xunguo
幼苗
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x^2+3y^2=3
a=√3,c=√2
F1(-√2,0),F2(√2,0)
延长AF1交椭圆A'
∵F1A=5F2B
由椭圆的对称性得|F1A|=5|F1A'|
设 AA':x=ty-√2代入x^2+3y^2=3
得:(ty-√2)^2+3y^2-3=0
(t^2+3)y^2-2√2ty-1=0
令A(X1,Y1),A'(X2,Y2)
则由韦达定理得:
y1+y2=2√2t/ (t^2+3)
y1y2=-1/ (t^2+3)
∵|F1A|=5|F1A'|
∴y1=-5y2
-4y2= 2√2t/ (t^2+3) (1)
5y2^2 =1/(t^2+3)==> (2)
(1)^2/(2):
2/5 =t^2/(t^2+3)
∴ t^2=2
y^2=1/25,y1^2=1,y1=±1,x1=0
∴ A(0,±1)
1年前
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