Question

过椭圆x^2/9+y^2/4=1内一点D(1,0)引动弦AB,求AB中点M的轨迹方程

Answer 1

AB中点M(x,y)

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

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

x^2/9+y^2/4=1

4x^2+9y^2=36

4(xA)^2+9(yA)^2=36.(1)

4(xB)^2+9(yB)^2=36.(2)

(1)-(2):

4(xA+xB)*(xA-xB)+9(yA+yB)*(yA-yB)=0

4(xA+xB)+9(yA+yB)*(yA-yB))/(xA-xB)=0

4*2x+9*2y*y/(x-1)=0

AB中点M的轨迹方程:4(x-0.5)^2+9y^2=1