【答案】
分析:(1)把點A、B、C的坐標(biāo)分別代入已知拋物線的解析式列出關(guān)于系數(shù)的三元一次方程組
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/0.png)
,通過解該方程組即可求得系數(shù)的值;
(2)由(1)中的拋物線解析式易求點M的坐標(biāo)為(0,1).所以利用待定系數(shù)法即可求得直線AM的關(guān)系式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/1.png)
x+1.由題意設(shè)點D的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/2.png)
),則點F的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/3.png)
).易求DF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/5.png)
.根據(jù)二次函數(shù)最值的求法來求線段DF的最大值;
(3)需要對點P的位置進(jìn)行分類討論:點P分別位于第一、二、三、四象限四種情況.此題主要利用相似三角形的對應(yīng)邊成比例進(jìn)行解答.
解答:解:由題意可知
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/6.png)
.解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/7.png)
.
∴拋物線的表達(dá)式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/8.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/images9.png)
(2)將x=0代入拋物線表達(dá)式,得y=1.∴點M的坐標(biāo)為(0,1).
設(shè)直線MA的表達(dá)式為y=kx+b,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/9.png)
.
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/10.png)
.
∴直線MA的表達(dá)式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/11.png)
x+1.
設(shè)點D的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/12.png)
),則點F的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/13.png)
).
DF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/15.png)
.
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/16.png)
時,DF的最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/17.png)
.
此時
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/18.png)
,即點D的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/19.png)
).
(3)存在點P,使得以點P、A、N為頂點的三角形與△MAO相似.設(shè)P(m,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/20.png)
).
在Rt△MAO中,AO=3MO,要使兩個三角形相似,由題意可知,點P不可能在第一象限.
①設(shè)點P在第二象限時,∵點P不可能在直線MN上,∴只能PN=3AN,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/21.png)
,即m
2+11m+24=0.解得m=-3(舍去)或m=-8.又-3<m<0,故此時滿足條件的點不存在.
②當(dāng)點P在第三象限時,∵點P不可能在直線MN上,∴只能PN=3AN,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/22.png)
,即m
2+11m+24=0.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/images24.png)
解得m=-3或m=-8.此時點P的坐標(biāo)為(-8,-15).
③當(dāng)點P在第四象限時,若AN=3PN時,則-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/23.png)
,即m
2+m-6=0.
解得m=-3(舍去)或m=2.
當(dāng)m=2時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/24.png)
.此時點P的坐標(biāo)為(2,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/25.png)
).
若PN=3NA,則-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/26.png)
,即m
2-7m-30=0.
解得m=-3(舍去)或m=10,此時點P的坐標(biāo)為(10,-39).
綜上所述,滿足條件的點P的坐標(biāo)為(-8,-15)、(2,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456282421/SYS201311011933384562824023_DA/27.png)
)、(10,-39).
點評:本題考查了二次函數(shù)綜合題.其中涉及到了待定系數(shù)法求二次函數(shù)解析式,相似三角形的性質(zhì)以及二次函數(shù)最值的求法.需注意分類討論,全面考慮點P所在位置的各種情況.