椭圆 x^2/a^2+y^2/b^2=1 的右顶点A到左右两个焦点F1,F2距离分别为8和2,求椭圆方程设动点P满足PF2^2-PA^2=4,求动点P的轨迹方程
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1AF1=8,AF2=2
AF1-AF2=F1F2=2c=8-2
c=3OF1=3AF1-OF1=a=8-3=5
b^2=a^2-c^2=4^2
x^2/5^2+y^2/4^2=1
2A(5,0) F2(3,0)
[(x-3)^2+y^2 ]- [(x-5)^2+y^2]=4
(x-3)^2-(x-5)^2=4
4x=20x=5