【答案】
分析:(1)已知A、B、C三点的坐标,利用待定系数法能确定直线AC与抛物线的解析式.
(2)首先表示出BP、BD、OD、OE四边的长,若四边形DEFP为矩形,那么必须满足的条件是∠PDE是直角,此时△PBD、△DOE相似,可据此求出t的值,在求出BP的长以及点P的坐标后,利用待定系数法即可求出直线PQ的解析式(直线PQ与直线AC平行,那么它们的斜率相同,在设直线解析式时可利用这个特点).
(3)方法同(2),不过由四边形DEFP为正方形得出的条件变为△PBD、△DOE全等,首先由BD=OE求出t的值,再由OD=BP求出a的值;进一步能得到DP、DE的长,由此求得正方形的面积.
(4)此题需要注意两方面:
①线段MN是底边(此时线段MN的长是点M纵坐标的2倍);②线段MN为腰(此时线段MN的长等于点M的纵坐标);
解法大致相同,首先设出点M或N的纵坐标,利用△CMN、△CAO相似,求出这个纵坐标,再利用直线OC、直线AC解析式确定出点M、N的坐标后,即可得到点P的坐标.
解答:解:(1)设直线AC的解析式为:y=kx+b,依题意,有:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/0.png)
,解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/1.png)
∴直线AC:y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/2.png)
x+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/3.png)
.
设抛物线的解析式为:y=ax
2+bx+c,依题意,有:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/4.png)
,解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/5.png)
∴抛物线:y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/6.png)
x
2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/7.png)
x+c.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/images8.png)
(2)过点B作BS∥AC,交x轴于点S,则AS=BC=5,OR=3,∴tan∠OBS=tan∠ODE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/8.png)
.
BP=BC-CP=5-at=5-t,BD=t,OD=OB-BD=4-t,OE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/9.png)
OD=3-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/10.png)
t;
由题意,四边形DEFP是平行四边形,若四边形DEFP是矩形,所以∠PDE=90°;
∵∠PDB=∠DEO=90°-∠ODE,∠PBD=∠DOE=90°,
∴△PBD∽△DOE,得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/11.png)
即:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/12.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/13.png)
,解得 t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/14.png)
,则P(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/15.png)
,4);
由于直线PQ∥AC,设直线PQ:y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/16.png)
x+b,代入点P,得:
-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/17.png)
×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/18.png)
+b=4,解得 b=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/19.png)
∴若a=1,当t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/20.png)
时,四边形DEFP为矩形;此时直线PQ的解析式:y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/21.png)
x+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/22.png)
.
(3)同(2)可求得:△PBD≌△DOE,则 BD=OE,BP=OD;
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/23.png)
,解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/24.png)
由题意,此时a的值不在0<a≤1.25的范围内,所以不存在符合条件的a、t值.
(4)易求得:直线OC:y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/25.png)
x;直线AC:y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/26.png)
x+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/27.png)
.
设点M、N的纵坐标为m,分两种情况讨论:
(Ⅰ)线段MN为等腰Rt△MNR的底边,则 MN=2m;
由MN∥OA,得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/28.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/29.png)
,解得 m=2;
∴M(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/30.png)
,2)、N(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/31.png)
,2)
∴点R(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/32.png)
,0).
(Ⅱ)线段MN为等腰Rt△MNR的腰,则 MN=m;
由MN∥OA,得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/33.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/34.png)
,解得 m=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/35.png)
∴M(6,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/36.png)
)、N(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/37.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/38.png)
)
①当点N是直角顶点时,NR⊥x轴,点R(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/39.png)
,0);
②当点M是直角顶点时,MR⊥x轴,点R(6,0);
综上,存在符合条件的点R,且坐标为(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/40.png)
,0)、(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190557583147228/SYS201311011905575831472023_DA/41.png)
,0)、(6,0).
点评:此题考查的是动点函数问题,主要涉及了利用待定系数法确定函数解析式、矩形和正方形的性质以及等腰直角三角形的判定和性质;其中还穿插了全等、相似三角形的性质以及解直角三角形的应用;综合性很强.在解答这道题时,对图示的理解很重要,着重体现了数形结合的重要性.