Ellipse having foci (3,1) and (1,1) and passing through (1,3) has eccentricity
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e = 0.4142
Step-by-step explanation:
It is known that the sum of distance between a point on the ellipse and its focii is equal to 2 x a, where a is the length of the semi-major axis.
Now, distance between the two foci is 2 x a x e = 2 units.
And, the sum of the distance (S) between the foci and the point on the ellipse is:
S = 2 + (8)^(0.5)
Hence, we calculate e from the above two equations to be 0.4142.
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