Find the equation of the ellipse whose focus | vedclass

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(B) Let $P(x, y)$ be any point on the ellipse.Given focus $S = (-1, 1)$ and directrix $L: x - y + 3 = 0$.Let $PM$ be the perpendicular distance from $

Vedclass · Accepted Answer

$7x^{2} + 7y^{2} + 2xy + 10x - 10y + 7 = 0$

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(B) Let $P(x, y)$ be any point on the ellipse.Given focus $S = (-1, 1)$ and directrix $L: x - y + 3 = 0$.Let $PM$ be the perpendicular distance from $P(x, y)$ to the directrix $x - y + 3 = 0$.By the definition of an ellipse,$SP = e \cdot PM$,where $e = 1/2$.Squaring both sides,we get $SP^{2} = e^{2} \cdot PM^{2}$.$(x + 1)^{2} + (y - 1)^{2} = (1/2)^{2} \cdot \left( \frac{x - y + 3}{\sqrt{1^{2} + (-1)^{2}}} \right)^{2}$$(x^{2} + 2x + 1 + y^{2} - 2y + 1) = \frac{1}{4} \cdot \frac{(x - y + 3)^{2}}{2}$$8(x^{2} + y^{2} + 2x - 2y + 2) = (x - y + 3)^{2}$$8x^{2} + 8y^{2} + 16x - 16y + 16 = x^{2} + y^{2} + 9 - 2xy + 6x - 6y$$7x^{2} + 7y^{2} + 2xy + 10x - 10y + 7 = 0$Thus,the equation of the ellipse is $7x^{2} + 7y^{2} + 2xy + 10x - 10y + 7 = 0$.

Find the equation of the ellipse whose focus is $(-1, 1)$,eccentricity is $1/2$,and directrix is $x - y + 3 = 0$.

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