求过Q(8,2)的直线被椭圆截得的弦中点的轨迹方程请用韦达定理,要完整的过程,

Question

求过Q(8,2)的直线被椭圆截得的弦中点的轨迹方程
请用韦达定理,要完整的过程,

Answer 1

椭圆:x^2/a^2+y^2/b^2=1
弦AB中点M(x,y)
k(AB)=k(QM),xA+xB=2xM=2x,yA+yB=2y
(yA-yB)/(xA-xB)=(y-2)/(x-8)
[(xA)^2/a^2+(yA)^2/b^2]-[(xB)^2/a^2+(yB)^2/b^2]=1-1
b^2*(xA+xB)*(xA-xB)+a^2*(yA+yB)*(yA-yB)=0
b^2*2x+a^2*2y*[(yA-yB)/(xA-xB)]=0
b^2*x+a^2*y*(y-2)/(x-8)=0
b^2*(x^2-8x)+a^2*(y^2-2y)=0
椭圆:(x-4)^2/[(a^2+16b^2)/b^2]+(y-1)^2/[(a^2+16b^2)/a^2]=1