Question

已知直线x+根号3y-m=0与圆x2+y2=1交于A,B两点,则向量OA+向量OB共线的向量为?

Answer 1

x^2+y^2=1
x=m-√3y，即：(m-√3y)^2+y^2=1
即：4y^2-2√3my+m^2-1=0
即：y1=(√3m+sqrt(4-m^2))/4，y2=(√3m-sqrt(4-m^2))/4
故：x1=m-√3y1，x2=m-√3y2
即AB中点坐标：x0=(x1+x2)/2=m-√3(y1+y2)/2
=m-3m/4=m/4
y0=(y1+y2)/2=√3m/4
即：AB中点坐标(m/4,√3m/4)
OA+OB=2(m/4,√3m/4)=(m/2,√3m/2)
与OA+OB共线的向量：k(m/2,√3m/2)
-2<m<2