Y^2=x-c......we get a = 1/4 and Center (c,0)and Directrix is x-c+a=0...Putting x =0.c=a ....c=1/4

To find the value of c in the equation of the parabola x = y^2, we need to determine the condition for the tangents to the parabola from the origin to be perpendicular.

Let's start by finding the equation of the tangent to the parabola at any point (t^2, t).

The slope of the tangent is given by the derivative of the equation of the parabola:

dy/dx = 2t

The slope of the tangent is perpendicular to the line connecting the origin and the point of tangency. The slope of this line is given by:

m = (t - 0) / (t^2 - 0) = 1/t

Since the slopes of two perpendicular lines are negative reciprocals of each other, we can write:

(1/t) * 2t = -1

Simplifying the equation, we get:

2 = -t^2

Dividing both sides by -1, we have:

t^2 = -2

Since t^2 is always positive, there are no real values of t that satisfy this equation. Therefore, there are no tangents to the parabola x = y^2 from the origin that are perpendicular.

However, if we consider the parabola x = -y^2, the equation of the tangent at any point (t^2, -t) is given by:

dy/dx = -2t

The slope of the line connecting the origin and the point of tangency is:

m = (-t - 0) / (t^2 - 0) = -1/t

Setting the product of the slopes of the two lines equal to -1, we get:

(-1/t) * (-2t) = -1

Simplifying the equation, we have:

2 = -t^2

Dividing both sides by -1, we get:

t^2 = 2

Therefore, the tangents to the parabola x = -y^2 from the origin are perpendicular when c = 2. Hence, the correct answer is option 'C'.