`y = sqrt(3)x +lambda` is drawn through focus S of the parabola `y^(2)= 8x +16`. If two intersection points of the given line and the parabola are A and B such that perpendicular bisector of AB intersects the x-axis at P then length of PS is

Options

A. `8//7`
B. `7//17`
C. `8sqrt(3)`
D. `16//3`

Answer

D

Explanation

`y^(2) = 8x +16`
`rArry^(2) = 8 (x+2)`
Thus, focus is (0,0)
So, line is `y = sqrt(3)x`. Solving this line with parabola,
`3x^(2) - 8x -16 =0`
i.e., `x_(1) + x_(2) = (8)/(3)`
`rArr (x_(1)+x_(2))/(2) = (4)/(3)`
So, x-coordinate of E is `4//3`.
`/_PSE = 60^(@)`
`:. SP = 2SE`
`SE = SM sec 60^(@) = (8)/(3) :. SP = 16//3`.