A,B是y=[(2sqr5)/5]x和y= [-(2sqr5)/5]x上动点,abs(AB)=2sqr5,o为坐标原点,向量（op=oa+ob）,求p的轨迹方
问题描述：
A,B是y=[(2sqr5)/5]x和y= [-(2sqr5)/5]x上动点,abs(AB)=2sqr5,o为坐标原点,向量（op=oa+ob）,求p的轨迹方
答
设A(m,m(2*5^(1/2))/5),B(n,-n(2*5^(1/2))/5)
|AB|^2=(m-n)^2+(4/5)(m+n)^2=20-----------------(1)
OP=OA+OB=(m+n,(m-n)(2*5^(1/2))/5)
设P(x,y)
则:x=m+n
y=(m-n)(2*5^(1/2))/5,m-n=y(5^(1/2))/2
代入(1),得:
(5/4)y^2+(4/5)x^2=20
16x^2+25y^2=400
x^2/5^2+y^2/4^2=1