Question

已知椭圆x²/a²+y²/b²=1(a＞b＞0)的离心率=二分之根号三,连接椭圆

Answer 1

令F1M=m,F2M=n ,焦距为c

由题意:m+n=2a

4c^2=4a^2-4b^2=m^2+n^2-2mncosΦ=4a^2-2mn-2mncosΦ

所以mn=2b^2/(1+cosΦ)

S△F1MF2=mnsinΦ/2=b^2*sinΦ/(1+cosΦ)=b^2*tan(φ/2)