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Question

There are exactly two points on the ellipse x2a2+y2b2=1 whose distance from its center is the same and is equal to a2+2b22​​. Then the eccentricity of the ellipse is


A

12

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B

12

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C

13

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D

132

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Solution

The correct option is B

12


Explanation for correct option:

Eccentricity of the ellipse:

As we know that there are exactly two points on the ellipse x2a2+y2b2=1, whose distance from the center is same, the points would be either endpoints of the major axis or of the minor axis. a2+2b22>b, so the points must be the vertices of major axis.

a=a2+2b22

Squaring on both sides, we get,

a2=a2+2b222a2=a2+2b2a2=2b2

Now, eccentricity of ellipse x2a2+y2b2=1 is

=1-b2a2=1-b22b2=1-12=12

Hence, Option B = 12 is the correct answer.


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