已知椭圆x^2/4+y^2=1设A.B.M是椭圆上的三点OM=mOA+nOB,m^2+n^2=1求线段AB中点轨迹

Question

已知椭圆x^2/4+y^2=1
设A.B.M是椭圆上的三点OM=mOA+nOB,m^2+n^2=1求线段AB中点轨迹

Answer 1

设A(x1,y1),B(x2,y2),M(x3,y3),AB中点P(xo,yo)
x1^2/4+y1^2=1
x2^2/4+y2^2=1
又有x3=mx1+nx2,y3=my1+ny2
x3^2/4+y3^2=1
(m^2x1^2+2mnx1x2+n^2x2^2)/4+(m^2y1^2+2mny1y2+n^2y2^2)=1
m^2(x1^2/4+y1^2)+n^2(x2^2/4+y2^2)+2mnx1x2/4+2mny1y2=1
m^2+n^2+2mn*(x1x2/2+y1y2)=1
2mn(x1x2/2+y1y2)=0
如:2mn=0
故有m^2+n^2=1,可得到m=0,n=1或m=1,n=0,与题目矛盾,故有:
x1x2/2+y1y2=0
即有(x1^2+x2^2+2x1x2)/4+(y1^2+y2^2+2y1y2)=2
即有(x1+x2)^2/4+(y1+y2)^2=2
即有(2xo)^2/4+(2yo)^2=2
即有xo^2+4yo^2=2
即中点的轨迹方程是x^2/2+y^2/(1/2)=1