设P(x,y)
则PF1=Ia+exI
PF2=Ia-exI
已知PF1/PF2=e
所以Ia+exI=e*Ia-exI
(1) a+ex=e(a-ex) x=a(e-1)/(e+e^2)
因-a≤x≤a 则-a≤a(e-1)/(e+e^2)≤a
解得e≥√2-1
(2) a+ex=e(ex-a) x=a(1+e)/(e^2-1)
因-a≤x≤a 则-a≤a(1+e)/(e^2-1)≤a
解得0≤e≤√2+1
综上√2-1≤e≤√2+1
设P(x,y)
则PF1=Ia+exI
PF2=Ia-exI
已知PF1/PF2=e
所以Ia+exI=e*Ia-exI
(1) a+ex=e(a-ex) x=a(e-1)/(e+e^2)
因-a≤x≤a 则-a≤a(e-1)/(e+e^2)≤a
解得e≥√2-1
(2) a+ex=e(ex-a) x=a(1+e)/(e^2-1)
因-a≤x≤a 则-a≤a(1+e)/(e^2-1)≤a
解得0≤e≤√2+1
综上√2-1≤e≤√2+1