Question

一个消参求轨迹方程的问题一个椭圆方程x^2+(y/2)^2=1,C(0,1)过的直线l与椭圆交于A,B.Q为AB的中点,

Answer 1

Q(x,y)

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

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

x^2+(y/2)^2=1

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

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

(1)-(2):

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

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

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

4x^2+y^2-y=0

Q的轨迹方程是椭圆：x^2/0.25^2+(y-1)^2/0.5^2=1