The focal chord to $y^2=16x$ is tangent to $(x-6{)}^2+y^2=2$ then the possible values of the slope of this chord are

The correct option is A $-1and1$
$\begin{array}{c}Wehave\\ y=4ax\left(a,0\right)\\ y-0=m(x-4)\left(4,0\right)\\ \Rightarrow y=mx-4m\\ \Rightarrow y-mx+4m=0\\ Now,\\ {\left(x-6\right)}^2+y^2=2\left(6.0\right)r=\sqrt{2}\\ Then\\ \left|\frac{y_1-3x_1+4m}{\sqrt{1+m^2}}\right|=\sqrt{2}\\ \Rightarrow \frac{{\left|0-6m+4m\right|}^2}{{\left(\sqrt{1+m^2}\right)}^2}\\ ={\left(\sqrt{2}\right)}^2\\ Now\\ 4m^2=2\left(1+m^2\right)\\ \Rightarrow 2m^2=2\\ \Rightarrow m^2=1\\ \therefore m=\pm 1\end{array}$

Hence,option $A$is the correct answer.