Coordinates of points on curve `5x^(2) - 6xy +5y^(2) - 4 = 0` which are nearest to origin are

Options

A. `((1)/(2),(1)/(2))`
B. `(-(1)/(2),(1)/(2))`
C. `(-(1)/(2),-(1)/(2))`
D. `((1)/(2),-(1)/(2))`

Answer

B, D

Explanation

Put `x = r cos theta,y = r sin theta` in given equation
`5r^(2) - 3r^(2) sin 2 theta = 4`
`rArr r^(2) =(4)/(5-3 sin 2theta)`
`rArr r_(min)^(2) = 4//8 = 1//2` (when `sin 2 theta =- 1)`
`rArr r_(min) = (1)/(sqrt(2))` at `2 theta =(3pi)/(2),(7pi)/(2)`, as `[2theta in (0,4pi)]`
So `theta = (3pi)/(4),(7pi)/(4)`
So points are `((1)/(sqrt(2))cos theta,(1)/(sqrt(2))sin theta)`
`= (-(1)/(2),(1)/(2))` and `((1)/(2),-(1)/(2))`