An ellipse having foci at (3, 3) and (-4, 4) | vedclass

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(C) 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 fr

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$\frac{5}{7}$

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(C) 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,$2a$.$PS_1 = \sqrt{(3-0)^2 + (3-0)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.$PS_2 = \sqrt{(-4-0)^2 + (4-0)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$.Thus,$2a = PS_1 + PS_2 = 3\sqrt{2} + 4\sqrt{2} = 7\sqrt{2}$.The distance between the foci is $2ae = S_1S_2$.$S_1S_2 = \sqrt{(-4-3)^2 + (4-3)^2} = \sqrt{(-7)^2 + (1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$.Therefore,$2ae = 5\sqrt{2}$.Dividing the two equations: $e = \frac{2ae}{2a} = \frac{5\sqrt{2}}{7\sqrt{2}} = \frac{5}{7}$.

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

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