若拋物線y=x2+2x-1上有兩點A、B,且原點位于線段AB的三等分點處,則這兩點的坐標為 .
【答案】
分析:過A作AE⊥x軸于E,過B作BF⊥x軸于F,分兩種情況:
(1)當OA=2OB時,設B(a,b),(a>0,b>0)則A的坐標是(-2a,-2b),代入y=x
2+2x-1即可求出A、B的坐標;
(2)當2OA=OB時,與(1)方法類似即可求出A、B的坐標.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/images0.png)
解:過A作AE⊥x軸于E,過B作BF⊥x軸于F,
(1)當OA=2OB時,如圖
設B(a,b),(a>0,b>0)則A的坐標是(-2a,-2b),代入y=x
2+2x-1得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/0.png)
,
解得:a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/1.png)
,b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/2.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/3.png)
,
∴-2a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/4.png)
,-2b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/5.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/6.png)
,
∴A(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/7.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/8.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/9.png)
),B(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/11.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/12.png)
);
(2)當2OA=OB時,與(1)解法類似可求出A(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/13.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/14.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/15.png)
),B(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/17.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/18.png)
).
故答案為:(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/19.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/20.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/21.png)
),(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/23.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/24.png)
)或(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/25.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/26.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/27.png)
),(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/28.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/29.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190438405411041/SYS201311011904384054110015_DA/30.png)
).
點評:本題主要考查了二次函數(shù)圖象上點的特征,平行線分線段成比例定理,解二元二次方程組等知識點,解此題的關鍵是設出A和B的坐標,代入解析式能求出方程的解.