The equation of the conic with focus at (1,-1) directrix x-y+1=0 and eccentricity $\sqrt{2}$ is

$x y = 1$

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$2 x y + 4 y - 4 y - 1 = 0$

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$x^{2} - y^{2} = 1$

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$2 x y - 4 x + 4 y + 1 = 0$

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Solution

The correct option is D
$2 x y - 4 x + 4 y + 1 = 0$

$Let (x,y)be any point on the hyperbola Then,the distance of any point from the focus is eccentricity times the distance from the directrix.$

$∴ \sqrt{(x - 1)^{2} + (y + 1)^{2}} = \sqrt{2} ∣ ∣ ∣ \frac{x - y + 1}{\sqrt{2}} ∣ ∣ ∣$

$S q u a r i n g b o t h t h e s i d e s, w e g e t :$

$(x - 1)^{2} + (y + 1)^{2} = (x - y + 1)^{2}$

$x^{2} - 2 x + 1 + y^{2} + 1 + 2 y = x^{2} + y^{2} + 1 - 2 x y - 2 y + 2 x$

$2 x y - 4 x + 4 y + 1 = 0$

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The equation of the conic with focus at (1, –1), directrix along x – y + 1 = 0 and with eccentricity $\sqrt{2}$ is

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