Question

已知双曲线c:x^2/a^2-y^2/b^2=1(a>b>0),以右焦点F为圆心,|OF|为半径的圆交双曲线两渐近线于点

Answer 1

设M坐标是(xo,b/axo.),N 坐标是(xo,-b/axo)

MN=2b/axo=2根号3a

故有xo=根号3a^2/b

又有xo^2+(b/axo)^2=OF^2=c^2

即有3a^4/b^2+3a^2=c^2

3a^2/(c^2-a^2)+3=e^2

3/(e^2-1)+3=e^2

3+3e^2-3=e^4-e^2

e^4-4e^2=0

e^2=4

e=2