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(A) The given equation represents an ellipse with foci $F_1(3, 2)$ and $F_2(5, 4)$ and major axis length $2a = 6$,so $a = 3$.The distance between the

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$9\sqrt{2}$

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(A) The given equation represents an ellipse with foci $F_1(3, 2)$ and $F_2(5, 4)$ and major axis length $2a = 6$,so $a = 3$.The distance between the foci is $2ae = \sqrt{(5-3)^2 + (4-2)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.Thus,$ae = \sqrt{2}$,which gives the eccentricity $e = \frac{\sqrt{2}}{3}$.The distance between the directrices of an ellipse is given by $\frac{2a}{e}$.Substituting the values,we get $\frac{2(3)}{\frac{\sqrt{2}}{3}} = \frac{6 	imes 3}{\sqrt{2}} = \frac{18}{\sqrt{2}} = 9\sqrt{2}$.

The distance between the directrices of the conic $\sqrt{(x - 5)^2 + (y - 4)^2} + \sqrt{(x - 3)^2 + (y - 2)^2} = 6$ is

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