If the equation of pair of asymptote of a hyperbola H is $x y - 2 x - y + 2 = 0$ . Its conjugate hyperbola $H^{'}$ passes through (3,0) then

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Eccentricity of $H^{'}$ is $\sqrt{2}$

focus of $H^{'}$ is $(2 + \frac{1}{\sqrt{2}}, 2 + \frac{1}{\sqrt{2}})$

focus of H is $(\sqrt{2}, - \sqrt{2})$

equation of transverse axis of H is $x - y + 1 = 0$

answer is A, D.

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Detailed Solution

Equation of hyperbola is $x y - y - 2 x - 2 = 0$
$\Rightarrow (x - 1) (y - 2) = 4$
$X Y = 4$
Conjugate hyperbola XY = -4
$\frac{a^{2}}{2} = 4 \Rightarrow a = 2 \sqrt{2}$
Focus of with respect to X-Y system = $(- 2 \sqrt{2}, 2 \sqrt{2}) a n d (2 \sqrt{2}, - 2 \sqrt{2})$
So focus in x-y system $(1 - 2 \sqrt{2}, 2 + 2 \sqrt{2}) a n d (1 + 2 \sqrt{2}, 2 - 2 \sqrt{2})$
Equation of transverse axis Y= X
$y - 2 = x - 1 \Rightarrow x - y + 1 = 0$