设交点为(x1,y1),(x2,y2)
|AB|=√((x2-x1)²+(y2-y1)²) 将y2=kx2+1,y1=kx1+1 代入得
|AB|=√((x2-x1)²+(kx2-kx1)²)
=√(1+k²)|x2-x1|
将直线l:y=kx+1代入双曲线C:3x^2-y^2=1 得
3x²-(kx+1)²=1
整理得 3x²-k²x²-2kx-2=0
两根之差的绝对值为
|x2-x1|=√((x1+x2)²-4x1x2)=√(2k/(3-k²))²+8/(3-k²))
=√(2k+8(3-k²)/|3-k²|
=√(2k+24-8k²)/|3-k²|
|AB|=√(1+k²)*√(2k+24-8k²)/|3-k²|