【答案】
分析:(1)由拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/0.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/1.png)
x+m
2-3m+2與x軸的交點(diǎn)分別為原點(diǎn)O,令x=0,y=0,解得m的值,點(diǎn)B(2,n)在這條拋物線上,把該點(diǎn)代入拋物線方程,解得n.
(2)設(shè)直線OB的解析式為y=k
1x,求得直線OB的解析式為y=2x,由A點(diǎn)是拋物線與x軸的一個(gè)交點(diǎn),可求得A點(diǎn)的坐標(biāo),設(shè)P點(diǎn)的坐標(biāo)為(a,0),根據(jù)題意作等腰直角三角形PCD,如圖1.可求得點(diǎn)C的坐標(biāo),進(jìn)而求出OP的值,依題意作等腰直角三角形QMN,設(shè)直線AB的解析式為y=k
2x+b,求出直線AB的解析式,當(dāng)P點(diǎn)運(yùn)動(dòng)到t秒時(shí),兩個(gè)等腰直角三角形分別有一條邊恰好落在同一條直線上,有以下三種情況,解出各種情況下的時(shí)間t.
解答:解:(1)∵拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/2.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/3.png)
x+m
2-3m+2經(jīng)過原點(diǎn),
∴m
2-3m+2=0,
解得m
1=1,m
2=2,
由題意知m≠1,
∴m=2,
∴拋物線的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/4.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/5.png)
x,
∵點(diǎn)B(2,n)在拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/6.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/7.png)
x上,
∴n=4,
∴B點(diǎn)的坐標(biāo)為(2,4).
(2)設(shè)直線OB的解析式為y=k
1x,
求得直線OB的解析式為y=2x,
∵A點(diǎn)是拋物線與x軸的一個(gè)交點(diǎn),可求得A點(diǎn)的坐標(biāo)為(10,0),
設(shè)P點(diǎn)的坐標(biāo)為(a,0),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/images8.png)
則E點(diǎn)的坐標(biāo)為(a,2a),
根據(jù)題意作等腰直角三角形PCD,
如圖1,可求得點(diǎn)C的坐標(biāo)為(3a,2a),
由C點(diǎn)在拋物線上,
得:2a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/8.png)
´(3a)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/9.png)
´3a,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/10.png)
a
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/11.png)
a=0,
解得a
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/12.png)
,a
2=0(舍去),
∴OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/13.png)
.
依題意作等腰直角三角形QMN,設(shè)直線AB的解析式為y=k
2x+b,
由點(diǎn)A(10,0),點(diǎn)B(2,4),求得直線AB的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/14.png)
x+5,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/images16.png)
當(dāng)P點(diǎn)運(yùn)動(dòng)到t秒時(shí),兩個(gè)等腰直角三角形分別有一條邊恰好落在同一條直線上,有以下三種情況:
第一種情況:CD與NQ在同一條直線上.
如圖2所示.可證△DPQ為等腰直角三角形.此時(shí)OP、DP、AQ的長(zhǎng)可依次表示為t、4t、2t個(gè)單位.
∴PQ=DP=4t,
∴t+4t+2t=10,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/images17.png)
∴t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/15.png)
.
第二種情況:PC與MN在同一條直線上.如圖3所示.可證△PQM為等腰直角三
角形.此時(shí)OP、AQ的長(zhǎng)可依次表示為t、2t個(gè)單位.
∴OQ=10-2t,
∵F點(diǎn)在直線AB上,
∴FQ=t,
∴MQ=2t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/images19.png)
∴PQ=MQ=CQ=2t,
∴t+2t+2t=10,
∴t=2.
第三種情況:點(diǎn)P、Q重合時(shí),PD、QM在同一條直線上,如圖4所示.此時(shí)OP、
AQ的長(zhǎng)可依次表示為t、2t個(gè)單位.
∴t+2t=10,
∴t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/16.png)
.
綜上,符合題意的t值分別為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101552935920814/SYS201311031015529359208005_DA/17.png)
,2,
點(diǎn)評(píng):本題是二次函數(shù)的綜合題,要會(huì)求拋物線的解析式,討論分類情況,此題比較繁瑣,做題多加用心.