曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)OP垂直OQ曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)O为原点,OP垂直OQ求直线PQ的方程.注意:O为原点
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![曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)OP垂直OQ曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)O为原点,OP垂直OQ求直线PQ的方程.注意:O为原点](/uploads/image/z/5190336-0-6.jpg?t=%E6%9B%B2%E7%BA%BFx%5E2%2By%5E2%2Bx-6y%2B3%3D0%E4%B8%8A%E4%B8%A4%E7%82%B9P%2CQ%E6%BB%A1%E8%B6%B3%EF%BC%9A%EF%BC%881%EF%BC%89%E5%85%B3%E4%BA%8E%E7%9B%B4%E7%BA%BFkx-y%2B4%3D0%E5%AF%B9%E7%A7%B0%EF%BC%882%EF%BC%89OP%E5%9E%82%E7%9B%B4OQ%E6%9B%B2%E7%BA%BFx%5E2%2By%5E2%2Bx-6y%2B3%3D0%E4%B8%8A%E4%B8%A4%E7%82%B9P%2CQ%E6%BB%A1%E8%B6%B3%EF%BC%9A%EF%BC%881%EF%BC%89%E5%85%B3%E4%BA%8E%E7%9B%B4%E7%BA%BFkx-y%2B4%3D0%E5%AF%B9%E7%A7%B0%EF%BC%882%EF%BC%89O%E4%B8%BA%E5%8E%9F%E7%82%B9%2COP%E5%9E%82%E7%9B%B4OQ%E6%B1%82%E7%9B%B4%E7%BA%BFPQ%E7%9A%84%E6%96%B9%E7%A8%8B.%E6%B3%A8%E6%84%8F%EF%BC%9AO%E4%B8%BA%E5%8E%9F%E7%82%B9)
曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)OP垂直OQ曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)O为原点,OP垂直OQ求直线PQ的方程.注意:O为原点
曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)OP垂直OQ
曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)O为原点,OP垂直OQ
求直线PQ的方程.
注意:O为原点
曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)OP垂直OQ曲线x^2+y^2+x-6y+3=0上两点P,Q满足:(1)关于直线kx-y+4=0对称(2)O为原点,OP垂直OQ求直线PQ的方程.注意:O为原点
曲线x^2+y^2+x-6y+3=0为圆,标准方程为:(x+1/2)^2+(y-3)^2=25/4
圆心(-1/2,3)
半径5/2
直线kx-y+4=0过点(0,4),则这点到点P和点Q的距离相等;
另外,圆心到点P与点Q的距离也相等,所以点(0,4)与圆心的连线是PQ的垂直平分线,斜率为2,就是直线kx-y+4=0,则
2x-y+4=0
那么PQ直线的斜率就是-1/2,设方程为y=-x/2+m,代入圆的方程得:
x^2+(-x/2+m)^2+x-6(-x/2+m)+3=0
x^2+x^2/4-mx+m^2+x+3x-6m+3=0
5x^2/4-mx+4x+m^2-6m+3=0
x1+x2=4(4-m)/5
x1*x2=4(m^2-6m+3)/5
再根据PQ中点到O的距离与到点P点Q距离相等可解出x1,x2来.