Given equation of hyperbola is $\frac{x^2}{\frac{6}{k}}-\frac{y^2}{6}=1$

Equation of directrix will be $x=\pm \frac{a}{e}$,

Where $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{6}{{\displaystyle \frac{6}{k}}}}=\sqrt{1+k}$

i.e. $k\ge -1$

 $\Rightarrow x=\pm \frac{\sqrt{6}}{\sqrt{k}\left(\sqrt{1+k}\right)}$ is the equation of directrix

i.e. $\pm \frac{\sqrt{6}}{\sqrt{k}\left(\sqrt{1+k}\right)}=1$

$\therefore {\left(\sqrt{k}\sqrt{1+k}\right)}^2=6\Rightarrow k^2+k-6=0$

$\Rightarrow k=2$ as $k=-3$ is rejected

So equation of hyperbola will be $H\equiv \frac{x^2}{3}-\frac{y^2}{6}=1$

$\therefore \left(\sqrt{5},-2\right)$ satisfy the given hyperbola