Correct Answer - D
Let `P (t^(2), 2t)` be a point on the curve `y^(2) = 4x`, whose image is Q(x,y) on x + y + 4 = 0, then
`(x - t^(2))/(1) = (y - 2t)/(1) = (-2(t^(2) + 2t + 4))/(1^(2) + 1^(2))`
`implies x = - 2t - 4`
and `y = - t^(2) - 4`
Now, the straight line y = - 5 meets the mirror image.
`:. - t^(2) - 4 = - 5`
`implies t^(2) = 1`
`implies t == +- 1`
Thus, points of intersection of A and B are -6, - 5) and (-2,-5).
`:.` Distance, `AB = sqrt((-2 + 6)^(2) + (-5 + 5)^(2)) = 4`