【答案】
分析:(1)本題要根據(jù)圖2的分段函數(shù)進行求解.當(dāng)0<z≤2時,P在OA上運動,因此S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/0.png)
OC•z=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/1.png)
mz.當(dāng)2<z≤3時,P在AB上運動,因此S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/2.png)
OC•OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/3.png)
mn.由此可得出當(dāng)P從A運動到B時,S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/4.png)
mn=m,因此n=2.而z的值是由2逐漸增大到3因此AB=1,因此B點的坐標應(yīng)該是(1,2).
(2)求四邊形OABC的面積,關(guān)鍵是確定m的值.(由于P不可能與O,D重合)可分三種情況進行討論:
①當(dāng)P在OA上時,此時P,O,C不可能構(gòu)成拋物線.因此這種情況不成立.
②當(dāng)P在AB上時,可先根據(jù)O,C的坐標來列出拋物線的解析式.此時P的縱坐標為2,然后可根據(jù)拋物線的解析式表示出P的橫坐標,然后將得出的P的坐標代入雙曲線中即可得出m的值.
③當(dāng)P在BC上時,也要先得出P點的縱坐標,具體思路是過B,P作x軸的垂線,通過相似三角形來求出P點的縱坐標,然后按①的方法求出m的值.
綜合上述的情況即可得出m的值,也就能確定OC的長,即可求出梯形OABC的面積.
解答:解:(1)從圖1中可知,當(dāng)P從O向A運動時,△POC的面積S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/5.png)
mz,z由0逐步增大到2,則S由0逐步增大到m,
故OA=2,n=2.
同理,AB=1,故點B的坐標是(1,2).
(2)∵拋物線y=ax
2+bx+c經(jīng)過點O(0,0),C(m,0)
∴c=0,b=-am,
∴拋物線為y=ax
2-amx,頂點坐標P為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/6.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/7.png)
am
2).
∵m>1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/images8.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/8.png)
>0,且
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/9.png)
≠m,
∴P不在邊OA上且不與C重合.
∵P在雙曲線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/10.png)
上,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/11.png)
×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/12.png)
am
2)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/13.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/14.png)
.
①當(dāng)1<m≤2時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/15.png)
<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/16.png)
≤1,如圖2,分別過B,P作x軸的垂線,
M,N為垂足,此時點P在線段AB上,且縱坐標為2,
∴-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/17.png)
am
2=2,即a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/18.png)
.
又∵a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/19.png)
,
∴-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/20.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/21.png)
,m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/22.png)
>2,而1<m≤2,不合題意,舍去.
②當(dāng)m≥2時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/23.png)
>1,如圖3,分別過B,P作x軸的垂線,M,N為垂足,ON>OM,
此時點P在線段CB上,易證Rt△BMC∽Rt△PNC,
∴BM:PN=MC:NC,即2:PN=(m-1):
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/24.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/images26.png)
∴PN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/25.png)
而P的縱坐標為-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/26.png)
am
2,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/27.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/28.png)
am
2,即a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/29.png)
.
而a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/30.png)
,
∴-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/32.png)
化簡得:5m
2-22m+22=0.
解得:m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/33.png)
,
但m≥2,所以m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/34.png)
舍去,
取m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/35.png)
.
由以上,這時四邊形OABC的面積為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/36.png)
(AB+OC)×OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/37.png)
(1+m)×2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103418894252984/SYS201311031034188942529026_DA/38.png)
.
點評:本題著重考查了二次函數(shù)以及反比例函數(shù)的相關(guān)知識、三角形相似等知識點,綜合性強,能力要求較高.考查學(xué)生分類討論,數(shù)形結(jié)合的數(shù)學(xué)思想方法.