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Q.

There are exactly two points on the ellipse x2a2+y2b2=1 whose distances from its centre are the same and are equal to a2+2b22 then the eccentricity of the ellipse is

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a

12

b

132

c

12

d

13

answer is C.

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Detailed Solution

Given ellipse x2a2+y2b2=1

Since there are exactly two points on the ellipse whose distance from the centre is same, the points would be either end points of the major axis (or) of the minor axis.

But a2+2b22>b, so the points are the vertices of the major axis.

Hence a=a2+2b22

squaring on both sides we get

 2a2=a2+2b2  a2=2b2  b2a2=12

Now e=1-b2a2=1-12=12

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