The equation of the conic with focus at (1, - | vedclass

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(C) Let $P(x, y)$ be any point on the conic.By the definition of a conic,the distance from the focus $S(1, -1)$ to $P(x, y)$ is $SP = e \cdot PM$,wher

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $P(x, y)$ be any point on the conic.By the definition of a conic,the distance from the focus $S(1, -1)$ to $P(x, y)$ is $SP = e \cdot PM$,where $PM$ is the perpendicular distance from $P$ to the directrix $x - y + 1 = 0$.$SP = \sqrt{(x - 1)^2 + (y + 1)^2}$$PM = \frac{|x - y + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 1|}{\sqrt{2}}$Given $e = \sqrt{2}$,we have $SP^2 = e^2 \cdot PM^2$.$(x - 1)^2 + (y + 1)^2 = 2 \cdot \frac{(x - y + 1)^2}{2}$$(x^2 - 2x + 1) + (y^2 + 2y + 1) = (x - y + 1)^2$$x^2 + y^2 - 2x + 2y + 2 = x^2 + y^2 + 1 - 2xy + 2x - 2y$Rearranging the terms:$2xy - 4x + 4y + 1 = 0$.

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

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