If the line x - 1 = 0 , is a directrix of the hyperbola k x ^ 2 - y...

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

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

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

i.e. $k \ge - 1$

$⇒ x = \pm \dfrac{\sqrt{6}}{\sqrt{k}\left(\sqrt{1 + k}\right)}$$is the equation of directrix</p><p>i.e.$\pm \dfrac{\sqrt{6}}{\sqrt{k}\left(\sqrt{1 + k}\right)} = 1$</p><p>$∴ {\left(\sqrt{k}\sqrt{1 + k}\right)}^{2} = 6 ⇒ {k}^{2} + k - 6 = 0$</p><p>$⇒ k = 2$as$k = - 3$is rejected</p><p>So equation of hyperbola will be$H ≡ \dfrac{{x}^{2}}{3} - \dfrac{{y}^{2}}{6} = 1$</p><p>$∴ \left(\sqrt{5}, - 2\right) satisfy the given hyperbola