A line from `(-1,0)` intersects the parabola `x^(2)= 4y` at A and B. Then the locus of centroid of `DeltaOAB` is (where O is origin) (a) `3x^(2) -2x =4y` (b) `3y^(2) -2y = 4x` (c) `3x^(2) +2x=4y` (d) none of these

Options

A. `3x^(2) -2x =4y`
B. `3y^(2) -2y = 4x`
C. `3x^(2) +2x=4y`
D. none of these

Answer

C

Explanation

Let `A(2t_(1),t_(1)^(2))` and `B(2t_(2),t_(2)^(2))`
`P(-1,0),A` and B are collinear.
`rArr(t_(2)^(2)-t_(1)^(2))/(2(t_(2)-t_(1))) =(t_(2)^(2))/(2t_(2)+1)` (i)
`rArr (t_(1)+t_(2)) =- 2t_(1)t_(2)`
Let centroid be (h,k)
`:. h =(2(t_(1)+t_(2)))/(3)` (ii)
`:. k = (t_(1)^(2)+t_(2)^(3))/(3)`
From Eqs. (i),(ii), and (iii) eliminating `t_(1)` and `t_(2)`, we get
`3h^(2) +2h =4k`
Hence, locus is `3x^(2) +2x =4y`.