The equation of an ellipse whose focus is (-1 | vedclass

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Question

(A) 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

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

Vedclass · Answer

(A) 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 $L: x - y + 3 = 0$.$SP^2 = e^2 \cdot d(P, L)^2$$(x + 1)^2 + (y - 1)^2 = \left(\frac{1}{2}\right)^2 \cdot \frac{(x - y + 3)^2}{1^2 + (-1)^2}$$(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 + 2xy + 7y^2 + 10x - 10y + 7 = 0$.

The equation of an ellipse whose focus is $(-1, 1)$,whose directrix is $x - y + 3 = 0$,and whose eccentricity is $e = \frac{1}{2}$,is given by

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