If the line x - 1=0 is a directrix of the hyperbola kx²-y²=6. Then the hyperbola passes through the point

$\frac{{\mathrm{x}}^2}{6/\mathrm{k}}-\frac{{\mathrm{y}}^2}{6}=1....\left(1\right)$
take ${\mathrm{e}}^2=1+{\mathrm{b}}^2/{\mathrm{a}}^2=1+\frac{6}{{\displaystyle \frac{6}{k}}}$
$\mathrm{e}=\sqrt{1+\mathrm{k}};\mathrm{a}=\sqrt{\frac{6}{\mathrm{k}}}$
Equation of directrix $\mathrm{x} = \frac{\mathrm{a}}{\mathrm{e}}$
$\Rightarrow \mathrm{x}=\sqrt{\frac{6}{\mathrm{k}(\mathrm{k}+1)}}\Rightarrow \frac{6}{\mathrm{k}(\mathrm{k}+1)}=1\Rightarrow \mathrm{k}=2$
from(1) we get $2{\mathrm{x}}^2-{\mathrm{y}}^2=6$
by verification 3rd option $(\sqrt{5},-2)$ lies on hyperbola.