1.

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))`

Answer» 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)))`


Discussion

No Comment Found

Related InterviewSolutions