Question

椭圆题!已知F(c,0)是椭圆C:x2/a2+y2/b2=1(a大于b大于0)的右焦点,B为椭圆C与y轴的正半轴的交点,

Answer 1

题意可知,

BF=EF=DF＝2c

所以,三角形BDE是直角三角形

BE^2=5C^2,BD^2=DE^2-BE^2=11C^2

DB^2=a^2+b^2

a^2+b^2=11C^2

2a^2-c^2=11c^2

e^2=1/6

e=√6/6

直线AB与圆F相切