直线Y=k(X-√2)与曲线X^2-Y^2=1(X>0),相交与A、B两点,求直线的倾斜角范围

Question

直线Y=k(X-√2)与曲线X^2-Y^2=1(X>0),相交与A、B两点,求直线的倾斜角范围

Answer 1

y=kx-√2k
x^2-k^2x^2+2√2k^2x-2k^2=1
(1-k^2)x^2+2√2k^2x-(2k^2+1)=0
x>0
有两个交点即方程有两个正跟
x1+x2=-2√2k^2/(1-k^2)>0
x1*x2=-(2k^2+1)/(1-k^2)>0
-2√2k^2/(1-k^2)>0
k^2/(k^2-1)>0
k^2(k^2-1)>0
k^2>1,k^21
-(2k^2+1)/(1-k^2)>0
(2k^2+1)/(k^2-1)>0
(2k^2+1)(k^2-1)>0
k^2>1
所以k>1,k