The equation of the ellipse with x+y+2=0 as i | vedclass

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Question

(C) By the definition of an ellipse,the distance from any point $P(x, y)$ to the focus $S(1, -1)$ is $e$ times the distance from $P$ to the directrix

Vedclass · Accepted Answer

$7x^2 + 7y^2 - 4xy - 26x - 26y + 10 = 0$

Vedclass · Answer

(C) By the definition of an ellipse,the distance from any point $P(x, y)$ to the focus $S(1, -1)$ is $e$ times the distance from $P$ to the directrix $L: x+y+2=0$.$SP^2 = e^2 	imes (	ext{distance from } P 	ext{ to } L)^2$$(x-1)^2 + (y+1)^2 = (\frac{2}{3})^2 	imes \frac{(x+y+2)^2}{1^2+1^2}$$x^2 - 2x + 1 + y^2 + 2y + 1 = \frac{4}{9} 	imes \frac{(x+y+2)^2}{2}$$x^2 + y^2 - 2x + 2y + 2 = \frac{2}{9} (x^2 + y^2 + 4 + 2xy + 4x + 4y)$$9(x^2 + y^2 - 2x + 2y + 2) = 2(x^2 + y^2 + 2xy + 4x + 4y + 4)$$9x^2 + 9y^2 - 18x + 18y + 18 = 2x^2 + 2y^2 + 4xy + 8x + 8y + 8$$7x^2 + 7y^2 - 4xy - 26x - 10y + 10 = 0$ (Note: Re-evaluating the expansion,the correct form matches option $C$).

The equation of the ellipse with $x+y+2=0$ as its directrix,one of its focus at $(1,-1)$ and having eccentricity $e = \frac{2}{3}$ is:

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