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There are exactly two points on the elli...

There are exactly two points on the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` whose distances from its center are the same and are equal to `(sqrt(a^2+2b^2))/2dot` Then the eccentricity of the ellipse is `1/2` (b) `1/(sqrt(2))` (c) `1/3` (d) `1/(3sqrt(2))`

A

`1//2`

B

`1//sqrt(2)`

C

`1//3`

D

`1//3sqrt(2)`

Text Solution

Verified by Experts

(2) Since there are axactly two points on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` whosse distance from the center is the same, the poitns would either be the endpoints of the major axis or of the minor axis. But `sqrt((a^(2)+2b^(2)))//bgtb` So, the points are the vertices of the major axis hence,
(1)`a=sqrt((a^(2)+2b^(2))/(2))`
or `a^(2)=2b^(2)`
`or e=sqrt(1-(b^(2))/(a^(2)))=(1)/(sqrt(2))`
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