Line `3x + by = k` intersect the curve `2x^2 + 3y^2 + 2xy=1` at points A and B. Thecircle on AB as diameter passes through origin,then sum of all possible values of 'k' is

Options

A. 3
B. 4
C. `-4`
D. `-3`

Answer

A, D

Explanation

We have `(3x +6y)/(k) = 1` (1)
`2x^(2) +2xy +3y^(2) -1 = 0` (2)
Now homogenising (2) with the help of (1), we get
`rArr 2x^(2) +2xy +3y^(2) -((3x +6y)/(k))^(2) =0`
`rArr k^(2)(2x^(2)+3x +3y^(2)) -(3x +6y)^(2) =0`
Now, coefficient of `x^(2)+` coefficient of `y^(2) =0`
`rArr (2k^(2)-9) +(3k^(2) -36) =0`
`rArr 5k^(2) =45 rArr k^(2) = 9 rArr k = 3` or `-3`