Given hyperbola : $${{{x^2}} \over {6/k}} - {{{y^2}} \over 6} = 1$$

Eccentricity $$ = e = \sqrt {1 + {6 \over {6/k}}} = \sqrt {1 + k} $$

Directrices : $$x = \, \pm \,{a \over e} \Rightarrow x = \, \pm \,{{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }}$$

As given : $${{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }} = 1$$

$$ \Rightarrow k = 2$$

Here hyperbola is $${{{x^2}} \over 3} - {{{y^2}} \over 6} = 1$$

Checking the option gives $$\left( {\sqrt 5 , - 2} \right)$$ satisfies it.