【答案】
分析:(1)根據(jù)A、B兩點(diǎn)坐標(biāo)求直線AB的解析式,令x=0,可求E點(diǎn)坐標(biāo);
(2)設(shè)拋物線解析式為y=ax
2+bx+c,將A(-2,2),B(6,6),O(0,0)三點(diǎn)坐標(biāo)代入,列方程組求a、b、c的值即可;
(3)依題意,得直線OB的解析式為y=x,設(shè)過(guò)N點(diǎn)且與直線OB平行的直線解析式為y=x+m,與拋物線解析式聯(lián)立,得出關(guān)于x的一元二次方程,當(dāng)△=0時(shí),△BON面積最大,由此可求m的值及N點(diǎn)的坐標(biāo);
(4)根據(jù)三角形相似的性質(zhì)得到BO:OA=OP:AN=BP:ON,然后根據(jù)勾股定理分別計(jì)算出BO=6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/0.png)
,OA=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/1.png)
,AN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/2.png)
,ON=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/3.png)
,這樣可求出OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/4.png)
,BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/5.png)
,設(shè)P點(diǎn)坐標(biāo)為(x,y),再利用勾股定理得到關(guān)于x,y的方程組,解方程組即可.
解答:解:(1)設(shè)直線AB解析式為y=kx+b,
將A(-2,2),B(6,6)代入,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/6.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/7.png)
,
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/8.png)
x+3,令x=0,
∴E(0,3);
(2)設(shè)拋物線解析式為y=ax
2+b′x+c,
將A(-2,2),B(6,6),O(0,0)三點(diǎn)坐標(biāo)代入,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/9.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/10.png)
,
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/11.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/12.png)
x
(3)依題意,得直線OB的解析式為y=x,設(shè)過(guò)N點(diǎn)且與直線OB平行的直線解析式為y=x+m,
聯(lián)立
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/13.png)
,得x
2-6x-4m=0,當(dāng)△=36+16m=0時(shí),過(guò)N點(diǎn)與OB平行的直線與拋物線有唯一的公共點(diǎn),則點(diǎn)N到BO的距離最大,所以△BON面積最大,
解得m=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/14.png)
,x=3,y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/15.png)
,即N(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/16.png)
);
此時(shí)△BON面積=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/17.png)
×6×6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/18.png)
(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/19.png)
+6)×3-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/20.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/21.png)
×3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/22.png)
;
(4)過(guò)點(diǎn)A作AS⊥GQ于S,
∵A(-2,2),B(6,6),N(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/23.png)
),
∵∠AOE=∠OAS=∠BOH=45°,
OG=3,NG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/24.png)
,NS=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/25.png)
,AS=5,
在Rt△SAN和Rt△NOG中,
∴tan∠SAN=tan∠NOG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/26.png)
,
∴∠SAN=∠NOG,
∴∠OAS-∠SAN=∠BOG-∠NOG,
∴∠OAN=∠NOB,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/images27.png)
∴ON的延長(zhǎng)線上存在一點(diǎn)P,使得△BOP∽△OAN,
∵A(-2,2),N(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/27.png)
),
∵△BOP與△OAN相似(點(diǎn)B、O、P分別與點(diǎn)O、A、N對(duì)應(yīng)),即△BOP∽△OAN,
∴BO:OA=OP:AN=BP:ON
又∵A(-2,2),N(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/28.png)
),B(6,6),
∴BO=6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/29.png)
,OA=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/30.png)
,AN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/31.png)
,ON=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/32.png)
,
∴OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/33.png)
,BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/34.png)
,
設(shè)P點(diǎn)坐標(biāo)為(4x,x),
∴16x
2+x
2=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/35.png)
)
2,
解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/36.png)
,4x=15,
∵P、P′關(guān)于直線y=x軸對(duì)稱,
∴P點(diǎn)坐標(biāo)為(15,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/37.png)
)或(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190557582772419/SYS201311011905575827724027_DA/38.png)
,15).
點(diǎn)評(píng):本題考查了二次函數(shù)的綜合運(yùn)用.根據(jù)已知條件求直線、拋物線解析式,再根據(jù)圖形特點(diǎn),將問(wèn)題轉(zhuǎn)化為列方程組,利用代數(shù)方法解題.