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)