求滿足條件的P
1點(diǎn)的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/31.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/32.png)
).(10分)
(ⅱ)當(dāng)k>0時(shí),據(jù)圓的對稱性知P
2點(diǎn)是P
1點(diǎn)關(guān)于直線y=2對稱的點(diǎn),從而可得P
2點(diǎn)的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/33.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164926978953975/SYS201310221649269789539023_DA/34.png)
).(12分)
點(diǎn)評:此題難度很大,把一次函數(shù),圓,三角形的知識結(jié)合起來,綜合性很強(qiáng),解答此題的關(guān)鍵是根據(jù)題意畫出圖形,利用數(shù)形結(jié)合求出結(jié)論.