Focal chord of  $y^2=16x$ is a tangent to  ${(x-6)}^2+y^2=2$  then possible values of the slopes are

Equation of the focal chord having slope ‘m’ is
$y-0=m(x-4)\Rightarrow mx-y-4m=0$
$\Rightarrow$it is tangent to circle  ${(x-6)}^2+y^2=2$
$\Rightarrow \left|\frac{6m-0-4m}{\sqrt{m^2+1}}\right|=\sqrt{2}\Rightarrow m^2=1\Rightarrow m=\pm 1$