作业帮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
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![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](/uploads/image/z/5285107-19-7.jpg?t=A%2CB%E6%98%AFy%3D%5B%282sqr5%29%2F5%5Dx%E5%92%8Cy%3D+%5B-%282sqr5%29%2F5%5Dx%E4%B8%8A%E5%8A%A8%E7%82%B9%2Cabs%28AB%29%3D2sqr5%2Co%E4%B8%BA%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%2C%E5%90%91%E9%87%8F%EF%BC%88op%3Doa%2Bob%EF%BC%89%2C%E6%B1%82p%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9)
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
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
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
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
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