【答案】
分析:(1)設(shè)拋物線的解析式是y=ax
2+bx+c,求出A、B、D的坐標(biāo)代入即可;
(2)①由勾股定理即可求出,②假設(shè)存在點(diǎn)R,可構(gòu)成以P、B、R、Q為頂點(diǎn)的平行四邊形,求出P、Q的坐標(biāo),再分為兩種種情況:A、B、C即可根據(jù)平行四邊形的性質(zhì)求出R的坐標(biāo).
(3)A關(guān)于拋物線的對(duì)稱(chēng)軸的對(duì)稱(chēng)點(diǎn)為B,過(guò)B、D的直線與拋物線的對(duì)稱(chēng)軸的交點(diǎn)為所求M,求出直線BD的解析式,把拋物線的對(duì)稱(chēng)軸x=1代入即可求出M的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/images0.png)
解:(1)設(shè)拋物線的解析式是y=ax
2+bx+c,
∵正方形的邊長(zhǎng)2,
∴B的坐標(biāo)(2,-2)A點(diǎn)的坐標(biāo)是(0,-2),
把A(0,-2),B(2,-2),D(4,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/0.png)
)代入得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/1.png)
,
解得a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/2.png)
,b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/3.png)
,c=-2,
∴拋物線的解析式為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/4.png)
,
答:拋物線的解析式為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/5.png)
.
(2)①由圖象知:PB=2-2t,BQ=t,
∴S=PQ
2=PB
2+BQ
2,
=(2-2t)
2+t
2,
即S=5t
2-8t+4(0≤t≤1).
答:S與運(yùn)動(dòng)時(shí)間t之間的函數(shù)關(guān)系式是S=5t
2-8t+4,t的取值范圍是0≤t≤1.
②解:假設(shè)存在點(diǎn)R,可構(gòu)成以P、B、R、Q為頂點(diǎn)的平行四邊形.
∵S=5t
2-8t+4(0≤t≤1),
∴當(dāng)S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/6.png)
時(shí),5t
2-8t+4=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/7.png)
,得20t
2-32t+11=0,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/8.png)
,t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/9.png)
(不合題意,舍去),
此時(shí)點(diǎn)P的坐標(biāo)為(1,-2),Q點(diǎn)的坐標(biāo)為(2,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/10.png)
),
若R點(diǎn)存在,分情況討論:
(i)假設(shè)R在BQ的右邊,如圖所示,這時(shí)QR=PB,RQ∥PB,
則R的橫坐標(biāo)為3,R的縱坐標(biāo)為-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/11.png)
,
即R(3,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/12.png)
),
代入
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/13.png)
,左右兩邊相等,
∴這時(shí)存在R(3,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/14.png)
)滿足題意;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/images16.png)
(ii)假設(shè)R在QB的左邊時(shí),這時(shí)PR=QB,PR∥QB,
則R(1,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/15.png)
)代入,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/16.png)
,
左右不相等,∴R不在拋物線上.(1分)
綜上所述,存點(diǎn)一點(diǎn)R(3,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/17.png)
)滿足題意.
則存在,R點(diǎn)的坐標(biāo)是(3,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/18.png)
);
(3)如圖,M′B=M′A,
∵A關(guān)于拋物線的對(duì)稱(chēng)軸的對(duì)稱(chēng)點(diǎn)為B,過(guò)B、D的直線與拋物線的對(duì)稱(chēng)軸的交點(diǎn)為所求M,
理由是:∵M(jìn)A=MB,若M不為L(zhǎng)與DB的交點(diǎn),則三點(diǎn)B、M、D構(gòu)成三角形,
∴|MB|-|MD|<|DB|,
即交點(diǎn)時(shí)差為|DB|為最大,
設(shè)直線BD的解析式是y=kx+b,把B、D的坐標(biāo)代入得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/19.png)
,
解得:k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/20.png)
,b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/21.png)
,
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/22.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/23.png)
,
拋物線
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/24.png)
的對(duì)稱(chēng)軸是x=1,
把x=1代入得:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/25.png)
∴M的坐標(biāo)為(1,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/26.png)
);
答:M的坐標(biāo)為(1,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155306727260075/SYS201310221553067272600027_DA/27.png)
).
點(diǎn)評(píng):本題主要考查了用待定系數(shù)法求一次函數(shù)和二次函數(shù)的解析式,勾股定理,平行四邊形的性質(zhì),二次函數(shù)圖象上點(diǎn)的坐標(biāo)特征等知識(shí)點(diǎn),解此題的關(guān)鍵是綜合運(yùn)用這些知識(shí)進(jìn)行計(jì)算.此題綜合性強(qiáng),是一道難度較大的題目.