【答案】
分析:(1)首先過C作CD⊥x軸于G.構(gòu)造△OAB的中位線CG,根據(jù)A、B點(diǎn)的坐標(biāo)及三角形中位線的性質(zhì)不難求得點(diǎn)C的坐標(biāo).由于△ABO∽△ADC,利用相似三角形的性質(zhì)解得AD的長,那么D點(diǎn)的坐標(biāo)也就確定.
(2)運(yùn)用待定系數(shù)法求解.假設(shè)過B(0,4),C(1,2),D(-3,0)的拋物線的關(guān)系式為y=ax
2+bx+c,將三點(diǎn)坐標(biāo)值代入聯(lián)立組成三元一次方程組解得a、b、c的值.
(3)設(shè)點(diǎn)P的坐標(biāo)為(x,y)連BD,過點(diǎn)P作PH⊥x軸于H,交BD于E.觀察圖象發(fā)現(xiàn)S
四邊形PBCD=S
△BCD+S
△PBD,
因?yàn)镾
△BCD=S
△ACD為定值,所以要使四邊形PBCD的面積最大就是使△PBD的面積最大.再分別就①當(dāng)P在BD間的拋物線上時(shí)(即-3<x<0);②當(dāng)P在BC間的拋物線上時(shí)(即0<x<1)時(shí),討論x的取值,進(jìn)而得到P點(diǎn)的坐標(biāo),并驗(yàn)證結(jié)果的合理性.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/images0.png)
解:(1)過C作CD⊥x軸于G,
∵點(diǎn)C為線段AB的中點(diǎn),
∴CG是△OAB的中位線,
∴點(diǎn)C的坐標(biāo)是(1,2),┅┅┅┅┅┅┅┅(1分)
又∵OA=2,OB=4,
∴AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/0.png)
,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/1.png)
,
顯然△ABO∽△ADC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/2.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/3.png)
,┅┅┅┅┅┅┅┅┅┅┅(2分)
∴AD=5OD=AD-OA=3,
∴點(diǎn)D的坐標(biāo)是(-3,0);┅┅┅┅┅┅┅┅┅(3分)
(2)解:設(shè)過B(0,4),C(1,2),D(-3,0)的拋物線的關(guān)系式為y=ax
2+bx+c,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/4.png)
,┅┅┅┅┅┅(4分)
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/5.png)
,┅┅┅┅┅┅┅┅┅┅┅┅(5分)
∴拋物線的關(guān)系式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/6.png)
;┅┅┅┅┅┅┅┅┅(6分)
(3)解:設(shè)點(diǎn)P的坐標(biāo)為(x,y)連BD,過點(diǎn)P作PH⊥x軸于H,交BD于E
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/images8.png)
,
S
四邊形PBCD=S
△BCD+S
△PBD,
∵S
△BCD=S
△ACD為定值,
∴要使四邊形PBCD的面積最大就是使△PBD的面積最大,
①當(dāng)P在BD間的拋物線上時(shí),即-3<x<0,
S
△PBD=S
△PBE+S
△PED=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/7.png)
PE×DH+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/8.png)
PE×OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/9.png)
PE×OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/10.png)
PE,
∵PE=PH-EH=y
P-y
E,┅┅┅┅┅┅┅┅(7分)
直線BD的關(guān)系式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/11.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/images14.png)
∴PE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/12.png)
,
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/13.png)
,
當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/14.png)
時(shí),PE最大為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/15.png)
,
∴點(diǎn)P的坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/17.png)
),┅┅┅┅┅┅┅┅┅┅(8分)
②當(dāng)P在BC間的拋物線上時(shí),即0<x<1,
同理可求出四邊形PBCD的面積,
很顯然,此時(shí)四邊形PBCD的面積要小于點(diǎn)P在BD間的拋物線上時(shí)的四邊形PBCD的面積,
故P點(diǎn)的坐標(biāo)是(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022154432416093207/SYS201310221544324160932023_DA/19.png)
).┅┅┅┅┅┅┅┅┅(9分)
點(diǎn)評(píng):本題是二次函數(shù)的綜合題型,其中涉及到的知識(shí)點(diǎn)有利用待定系數(shù)法求拋物線的解析式和三角形的面積求法.在求有關(guān)動(dòng)點(diǎn)問題時(shí)要注意分析題意分情況討論結(jié)果;并有效利用了坐標(biāo)與線段的數(shù)形結(jié)合.