【答案】
分析:(1)利用當(dāng)P點(diǎn)運(yùn)動(dòng)到A點(diǎn)時(shí),△POC的面積為12,求出斜邊AO可得出m的值,圖1中四邊形ODEF是等腰梯形,點(diǎn)D的坐標(biāo)為D(m,12),得出y
E=y
D=12,此時(shí)圖1中點(diǎn)P運(yùn)動(dòng)到與點(diǎn)B重合,利用三角形面積求出OB的長(zhǎng),進(jìn)而得出B點(diǎn)坐標(biāo),以及利用△ABM≌△CON得出C點(diǎn)坐標(biāo)和利用勾股定理求出OF的長(zhǎng);
(2)①先求出點(diǎn)P的坐標(biāo),然后根據(jù)O、B可得出拋物線解析式;
②根據(jù)當(dāng)點(diǎn)P恰為經(jīng)過O,B兩點(diǎn)的拋物線的頂點(diǎn)時(shí),i)當(dāng)BP為以B,P,Q,R為頂點(diǎn)的菱形的邊時(shí),ii)當(dāng)BP為以B,P,Q,R為頂點(diǎn)的菱形的對(duì)角線時(shí),分別分析得出即可.
解答:解:(1)根據(jù)圖中得出:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/images0.png)
∵當(dāng)P點(diǎn)運(yùn)動(dòng)到A點(diǎn)時(shí),△POC的面積為12,
∴AO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/1.png)
,
∴m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/2.png)
;
∵圖1中四邊形ODEF是等腰梯形,點(diǎn)D的坐標(biāo)為D(m,12),
∴y
E=y
D=12,此時(shí)圖2中點(diǎn)P運(yùn)動(dòng)到與點(diǎn)B重合,
∵點(diǎn)B在x軸的正半軸上,
∴S
△POC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/3.png)
×OB×|yC|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/4.png)
×OB×3=12.
解得:OB=8,點(diǎn)B的坐標(biāo)為(8,0).
此時(shí)作AM⊥OB于點(diǎn)M,CN⊥OB于點(diǎn)N.
(如圖2).
∵點(diǎn)C的坐標(biāo)為C(n,-3),
∴點(diǎn)C在直線y=-3上.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/images6.png)
又∵由圖1中四邊形ODEF是等腰梯形可知圖2中的點(diǎn)C在過點(diǎn)O與AB平行的直線l上,
∴點(diǎn)C是直線y=-3與直線l的交點(diǎn),且∠ABM=∠CON.
又∵|y
A|=|y
C|=3,即AM=CN,
可得△ABM≌△CON.
∴ON=BM=6,點(diǎn)C的坐標(biāo)為C(6,-3).
∵圖2中 AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/6.png)
=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/7.png)
.
∴圖1中DE=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/8.png)
,OF=2x
D+DE=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/9.png)
+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/10.png)
.
(2)①當(dāng)點(diǎn)P恰為經(jīng)過O,B兩點(diǎn)的拋物線的頂點(diǎn)時(shí),作PG⊥OB于點(diǎn)G.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/images13.png)
(如圖3)
∵O,B兩點(diǎn)的坐標(biāo)分別為O(0,0),B(8,0),
∴由拋物線的對(duì)稱性可知點(diǎn)P的橫坐標(biāo)為4,即OG=BG=4.由tan∠ABM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/13.png)
,可得PG=2.
∴點(diǎn)P的坐標(biāo)為P(4,2),
設(shè)拋物線W的解析式為y=ax(x-8)(a≠0).
∵拋物線過點(diǎn)P(4,2),
∴4a(4-8)=2.
解得:a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/14.png)
,
∴拋物線W的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/15.png)
x2+x.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/images19.png)
②如圖4.
i)當(dāng)BP為以B,P,Q,R為頂點(diǎn)的菱形的邊時(shí),
∵點(diǎn)Q在直線y=-1上方的拋物線W 上,點(diǎn)P為拋物線W的頂點(diǎn),
結(jié)合拋物線的對(duì)稱性可知點(diǎn)Q只有一種情況,點(diǎn)Q與原點(diǎn)重合,其坐標(biāo)為Q
1(0,0).
ii)當(dāng)BP為以B,P,Q,R為頂點(diǎn)的菱形的對(duì)角線時(shí),可知BP的中點(diǎn)的坐標(biāo)為(6,1),BP的中垂線的解析式為y=2x-11.
∴點(diǎn)Q
2的橫坐標(biāo)是方程-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/16.png)
x2+x=2x-11的解.
將該方程整理得:x
2+8x-88=0.
解得x=-4±2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/17.png)
.
由點(diǎn)Q在直線y=-1上方的拋物線W上,結(jié)合圖4可知點(diǎn)Q
2的橫坐標(biāo)為2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/18.png)
-4.
∴點(diǎn)Q
2的坐標(biāo)是Q
2(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/19.png)
-4,4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/20.png)
-19).
綜上所述,符合題意的點(diǎn)Q的坐標(biāo)是Q
1(0,0),Q
2(2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/21.png)
-4,4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191918195030764/SYS201311011919181950307026_DA/22.png)
-19).
點(diǎn)評(píng):此題主要考查了二次函數(shù)的綜合應(yīng)用以及菱形性質(zhì)和等腰梯形性質(zhì)等知識(shí),根據(jù)數(shù)形結(jié)合得出梯形面積進(jìn)而得出B,C點(diǎn)的坐標(biāo)是解題關(guān)鍵,難度較大.