解:(1)设抛物线的解析式为y=a(x-1)
2-3
将A(-1,0)代入:0=a(-1-1)
2-3,
解得a=
![](http://thumb.1010pic.com/pic5/latex/365.png)
所以,抛物线的解析式为y=
![](http://thumb.1010pic.com/pic5/latex/365.png)
(x-1)
2-3,即y=
![](http://thumb.1010pic.com/pic5/latex/365.png)
x
2-
![](http://thumb.1010pic.com/pic5/latex/33.png)
x-
![](http://thumb.1010pic.com/pic5/latex/2321.png)
(2)是定值,
![](http://thumb.1010pic.com/pic5/latex/373620.png)
=1
∵AB为直径,
∴∠AEB=90°,
∵PM⊥AE,
∴PM∥BE,
∴△APM∽△ABE,
所以
![](http://thumb.1010pic.com/pic5/latex/373622.png)
①
同理:
![](http://thumb.1010pic.com/pic5/latex/373623.png)
②
①+②:
![](http://thumb.1010pic.com/pic5/latex/373624.png)
(3)∵直线EC为抛物线对称轴,
∴EC垂直平分AB,
![](http://thumb.1010pic.com/pic5/upload/201311/5285ada963770.png)
∴EA=EB,
∵∠AEB=90°,
∴△AEB为等腰直角三角形,
∴∠EAB=∠EBA=45°
如图,过点P作PH⊥BE于H,
由已知及作法可知,四边形PHEM是矩形.
∴PH=ME且PH∥ME.
在△APM和△PBH中,
∵∠AMP=∠PHB=90°,∠EAB=∠BPH=45°,
∴PH=BH,且△APM∽△PBH,
∴
![](http://thumb.1010pic.com/pic5/latex/373625.png)
,
∴
![](http://thumb.1010pic.com/pic5/latex/373626.png)
①
在△MEP和△EGF中,
∵PE⊥FG,
∴∠FGE+∠SEG=90°,
∵∠MEP+∠SEG=90°,
∴∠FGE=∠MEP,
∵∠PME=∠FEG=90°,
∴△MEP∽△EGF,
∴
![](http://thumb.1010pic.com/pic5/latex/373627.png)
②
由①、②知:
![](http://thumb.1010pic.com/pic5/latex/373621.png)
(本题若按分类证明,只要合理,可给满分)
分析:(1)已知抛物线的顶点坐标就可以利用顶点式求函数的解析式.
(2)AB是圆的直径,因而∠ADB=∠AEB=90°,得到PN∥AD,得到
![](http://thumb.1010pic.com/pic5/latex/373628.png)
=
![](http://thumb.1010pic.com/pic5/latex/203317.png)
,同理
![](http://thumb.1010pic.com/pic5/latex/373629.png)
=
![](http://thumb.1010pic.com/pic5/latex/118649.png)
,这样就可以求出
![](http://thumb.1010pic.com/pic5/latex/373620.png)
的值.
(3)易证△AEB为等腰直角三角形,过点P作PH⊥BE与H,四边形PHEM是矩形,易证△APM∽△PBH,则
![](http://thumb.1010pic.com/pic5/latex/373626.png)
,再证明△MEP∽△EGF,则
![](http://thumb.1010pic.com/pic5/latex/373627.png)
因而
![](http://thumb.1010pic.com/pic5/latex/373621.png)
可证.
点评:本题主要考查了待定系数法求二次函数的解析式,以及相似三角形的对应边的比相等.