If (1, -2) is the focus and x+y-2=0 is the di | vedclass

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(C) The definition of an ellipse is the locus of a point $P(x, y)$ such that the distance from the focus $S(1, -2)$ is $e$ times the distance from the

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$\frac{1}{3}$

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(C) The definition of an ellipse is the locus of a point $P(x, y)$ such that the distance from the focus $S(1, -2)$ is $e$ times the distance from the directrix $x+y-2=0$. $(x-1)^2 + (y+2)^2 = e^2 \frac{(x+y-2)^2}{1^2+1^2}$ $2(x^2 - 2x + 1 + y^2 + 4y + 4) = e^2(x^2 + y^2 + 4 + 2xy - 4x - 4y)$ $(2-e^2)x^2 - 2e^2xy + (2-e^2)y^2 + (4e^2-4)x + (8+4e^2)y + (10-4e^2) = 0$ Comparing this with the given equation $17x^2 - 2xy + 17y^2 - 32x + 76y + 86 = 0$: $\frac{2-e^2}{17} = \frac{-2e^2}{-2} = \frac{4e^2-4}{-32}$ From $\frac{2-e^2}{17} = e^2$: $2 - e^2 = 17e^2$ $\Rightarrow 18e^2 = 2$ $\Rightarrow e^2 = \frac{1}{9}$ $\Rightarrow e = \frac{1}{3}$.

If $(1, -2)$ is the focus and $x+y-2=0$ is the directrix of the ellipse $17x^2 - 2xy + 17y^2 - 32x + 76y + 86 = 0$,then its eccentricity is

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