椭圆x/9+y2=1上有动点P,F1、F2是椭圆的两个焦点,求△PF1F2的重心M的轨迹方程
人气:328 ℃ 时间:2019-08-18 20:03:55
解答
解设M(x,y),P(x',y')
由椭圆x/9+y2=1
知c^2=9-1=8
即c=2√2
故F1(2√2,0) F2(-2√2,0) P(x',y')
又由M是△PF1F2的重心
则x=(x'+2√2-2√2)/3
y=(y'+0-0)/3
即得x'=3x
y'=3y
由P(x',y')在
椭圆x^2/9+y2=1上
则x'^2/9+y'2=1上
故(3x)^2/9+(3y)2=1
故
△PF1F2的重心M的轨迹方程
为x^2+y2/(1/9)=1(y≠0).
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