Question

已知椭圆C的方称为x^2+y^2/2=1,过点P(1/2,1)的直线l交椭圆于A、B两点,求中点Q的轨迹方程

Answer 1

Q(x,y)

xA+xB=2x,yA+yB=2y

k(AB)=k(PQ)

(yA-yB)/(xA-xB)=(y-1)/(x-1/2)

[(xA)^2+(yA)^2/2]-[(xB)^2+(yB)^2/2]=1-1

2(xA+xB)*(xA-xB)+(yA+yB)*(yA-yB)=0

2*2x+2y*(yA+yB)/(xA-xB)=0

2x+y*(y-1)/(x-1/2)=0

Q:(x-1/4)^2/(3/16)+(y-1/2)^2/(3/8)=1