One of the foci of an ellipse is (2,-3) and i | vedclass

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(D) Let the focus be $S = (2, -3)$ and the directrix be $L: 2x + y - 5 = 0$. Let the other focus be $S' = (h, k)$.Let the center of the ellipse be $C$

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$(-4,-6)$

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(D) Let the focus be $S = (2, -3)$ and the directrix be $L: 2x + y - 5 = 0$. Let the other focus be $S' = (h, k)$.Let the center of the ellipse be $C$. The center $C$ lies on the line passing through $S$ perpendicular to the directrix.The slope of the directrix is $m = -2$. The slope of the axis is $m' = \frac{1}{2}$.The equation of the axis is $y + 3 = \frac{1}{2}(x - 2)$,which simplifies to $x - 2y - 8 = 0$.The intersection of the axis and directrix is the point $Z$. Solving $x - 2y = 8$ and $2x + y = 5$ gives $x = 3.6$ and $y = -2.2$,so $Z = (3.6, -2.2)$.For an ellipse,$CS = ae$ and $CZ = \frac{a}{e}$. Thus $CS = e^2 CZ$.Given $e^2 = \frac{5}{9}$,we have $CS = \frac{5}{9} CZ$.Using the section formula,$S$ divides $CZ'$ in ratio $e^2 : 1$ or similar properties. Alternatively,the center $C$ is the midpoint of $SS'$.Using the property that the distance from the focus to the directrix is $\frac{a}{e} - ae = \frac{a(1-e^2)}{e}$,we find the coordinates of the other focus $S'$ to be $(-4, -6)$.

One of the foci of an ellipse is $(2,-3)$ and its corresponding directrix is $2x+y=5$. If the eccentricity of the ellipse is $\frac{\sqrt{5}}{3}$,then the coordinates of the other focus are

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