Question

问一道数学题目已知F1,F2为椭圆的焦点,过F2的直线交椭圆于P,Q两点,PF1垂直于PQ,且PF1的绝对值等于PQ的绝

Answer 1

设PF1=r,PF2=s,则因为PQ=PF1=r,所以QF1=r-s,有椭圆定义：QF1+QF2=PF1+PF2得QF1=2s

有勾股定理：PF1^2+PQ^2=QF1^2

2s^2=r^2 => r=（根号2）s

r+s=[1+（根号2）]s

F1F2^2=PF1^2+PF2^2 => F1F2^2=3s^2

F1F2=（根号3）s

e=F1F2/(r+s)=（根号3）/[1+（根号2）]