数学题己知a=(x,0)b=(1,y)且(根号3a+b)垂直(根号3a-b)求点p(x,y)的轨迹C

√3*a+b=(√3*x+1,y),
√3*a-b)=(√3*x-1,0-y),
因为:(根号3a+b)垂直(根号3a-b),则有
(√3*X+1)(√3*X-1)+Y*(-Y)=0,
3X^2-Y^2=1.
则,点p(x,y)的轨迹C为:3X^2-Y^2=1.
2.直线l:y=kx-1与曲线C交于两点A,B并且A,B在Y轴的同一侧,则有,
⊿>0,
3X^2-Y^2=1,Y+KX-1,
X^2(3-K^2)+2KX-2=0.
(2K)^2-4(3-K^2)*(-2)>0,且K≠0,
K^2