The equation of the ellipse whose focus is (6 | vedclass

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Question

(B) By the definition of an ellipse,the distance from any point $P(x, y)$ on the ellipse to the focus $S(6, 7)$ is $e$ times the distance from $P$ to

Vedclass · Accepted Answer

$5x^2 - 2xy + 5y^2 - 76x - 88y + 506 = 0$

Vedclass · Answer

(B) By the definition of an ellipse,the distance from any point $P(x, y)$ on the ellipse to the focus $S(6, 7)$ is $e$ times the distance from $P$ to the directrix $L: x + y + 2 = 0$.$PS^2 = e^2 	imes (	ext{distance from } P 	ext{ to } L)^2$$(x - 6)^2 + (y - 7)^2 = (1/\sqrt{3})^2 	imes \frac{(x + y + 2)^2}{1^2 + 1^2}$$(x^2 - 12x + 36 + y^2 - 14y + 49) = \frac{1}{3} 	imes \frac{(x + y + 2)^2}{2}$$6(x^2 + y^2 - 12x - 14y + 85) = (x + y + 2)^2$$6x^2 + 6y^2 - 72x - 84y + 510 = x^2 + y^2 + 4 + 2xy + 4x + 4y$$5x^2 - 2xy + 5y^2 - 76x - 88y + 506 = 0$Thus,the correct option is $B$.

The equation of the ellipse whose focus is $(6, 7)$,directrix is $x + y + 2 = 0$,and eccentricity $e = 1/\sqrt{3}$ is:

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