Question

已知直线l和圆M：x²+y²+2x=0相切于点T,且与双曲线C：x²－y²=1相

Answer 1

设A(x1,y1),B(x2,y2)

则：T(x3,y3)满足：2x3=x1+x2,2y3=y1+y2

x1^2-y1^2=1,x2^2-y2^2=1

(x1^2-x2^2)-(y1^2-y2^2)=0

(x1-x2)(x1+x2)=(y1-y2)(y1+y2)

Kab=(y1-y2)/(x1-x2)=(x1+x2)/(y1+y2)=x3/y3

Kto=y3/(x3+1)

因为：AB⊥OT

所以,Kab*Kto=x3/(x3+1)=-1

x3=-1/2

代人x^2+y^2+2x=0得：1/4+y3^2-1=0,y3=±√3/2

Kab=x3/y3=±√3/3

所以,直线L的方程是

y+√3/2=√3/3*(x+1/2),即：√3y-x+1=0

或,y-√3/2=-√3/3*(x+1/2),即：√3y+x-1=0