【答案】
分析:(1)首先求得m的值和直線的解析式,根據(jù)拋物線對稱性得到B點坐標,根據(jù)A、B點坐標利用交點式求得拋物線的解析式;
(2)存在點E使得以A、C、E、F為頂點的四邊形是平行四邊形.如答圖1所示,過點E作EG⊥x軸于點G,構(gòu)造全等三角形,利用全等三角形和平行四邊形的性質(zhì)求得E點坐標和平行四邊形的面積.注意:符合要求的E點有兩個,如答圖1所示,不要漏解;
(3)本問較為復雜,如答圖2所示,分幾個步驟解決:
第1步:確定何時△ACP的周長最�。幂S對稱的性質(zhì)和兩點之間線段最短的原理解決;
第2步:確定P點坐標P(1,3),從而直線M
1M
2的解析式可以表示為y=kx+3-k;
第3步:利用根與系數(shù)關(guān)系求得M
1、M
2兩點坐標間的關(guān)系,得到x
1+x
2=2-4k,x
1x
2=-4k-3.這一步是為了后續(xù)的復雜計算做準備;
第4步:利用兩點間的距離公式,分別求得線段M
1M
2、M
1P和M
2P的長度,相互比較即可得到結(jié)論:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/0.png)
=1為定值.這一步涉及大量的運算,注意不要出錯,否則難以得出最后的結(jié)論.
解答:解:(1)∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/1.png)
經(jīng)過點(-3,0),
∴0=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/2.png)
+m,解得m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/3.png)
,
∴直線解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/4.png)
,C(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/5.png)
).
∵拋物線y=ax
2+bx+c對稱軸為x=1,且與x軸交于A(-3,0),∴另一交點為B(5,0),
設(shè)拋物線解析式為y=a(x+3)(x-5),
∵拋物線經(jīng)過C(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/6.png)
),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/7.png)
=a•3(-5),解得a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/8.png)
,
∴拋物線解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/9.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/10.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/11.png)
;
(2)假設(shè)存在點E使得以A、C、E、F為頂點的四邊形是平行四邊形,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/images12.png)
則AC∥EF且AC=EF.如答圖1,
(i)當點E在點E位置時,過點E作EG⊥x軸于點G,
∵AC∥EF,∴∠CAO=∠EFG,
又∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/12.png)
,
∴△CAO≌△EFG,
∴EG=CO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/13.png)
,即y
E=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/14.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/16.png)
x
E2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/17.png)
x
E+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/18.png)
,解得x
E=2(x
E=0與C點重合,舍去),
∴E(2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/19.png)
),S
?ACEF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/20.png)
;
(ii)當點E在點E′位置時,過點E′作E′G′⊥x軸于點G′,
同理可求得E′(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/21.png)
+1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/22.png)
),S
?ACF′E′=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/23.png)
.
(3)要使△ACP的周長最小,只需AP+CP最小即可.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/images25.png)
如答圖2,連接BC交x=1于P點,因為點A、B關(guān)于x=1對稱,根據(jù)軸對稱性質(zhì)以及兩點之間線段最短,可知此時AP+CP最�。ˋP+CP最小值為線段BC的長度).
∵B(5,0),C(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/24.png)
),∴直線BC解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/25.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/26.png)
,
∵x
P=1,∴y
P=3,即P(1,3).
令經(jīng)過點P(1,3)的直線為y=kx+b,則k+b=3,即b=3-k,
則直線的解析式是:y=kx+3-k,
∵y=kx+3-k,y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/27.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/28.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/29.png)
,
聯(lián)立化簡得:x
2+(4k-2)x-4k-3=0,
∴x
1+x
2=2-4k,x
1x
2=-4k-3.
∵y
1=kx
1+3-k,y
2=kx
2+3-k,∴y
1-y
2=k(x
1-x
2).
根據(jù)兩點間距離公式得到:
M
1M
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/30.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/32.png)
∴M
1M
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/33.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/34.png)
=4(1+k
2).
又M
1P=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/35.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/37.png)
;
同理M
2P=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/38.png)
∴M
1P•M
2P=(1+k
2)•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/39.png)
=(1+k
2)•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/40.png)
=(1+k
2)•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/41.png)
=4(1+k
2).
∴M
1P•M
2P=M
1M
2,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185136467334973/SYS201311011851364673349027_DA/42.png)
=1為定值.
點評:本題是難度很大的中考壓軸題,綜合考查了初中數(shù)學的諸多重要知識點:代數(shù)方面,考查了二次函數(shù)的相關(guān)性質(zhì)、一次函數(shù)的相關(guān)性質(zhì)、一元二次方程根與系數(shù)的關(guān)系以及二次根式的運算等;幾何方面,考查了平行四邊形、全等三角形、兩點間的距離公式、軸對稱-最短路線問題等.本題解題技巧要求高,而且運算復雜,因此對考生的綜合能力提出了很高的要求.