曲線x2+y2+x-6y+3=0上兩點P、Q關(guān)于直線kx-y+4=0對稱,求直線PQ的方程.
【答案】
分析:因為曲線方程為圓的方程,圓上的P與Q關(guān)于直線對稱得到直線過圓心,把圓心坐標(biāo)代入即可求出k,又因為PQ⊥直線kx-y+4=0得到直線PQ的斜率為-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/0.png)
,然后聯(lián)立直線與圓的方程求出P和Q的坐標(biāo),即可寫出直線的方程.
解答:解:曲線x
2+y
2+x-6y+3=0可變?yōu)椋?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/1.png">+(y-3)
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/2.png)
得到圓心(-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/3.png)
,3),半徑為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/4.png)
;
因為圓上有兩點P、Q關(guān)于直線對稱,得到圓心在直線上,
把(-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/5.png)
,3)代入到kx-y+4=0中求出k=2,且PQ與直線垂直,
所以直線PQ的斜率=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/6.png)
=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/7.png)
;
聯(lián)立得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/8.png)
解得:x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/9.png)
,y=3±
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/10.png)
,
任取一點坐標(biāo)得到(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/11.png)
,3+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/12.png)
)
所以直線PQ的方程為:y-(3+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/13.png)
)=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/14.png)
(x-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214623696922994/SYS201310232146236969229015_DA/15.png)
).
點評:考查學(xué)生理解圓的對稱軸為過直徑的直線,會根據(jù)兩直線垂直得到斜率乘積為-1,會根據(jù)條件寫出直線的一般式方程.