Question

点A(0,-1)作圆心直线y=kx+1与双曲线x²/3-y²=1交于QP两点,QP为圆上两点求K

Answer 1

点A(0,-1)
①y=kx+1
②x²/3-y²=1
解①，②交点：
x²/3-(kx+1)²=1
x(x/3-k²x-2k)=0
x1=0；y1=1

x2=6k/(3k²-1) ；y=6k²/(3k²-1) +1
A点到P，Q两点跑离相等：

[6k/(3k²-1)]²+[6k²/(3k²-1) +2]²=1+1
[6k/(3k²-1)]²+[6k²/(3k²-1) +2]²=2
36k²+36k⁴+2(3k²-1)²+24k²=0
后面自己解。