Let P(2,2) be a point on an ellipse whose foc | vedclass

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(B) The sum of the focal distances of any point on an ellipse is equal to the length of the major axis,$2a$.Given $P(2,2)$,$F_{1}(5,2)$,and $F_{2}(2,6

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$\frac{5}{7}$

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(B) The sum of the focal distances of any point on an ellipse is equal to the length of the major axis,$2a$.Given $P(2,2)$,$F_{1}(5,2)$,and $F_{2}(2,6)$.$PF_{1} = \sqrt{(5-2)^2 + (2-2)^2} = \sqrt{3^2 + 0^2} = 3$.$PF_{2} = \sqrt{(2-2)^2 + (6-2)^2} = \sqrt{0^2 + 4^2} = 4$.Therefore,$2a = PF_{1} + PF_{2} = 3 + 4 = 7$.The distance between the foci is $2ae = \sqrt{(5-2)^2 + (2-6)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.Since $2a = 7$ and $2ae = 5$,we have $e = \frac{2ae}{2a} = \frac{5}{7}$.

Let $P(2,2)$ be a point on an ellipse whose foci are $F_{1}(5,2)$ and $F_{2}(2,6)$. Find the eccentricity of the ellipse.

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