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)

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