Find the maxium distance of any point on the curve `x^2+2y^2+2xy=1` form the origin.

Answer

`(2)/(3-sqrt(5))`

Explanation

Let `x=r cos` and `y=r sin theta`
`x^(2)+2y^(2)+2xy=1`
`therefore r^(2)cos^(2)theta+2r^(2)sin^(2)theta+2r^(2)sin thetacostheta=1`
`therefore r^(2)=(1)/(cos^(2)theta+2sin^(2)theta+sin 2theta)`
`=(2)/(3-cos2theta+2sin 2theta)`
Now, `-sqrt(5)le-cos2theta+2sin 2thetalesqrt(5)`
`rArr 3-sqrt(5)le3-cos2theta+2sin 2thetale3+sqrt(5)`
`rArr r_("max")^(2)=(2)/(3-sqrt(5))`