Question

设F1,F2分别是椭圆x^2/a^2+y^2/b^2=1【a>b>0】的左右焦点,p是其右准线上纵坐标为根号3c【c为半焦径】的点,

且|F1F2|=|F2p|,则椭圆的离心率为多少

Answer 1

解：F1(-c,0),F2(c,0),右准线x=a^2/c,P(a^2/c,√3c)
|PF2|^2=[a^2/c-c]^2+(√3c)^2=|F1F2|^2=(2c)^2
所以a^2/c-c=c，a^2=2c^2，e^2=c^2/a^2=1/2，e=√2/2

追问

a^2/c-c=c　请问那步是怎么算的...

追答

[a^2/c-c]^2+(√3c)^2=(2c)^2,[a^2/c-c]^2+3c^2=4c^2,[a^2/c-c]^2=c^2,a^2/c-c=c