Question

43.9．椭圆x2/a2+y2/b2=1(a>b>0)的两个焦点是F1、F2,以|F1F2|为边作正三角...

Answer 1

左焦点连右边的交点,勾股定理(a+ex)^2+(a-ex)^2 =(2c)^2

x=0.5c

4a^4+c^4=8a^2c^2 同时除以a^4

e^4+4=8e^2

e^2=4-2根3

e=根3-1