If the line x-1=0 is a directrix of the hyper | vedclass

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(C) The given equation of the hyperbola is $kx^{2}-y^{2}=6$,which can be written as $\frac{x^{2}}{6/k} - \frac{y^{2}}{6} = 1$.Here,$a^{2} = \frac{6}{k

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$(\sqrt{5}, -2)$

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(C) The given equation of the hyperbola is $kx^{2}-y^{2}=6$,which can be written as $\frac{x^{2}}{6/k} - \frac{y^{2}}{6} = 1$.Here,$a^{2} = \frac{6}{k}$ and $b^{2} = 6$.The eccentricity $e$ is given by $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{6}{6/k} = 1 + k$.Thus,$e = \sqrt{1+k}$.The equation of the directrix for a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $x = \frac{a}{e}$.Given the directrix is $x = 1$,we have $1 = \frac{\sqrt{6/k}}{\sqrt{1+k}}$.Squaring both sides,$1 = \frac{6}{k(1+k)} \Rightarrow k^{2} + k - 6 = 0$.Solving the quadratic equation $(k+3)(k-2) = 0$,we get $k = 2$ (since $k$ must be positive for a hyperbola).The equation of the hyperbola is $2x^{2} - y^{2} = 6$.Checking the options: For option $C$,$2(\sqrt{5})^{2} - (-2)^{2} = 2(5) - 4 = 10 - 4 = 6$. Thus,the point $(\sqrt{5}, -2)$ lies on the hyperbola.

If the line $x-1=0$ is a directrix of the hyperbola $kx^{2}-y^{2}=6$,then the hyperbola passes through which of the following points?

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