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

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

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x y = 1

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

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

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Solution

The correct option is C $2 x y - 4 x + 4 y + 1 = 0$
Using Definition of hyperbola
$P S^{2} = e^{2} \cdot P M^{2}$
$(x - 1)^{2} + (y + 1)^{2} = 2 {(\frac{x - y + 1}{\sqrt{2}})}^{2}$
$(x^{2} + y^{2} - 2 x + 2 y + 2) = (x^{2} + y^{2} + 1 - 2 x y + 2 x - 2 y)$
$\Rightarrow 2 x y - 4 x + 4 y + 1 = 0$
Hence, option 'C' is correct.

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Similar questions

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

\sqrt{2}

(0, 0)

x - y + 1 = 0

The equation of the parabola whose focus is the point (0, 0) and the tangent at the vertex is x-y+1=0 is

(1, - 1)

x - y + 1 = 0

\sqrt{2}

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