【答案】
分析:(1)拋物線解析式中有兩個待定系數(shù)a,c,根據(jù)直線AC解析式求點(diǎn)A、C坐標(biāo),代入拋物線解析式即可;
(2)分析不難發(fā)現(xiàn),△ABP的直角頂點(diǎn)只可能是P,根據(jù)已知條件可證AC
2+BC
2=AB
2,故點(diǎn)C滿足題意,根據(jù)拋物線的對稱性,點(diǎn)C關(guān)于拋物線對稱軸的對稱點(diǎn)也符合題意;
(3)由于B,F(xiàn)是定點(diǎn),BF的長一定,實(shí)際上就是求BM+FM最小,找出點(diǎn)B關(guān)于直線AC的對稱點(diǎn)B',連接B'F,交AC于點(diǎn)M,點(diǎn)M即為所求,由(2)可知,BC⊥AC,延長BC到B',使BC=B'C,利用中位線的性質(zhì)可得B'的坐標(biāo),從而可求直線B'F的解析式,再與直線AC的解析式聯(lián)立,可求M點(diǎn)坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/images0.png)
解:(1)∵直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/0.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/1.png)
與x軸交于點(diǎn)A,與y軸交于點(diǎn)C
∴點(diǎn)A(-1,0),C(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/2.png)
)
∵點(diǎn)A,C都在拋物線上,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/3.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/4.png)
∴拋物線的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/5.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/7.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/8.png)
∴頂點(diǎn)F(1,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/9.png)
).
(2)存在:
p
1(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/10.png)
),p
2(2,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/11.png)
).
(3)存在
理由:
解法一:
延長BC到點(diǎn)B′,使B′C=BC,連接B′F交直線AC于點(diǎn)M,則點(diǎn)M就是所求的點(diǎn),
∵過點(diǎn)B′作B′H⊥AB于點(diǎn)H,
∵B點(diǎn)在拋物線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/12.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/13.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/14.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/15.png)
上,
∴B(3,0),
在Rt△BOC中,tan∠OBC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/16.png)
∴∠OBC=30°,BC=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/17.png)
在Rt△B′BH中,B′H=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/18.png)
BB′=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/19.png)
BH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/20.png)
B′H=6,∴OH=3,
∴B′(-3,-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/21.png)
).
設(shè)直線B′F的解析式為y=kx+b,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/22.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/23.png)
,
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/24.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/25.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/26.png)
,
∴M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/27.png)
)
∴在直線AC上存在點(diǎn)M,使得△MBF的周長最小,此時M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/28.png)
).
解法二:
過點(diǎn)F作AC的垂線交y軸于點(diǎn)H,則點(diǎn)H為點(diǎn)F關(guān)于直線AC的對稱點(diǎn),連接BH交AC于點(diǎn)M,則點(diǎn)M
即為所求.
過點(diǎn)F作FG⊥y軸于點(diǎn)G,則OB∥FG,BC∥FH,
∴∠BOC=∠FGH=90°,∠BCO=∠FHG
∴∠HFG=∠CBO
同方法一可求得B(3,0)
在Rt△BOC中,tan∠OBC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/29.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/images31.png)
∴∠OBC=30°,可求得GH=GC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/30.png)
∴GF為線段CH的垂直平分線,可證得△CFH為等邊三角形
∴AC垂直平分FH
即點(diǎn)H為點(diǎn)F關(guān)于AC對稱點(diǎn),
∴H(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/31.png)
)
設(shè)直線BH的解析式為y=kx+b,由題意得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/32.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/33.png)
,
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/34.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/35.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/36.png)
,
∴M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/37.png)
),
∴在直線AC上存在點(diǎn)M,使得△MBF的周長最小,此時M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103000756705273351/SYS201311030007567052733027_DA/38.png)
).
點(diǎn)評:考查代數(shù)幾何的綜合運(yùn)用能力,體現(xiàn)數(shù)學(xué)知識的內(nèi)在聯(lián)系和不可分割的特點(diǎn).