【答案】
分析:(1)因為直線l
1經(jīng)過點A(-2,0)和點B(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/0.png)
),可設(shè)直線l
1的解析式為y=kx+b,將A、B的坐標(biāo)代入,利用方程組即可求得該解析式,又因直線l
2的函數(shù)表達(dá)式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/1.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/2.png)
,l
1與l
2相交于點P,所以將兩函數(shù)解析式聯(lián)立得到方程組,解之即可得到交點P的坐標(biāo),過P作x軸的垂線段,垂足為H,由P的坐標(biāo)可知,AH=EH=3,PH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/3.png)
,所以∠PEA=∠PAE=30°,利用三角形的外角等于和它不相鄰的兩個內(nèi)角的和,可得到∠FPB是60°;
(2)當(dāng)C在射線PA的延長線上時,設(shè)⊙C和直線l
2相切時,D是切點,連接CD,則CD⊥PD.過點P作CM的垂線PG,垂足為G,因為∠CPG=∠CAB=30°,PC=PC,所以可證Rt△CDP≌Rt△PGC,所以PG=CD=R.當(dāng)點C在射線PA上,⊙C和直線l
2相切時,同理可證Rt△CDP≌Rt△PGC,所以PG=CD=R.取R=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/4.png)
-2時,C在AP的反向延長線上時,因為P的橫坐標(biāo)為1,所以a=1+R=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/5.png)
-1,C在PA上時,因為P的橫坐標(biāo)為1,所以a=-(R-1)=3-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/6.png)
;
(3)當(dāng)⊙C和直線l
2不相離時,由(2)知,分兩種情況討論:①當(dāng)0≤a≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/7.png)
-1時,四邊形是一個直角梯形,所以有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/9.png)
,利用二次函數(shù)的頂點公式即可求出S的最大值;
②當(dāng)=3-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/10.png)
≤a<0時,顯然⊙C和直線l
2相切即a=3-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/11.png)
時,S最大.此時
s
最大值=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/12.png)
,綜合以上①和②,即可求出答案.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/images13.png)
解:(1)設(shè)直線l
1的解析式為y=kx+b,
∵直線l
1經(jīng)過點A(-2,0)和點B(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/13.png)
),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/14.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/15.png)
,
∴直線l
1的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/16.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/17.png)
;
聯(lián)立l
1與l
2得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/18.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/19.png)
,
∴P(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/20.png)
);
過P作x軸的垂線段,垂足為H,
∵P(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/21.png)
),
∴AH=EH=3,PH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/22.png)
,
∴∠PEA=∠PAE=30°,
∴∠FPB=∠PEA+∠PAE=60°;
(2)設(shè)⊙C和直線l
2相切時的一種情況如圖1所示,D是切點,連接CD,則CD⊥PD.過點P作CM的垂線PG,垂足為G,則Rt△CDP≌Rt△PGC(∠PCD=∠CPG=30°,CP=PC),
∴PG=CD=R.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/images24.png)
當(dāng)點C在射線PA上,⊙C和直線l
2相切時,同理可證.
取R=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/23.png)
-2時,a=1+R=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/24.png)
-1,或a=-(R-1)=3-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/25.png)
;
(3)當(dāng)⊙C和直線l
2不相離時,由(2)知,分兩種情況討論:
①如圖2,當(dāng)0≤a≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/26.png)
-1時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/28.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/29.png)
時,(滿足a≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/30.png)
-1),S有最大值.此時s
最大值=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/31.png)
(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/32.png)
);
②當(dāng)=3-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/33.png)
≤a<0時,顯然⊙C和直線l
2相切即a=3-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/34.png)
時,S最大.此時
s
最大值=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/35.png)
.
綜合以上①和②,當(dāng)a=3或a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/36.png)
時,存在S的最大值,其最大面積為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163014085569595/SYS201310221630140855695027_DA/37.png)
.
點評:考查一次函數(shù)的解析式、圖象、性質(zhì)和圓的相關(guān)知識,及綜合應(yīng)用相關(guān)知識分析問題、解決問題的能力.
此題也較為新穎,符合新課標(biāo)的理念,揭示了求最值的一般方法,本題的難度設(shè)置也較為合適,使同學(xué)們都能有發(fā)揮自己能力的空間.