Question

以P(2,2)为圆心的圆与椭圆x2+2y2=m交于A B两点,求AB中点M的轨迹方程.

Answer 1

P(2,2)

M(x,y)

k(AB)=-1/k(PM)

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

xA+xB=2*2=4,yA+yB=2*2=4

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

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

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

4+2*4*(2-x)/(y-2)=0

2x-y-2=0