不论K取何值,直线y=k(x-2)+b与曲线x^2-y^2=1总有公共点,则实b的取值范围是?(求解法)

把y=k(x-2)+b代人x^2-y^2=1得：
x^2-[k(x-2)+b]^2=1
(1-k^2)x^2-2k(b-2k)x-(b-2k)^2-1=0
△=4k^2(b-2k)^2+4(1-k^2)[(b-2k)^2+1]
=4(1-k^2)+4(b-2k)^2
=4[3k^2-4bk+b^2+1]
=4[3(k-2b/3)^2+b^2+1-4b^2/3]
不论K取何值,△≥0
b^2+1-4b^2/3≥0
b^2/3≤1
b^2≤3
-√3≤b≤√3