"If the line $x-1=0$ is a directrix of the hyperbola $k x^{2}-y^{2}=6$, then the hyperbola passes through the point :"

Question

Accepted Answer

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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.

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If the line $x-1=0$ is a directrix of the hyperbola $k x^{2}-y^{2}=6$, then the hyperbola passes through the point :