【答案】
分析:(1)連接AM,在直角△AMO中,根據(jù)三角函數(shù)就可以求出OM,就可以得到M的坐標(biāo).
(2)根據(jù)三角函數(shù)就可以求出A,B的坐標(biāo),拋物線經(jīng)過點(diǎn)A、B、C,因而M一定是拋物線的頂點(diǎn).根據(jù)待定系數(shù)法就可以求出拋物線的解析式.
(3)四邊形ACBD的面積等于△ABC的面積+△ABP的面積,△ABC的面積一定,△ABP中底邊AB一定,P到AB的距離最大是三角形的面積最大,即當(dāng)P是圓與y軸的交點(diǎn)時(shí)面積最大.
(4)△PAB和△ABC相似,根據(jù)相似三角形的對應(yīng)邊的比相等,就可以求出P點(diǎn)的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/images0.png)
解:(1)如圖(1),
連接MA、MB,
則∠AMB=120°,
∴∠CMB=60°,∠OBM=30度.(2分)
∴OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/0.png)
MB=1,
∴M(0,1).(3分)
(2)由A,B,C三點(diǎn)的特殊性與對稱性,知經(jīng)過A,B,C三點(diǎn)的拋物線的解析式為y=ax
2+c.(4分)
∵OC=MC-MO=1,OB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/1.png)
,
∴C(0,-1),B(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/2.png)
,0).(5分)
∴c=-1,a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/3.png)
.
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/4.png)
x
2-1.(6分)
(3)∵S
四邊形ACBD=S
△ABC+S
△ABD,又S
△ABC與AB均為定值,(7分)
∴當(dāng)△ABD邊AB上的高最大時(shí),S
△ABD最大,此時(shí)點(diǎn)D為⊙M與y軸的交點(diǎn),如圖(1).(8分)
∴S
四邊形ACBD=S
△ABC+S
△ABD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/5.png)
AB•OC+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/6.png)
AB•OD
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/7.png)
AB•CD
=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/8.png)
cm
2.(9分)
(4)方法1:
如圖(2),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/images10.png)
∵△ABC為等腰三角形,∠ABC=30°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/9.png)
,
∴△ABC∽△PAB等價(jià)于∠PAB=30°,PB=AB=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/10.png)
,PA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/11.png)
PB=6.(10分)
設(shè)P(x,y)且x>0,則x=PA•cos30°-AO=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/12.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/13.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/14.png)
,y=PA•sin30°=3.(11分)
又∵P(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/15.png)
,3)的坐標(biāo)滿足y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/16.png)
x
2-1,
∴在拋物線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/17.png)
x
2-1上,存在點(diǎn)P(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/18.png)
,3),
使△ABC∽△PAB.
由拋物線的對稱性,知點(diǎn)(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/19.png)
,3)也符合題意.
∴存在點(diǎn)P,它的坐標(biāo)為(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/20.png)
,3)或(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/21.png)
,3).(12分)
說明:只要求出(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/22.png)
,3),(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/23.png)
,3),無最后一步不扣分.下面的方法相同.
方法2:
如圖(3),
當(dāng)△ABC∽△PAB時(shí),∠PAB=∠BAC=30°,又由(1)知∠MAB=30°,
∴點(diǎn)P在直線AM上.
設(shè)直線AM的解析式為y=kx+b,
將A(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/24.png)
,0),M(0,1)代入,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/images28.png)
∴直線AM的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/26.png)
x+1.(10分)
解方程組
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/27.png)
,
得P(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/28.png)
,3).(11分)
又∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/29.png)
,
∴∠PBx=60度.
∴∠P=30°,
∴△ABC∽△PAB.
∴在拋物線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/30.png)
x
2-1上,存在點(diǎn)(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/31.png)
,3),使△ABC∽△PAB.
由拋物線的對稱性,知點(diǎn)(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/32.png)
,3)也符合題意.
∴存在點(diǎn)P,它的坐標(biāo)為(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/33.png)
,3)或(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/34.png)
,3).(12分)
方法3:
如圖(3),
∵△ABC為等腰三角形,且
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/35.png)
,
設(shè)P(x,y),則△ABC∽△PAB等價(jià)于PB=AB=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/36.png)
,PA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/37.png)
AB=6.(10分)
當(dāng)x>0時(shí),得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/38.png)
,
解得P(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/39.png)
,3).(11分)
又∵P(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/40.png)
,3)的坐標(biāo)滿足y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/41.png)
x
2-1,
∴在拋物線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/42.png)
x
2-1上,存在點(diǎn)P(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/43.png)
,3),使△ABC∽△PAB.
由拋物線的對稱性,知點(diǎn)(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/44.png)
,3)也符合題意.
∴存在點(diǎn)P,它的坐標(biāo)為(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/45.png)
,3)或(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131211111102059289467/SYS201312111111020592894027_DA/46.png)
,3).(12分)
點(diǎn)評:本題主要考查了待定系數(shù)法求函數(shù)的解析式.并且本題考查了相似三角形的對應(yīng)邊的比相等.