Question

简单圆锥曲线题已知A（-2,0）,B（2,0）,点C、D满足|AC|=2,向量AD=0.5（向量AB+向量AC）,求点D

Answer 1

c:(x+2)^2+y^2=4

作图,知d为bc中点

c(x,y),d(m,n),b(2,0)

m=(x+2)/2

n=y/2

x=2m-2

y=2n

代入c,得m^2+n^2=1

即x^2 +y^2 =1

附别人的做法

向量|AC|=2 ==> C为以点A为圆心、半径=2的圆

点C(-2+cosT,sinT)

向量AC =(2*cosT,2*sinT),向量AB =(4,0)

向量AD =(向量AB+向量AC)/2 = (2+cosT,sinT)

点D =(cosT,sinT)

点D的轨迹方程:x^2 +y^2 =1 ...(1)