If an ellipse with foci at (3,3) and (-4,4) i | vedclass

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(A) Let the foci be $S_1 = (3,3)$ and $S_2 = (-4,4)$,and the point on the ellipse be $P = (0,0)$.By the definition of an ellipse,the sum of the distan

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$\frac{5}{7}$

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(A) Let the foci be $S_1 = (3,3)$ and $S_2 = (-4,4)$,and the point on the ellipse be $P = (0,0)$.By the definition of an ellipse,the sum of the distances from any point on the ellipse to the two foci is constant and equal to the length of the major axis,$2a$.$PS_1 + PS_2 = 2a$$\sqrt{(3-0)^2 + (3-0)^2} + \sqrt{(-4-0)^2 + (4-0)^2} = 2a$$3\sqrt{2} + 4\sqrt{2} = 2a$$7\sqrt{2} = 2a \Rightarrow a = \frac{7\sqrt{2}}{2}$.The distance between the foci is $2ae = S_1S_2$.$S_1S_2 = \sqrt{(-4-3)^2 + (4-3)^2} = \sqrt{(-7)^2 + (1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$.$2ae = 5\sqrt{2}$$2 \cdot \frac{7\sqrt{2}}{2} \cdot e = 5\sqrt{2}$$7\sqrt{2} \cdot e = 5\sqrt{2}$$e = \frac{5}{7}$.

If an ellipse with foci at $(3,3)$ and $(-4,4)$ is passing through the origin,then the eccentricity of that ellipse is

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