Official solution
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.