直线Y=KX+1与双曲线C:2x^2-y^2=1的右支交于不同的两点A,B,若双曲线C的右焦点F在以AB为直径的圆上,求K

a= √2/2,b=1,c=√6/2,
双曲线C的右焦点F在以AB为直径的圆上,
则AF⊥BF,
设A（x1,y1),B(x2,y2),
F(c,0),
向量AF=（x1-c,-y1),
向量BF=（x2-c,-y2),
∵向量AF⊥BF
∴向量AF·BF=x1x2-c(x1+x2)+c^2+y1y2=0,
y1=kx1+1,
y2=kx2+1,
y1y2=k^2x1x2+k(x1+x2)+1,
x1x2-c(x1+x2)+c^2+k^2x1x2+k(x1+x2)+1=0,
x1x2(1+k^2)+(x1+x2)(k-c)+1+c^2=0,
直线方程代入双曲线方程,
2x^2-(kx+1)^2=1,
(2-k^2)x^2-2kx-2=0,
根据韦达定理,
x1+x2=2k/(2-k^2),
x1x2=-2/(2-k^2),
(1+k^2)*[-2/(2-k^2)]+2k(k-c)/(2-k^2)+1-c^2=0,
-2-2k^2+2k^2-√6k+(1-3/2)(2-k*2)=0,
k^2-2√6k-6=0,
∴k=(√6±2√15)/2.