【答案】
分析:(1)易得點A(0,1),那么把A,B坐標(biāo)代入y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/0.png)
x
2+bx+c即可求得函數(shù)解析式;
(2)讓直線解析式與拋物線的解析式結(jié)合即可求得點E的坐標(biāo).△PAE是直角三角形,應(yīng)分點P為直角頂點,點A是直角頂點,點E是直角頂點三種情況探討;
(3)易得|AM-MC|的值最大,應(yīng)找到C關(guān)于對稱軸的對稱點B,連接AB交對稱軸的一點就是M.應(yīng)讓過AB的直線解析式和對稱軸的解析式聯(lián)立即可求得點M坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/images1.png)
解:(1)將A(0,1)、B(1,0)坐標(biāo)代入y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/1.png)
x
2+bx+c
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/2.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/3.png)
,
∴拋物線的解折式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/4.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/5.png)
x+1;(2分)
(2)設(shè)點E的橫坐標(biāo)為m,則它的縱坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/6.png)
m
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/7.png)
m+1,
即E點的坐標(biāo)(m,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/8.png)
m
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/9.png)
m+1),
又∵點E在直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/10.png)
x+1上,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/11.png)
m
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/12.png)
m+1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/13.png)
m+1
解得m
1=0(舍去),m
2=4,
∴E的坐標(biāo)為(4,3).(4分)
(Ⅰ)當(dāng)A為直角頂點時,
過A作AP
1⊥DE交x軸于P
1點,設(shè)P
1(a,0)易知D點坐標(biāo)為(-2,0),
由Rt△AOD∽Rt△P
1OA得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/14.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/15.png)
,
∴a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/16.png)
,
∴P
1(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/17.png)
,0).(5分)
(Ⅱ)同理,當(dāng)E為直角頂點時,過E作EP
2⊥DE交x軸于P
2點,
由Rt△AOD∽Rt△P
2ED得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/18.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/20.png)
,
∴EP
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/21.png)
,
∴DP
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/23.png)
∴a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/24.png)
-2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/25.png)
,
P
2點坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/26.png)
,0).(6分)
(Ⅲ)當(dāng)P為直角頂點時,過E作EF⊥x軸于F,設(shè)P
3(b、0),
由∠OPA+∠FPE=90°,得∠OPA=∠FEP,Rt△AOP∽Rt△PFE,
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/27.png)
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/28.png)
,
解得b
1=3,b
2=1,
∴此時的點P
3的坐標(biāo)為(1,0)或(3,0),(8分)
綜上所述,滿足條件的點P的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/29.png)
,0)或(1,0)或(3,0)或(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/30.png)
,0);
(3)拋物線的對稱軸為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/31.png)
,(9分)
∵B、C關(guān)于x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/32.png)
對稱,
∴MC=MB,
要使|AM-MC|最大,即是使|AM-MB|最大,
由三角形兩邊之差小于第三邊得,當(dāng)A、B、M在同一直線上時|AM-MB|的值最大.(10分)
易知直線AB的解折式為y=-x+1
∴由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/33.png)
,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/34.png)
,
∴M(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/35.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657925094/SYS201311011911586579250027_DA/36.png)
).(11分)
點評:一個三角形是直角三角形,應(yīng)分不同頂點為直角等多種情況進(jìn)行分析;
求兩條線段和或差的最值,都要考慮做其中一點關(guān)于所求的點在的直線的對稱點.