Q is any point on the circle `x^(2) +y^(2) = 9. QN` is perpendicular from Q to the x-axis. Locus of the point of trisection of QN is

Options

A. `4x^(2) +9y^(2) = 36`
B. `9x^(2) +4y^(2) = 36`
C. `9x^(2) +y^(2) = 9`
D. `x^(2) +9y^(2) = 9`

Answer

A, D

Explanation

Let `Q = (3 cos theta, 3 sin theta) rArr N =(3 cos theta,0)`
Point of trisection are `(3cos theta, sin theta)` or `(3 cos theta,2 sin theta)`.
Locus is `(x^(2))/(9) + y^(2) =1` or `(x^(2))/(9) + (y^(2))/(4) =1`