Question

过(0,1)的直线与3x^2-y^2=1,交与A,B两点,求直线斜率的范围

Answer 1

设直线为y=kx+1,代入3x^2-y^2=1中消去y得：（3-k²）x²-2kx-2=0.

△=4k²+8(3-k²)=24-4k²

1.若直线与双曲线有两个交点,必有△＞0,即24-4k²＞0.解得：-√6＜k＜√6.

2.若直线与双曲线有一个交点,则有△=0,即24-4k²=0.解得：k=+-√6.