Question

已知双曲线x2-y2=1 动直线条y=kx+m与圆x2+y2=1相切,且与双曲线左,右两支的交点分别为P1(x1,y1)

Answer 1

直线y=kx+m①与圆x^2+y^2=1相切,

所以|m|/√(k^2+1)=1,

所以m^2=k^2+1.

把①代入x^2-y^2=1得x^2-(k^2x^2+2kmx+m^2)=1,

(1-k^2)x^2-2kmx-m^2-1=0,

△=4k^2m^2+4(1-k^2)(m^2+1)=4(m^2+1-k^2)=8,

左右交点P1(x1,y1),P2(x2,y2)满足x1