很簡單,只要將題目的條件都轉(zhuǎn)化為代數(shù)式然后進(jìn)化化簡即得結(jié)果
設(shè)橢圓的長半軸是a1,雙曲線的實半軸是a2,它們的半焦距是c
并設(shè)PF1=m,PF2=n,m>n,根據(jù)橢圓的和雙曲線的定義可得
m+n=2a1
m-n=2a2
解得
m=a1+a2,n=a1-a2
又PF1⊥PF2,由勾股定理得
PF1²+PF2²=F1F2²
(a1+a2)²+(a1-a2)²=(2c)²
化簡可得
a1²+a2²=2c²
離心率e1=c/a1,e2=c/a2
(e1²+e2²)/(e1e2)²
=[(c/a1)²+(c/a2)²]/[(c/a1)(c/a2)]²
=[(c²/a1²)+(c²/a2)²]/[c²/(a1a2)]²
=[c²(a1²+a2²)/(a1a2)²]/[c⁴/(a1a2)²]
=c²×2c²/c⁴
=2