【答案】
分析:(1)已知了四個條件:abc=0 ①;a+b+c=3②;ab+bc+ca=-3③;a<b<c④.
根據(jù)①可知b=0或c=0(a≠0),那么本題可分兩種情況進行討論:
一:當(dāng)b=0,可聯(lián)立②③,求出a,c的值,然后根據(jù)④判斷出符合條件的a,c的值,進而可求出拋物線的解析式.
二:當(dāng)c=0時,方法同一.
綜合兩種情況可得出拋物線的解析式.
(2)①比較S
△APC和S
△AOC的大小實際就是比較△DPC和△AOD的面積.
△AOD中,根據(jù)OA,OD的長,可求出△AOD的面積.
△DPC中,可以CD為底邊,P點的縱坐標(biāo)為高,
過P作PG⊥x軸于G,OG就是△DPC的高.
可根據(jù)相似三角形ADO和APG,得出關(guān)于OD,PG,OA,OG的比例關(guān)系式.
設(shè)出P點的坐標(biāo),即可根據(jù)所得的比例關(guān)系式求出P點的坐標(biāo),從而可求出△DPC的面積.
然后比較△DPC和△AOD的面積即可得出S
△APC和S
△AOC的大�。�
②本題要分兩種情況進行討論:
當(dāng)P點在第一象限時,解法同①,只不過要設(shè)出P點的坐標(biāo)和OD的長,其他解法基本一樣,只是最后不是比較大小,而是得出一個等量關(guān)系.根據(jù)這個等量關(guān)系來求P點的坐標(biāo).
可分別過C,A作坐標(biāo)軸的平行線,可得出一個矩形,設(shè)兩條平行線的交點為Q,那么△AQC與△AOC的面積相等,而P在△ACQ內(nèi),因此△ACP的面積總小于△ACQ的面積.因此△ACP的面積不會和△ACO的面積相等.此種情況不成立.
解答:解:(1)∵a≠0,abc=0,
∴bc=0
當(dāng)b=0時:
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/0.png)
,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/1.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/2.png)
.
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/3.png)
∵a<b<c.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/4.png)
(不合題意,舍去)
∴a=-1,b=0,c=4
<2>當(dāng)c=0時
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/5.png)
,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/6.png)
解之得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/7.png)
.
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/8.png)
,
∵a<b<c;
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/9.png)
和
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/10.png)
,都不合題意,舍去.
∴所求的拋物線解析式為y=-x
2+4.
(2)①在y=-x
2+4中,當(dāng)y=0時,x=±2;當(dāng)x=0時,y=4.
∴A、B、C三點的坐標(biāo)分別為(-2,0),(2,0),(0,4)
過P作PG⊥x軸于G,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/images11.png)
設(shè)點P坐標(biāo)為(m,n)
∵點P是這條拋物線上第一象限內(nèi)的點
∴m>0,n>0,n=-m
2+4
∴PG=-m
2+4,OA=2,AG=m+2
∵OD∥PG,OD=1.5
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/12.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/14.png)
解得m
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/15.png)
,m
2=-2(不合題意,舍去)
∴OG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/16.png)
又CD=OC-OD=4-1.5=2.5
S
△PDC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/17.png)
•CD•OG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/18.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/19.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/21.png)
S
△AOD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/22.png)
•OA•OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/23.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/24.png)
×2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/25.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/26.png)
∴S
△PDC>S
△AOD又∵S
△APC=S
△PDC+S
△ADC,S
△AOC=S
△AOD+S
△ADC
∴S
△APC>S
△AOC②分兩種情況討論:
在第一象限內(nèi),設(shè)在拋物線上存在點Pn(m,n),使得S
△APnC=S
△AOC.
過Pn作PnM⊥x軸于點M,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/images28.png)
則m>0,n>0,n=-m
2+4
OM=m,PnM=-m
2+4,OA=2,AM=m+2
設(shè)APn交y軸于點Dn,設(shè)ODn=t
∵ODn∥PnM,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/28.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/29.png)
化簡為mt+2t=8-2m
2,DnC=OC-ODn=4-t
S
△AODn=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/30.png)
OA•ODn=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/31.png)
×2×t=t;
S
△PnCDn=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/32.png)
CDn•OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/33.png)
(4-t)×m;
∵S
△AOC=S
△APnC
∴S
△AODn=S
△PnCDn即t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/34.png)
(4-t)×m,mt+2t=4m
將mt+2t=4m代入mt+2t=8-2m
2中有8-2m
2=4m
整理得m
2+2m-4=0,m
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/35.png)
-1,m
2=-1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/36.png)
∵m>0,
∴m
2=-1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/37.png)
(不合題意,舍去)
∴m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/38.png)
-1,
此時n=-m
2+4=-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/39.png)
-1)
2+4=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/40.png)
-2
∴存在點Pn坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/41.png)
-1,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/42.png)
-2),
使得S
△APnC=S
△AOC在第二象限內(nèi),這條拋物線上任取一點Pnn,連接PnnA,PnnC,分別過點A作直線l
1垂直x軸,過點C作直線l
2垂直于y軸,l
1與l
2相交于Q點,則四邊形QAOC是矩形,S
△AQC=S
△AOC.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232403698072232/SYS201310212324036980722011_DA/images45.png)
設(shè)Pnn點坐標(biāo)為(mn,nn)
則有-2<mn<0
∵nn=-m
n2+4
∴0<nn<4
∴點Pn
n在矩形QAOC內(nèi),又易知Pnn在△AQC內(nèi)
∴S
△APnC<S
△AQC,S
△APnC<S
△AOC
∴在第二象限內(nèi)這條拋物線上不存在點Pnn,使S
△APnC=S
△AOC.
點評:本題結(jié)合三角形的相關(guān)知識考查了二次函數(shù)的綜合應(yīng)用,由于題中的數(shù)據(jù)較多,計算過程較復(fù)雜,因此細心求解是解題的關(guān)鍵.