x2

Aortic second sound

+

y2

b2

= 1(a>b>0)，|PF 1|=m，|PF2|=n。

In △PF 1F2, according to cosine theorem, 4c2 = M2+N2-2mncos 60.

∵m+n=2a,∴m2+n2=(m+n)2-2mn=4a2-2mn，

∴ 4c2 = 4a2-3mn。 That is 3mn = 4a2-4c2.

Mn≤(

m+n

2

) 2=a2 (take equal sign if and only if m=n),

∴4a2-4c2≤3a2,∴

c2

Aortic second sound

≥

1

four

E≥

1

2

.

The value range of ∴e is

1

2

, 1).

So the answer is [

1

2

, 1)