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If the maximum distance of any point on the ellipse `x^2+2y^2+2x y=1` from its center is `r ,` then `r` is equal to `3+sqrt(3)` (b) `2+sqrt(2)` `(sqrt(2))/(sqrt(3-sqrt(5)))` (d) `sqrt(2-sqrt(2))`

A

`3+sqrt(3)`

B

`2+sqrt(2)`

C

`sqrt(2)//sqrt(3-sqrt5)`

D

`sqrt(2-sqrt(2))`

Text Solution

Verified by Experts

Herem the center of the ellipse is (0,0)
Let `P(r cos theta, r sin theta)` be any point on the given ellipse. Then `r^(2) cos^(2) theta+2r^(2) sin ^(2) theta+2r^(2) sin theta cos theta=1`
`or r^(2)=(1)/(cos^(2)theta+2 sin^(2) theta+sin 2 theta)`
`or (1)/(sin^(2)theta+1sin 2theta)`
`=(2)/(1-cos 2 theta+2+2sin 2 theta)`
`(2)/(3-cos 2 theta+2 sin2 theta)`
`or r_("max")=(sqrt(2))/(sqrt(3-sqrt(5)))`
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