A point A(0, c) lies on the positive y-axis. The shortest distance from A to the parabola x^2 = 8y is 4 2 . Let P and Q be the points on the parabola that are closest to A . The area of the triangle OPQ , where O is the origin, is :

Step-by-step solution

Let the parametric coordinates of a point on the parabola x^2 = 8y be (4t, 2t^2) . The equation of the normal to the parabola at this point is: x + ty = 4t + 2t^3 Since the shortest distance is along the normal, the normal must pass through A(0, c) : 0 + tc = 4t + 2t^3 c = 4 + 2t^2 (for t 0 ) The square of the distance from A(0, c) to (4t, 2t^2) is: D^2 = (4t - 0)^2 + (2t^2 - c)^2 Substitute c = 4 + 2t^2 into the distance formula: D^2 = 16t^2 + (2t^2 - (4 + 2t^2))^2 = 16t^2 + 16 Given that the shortest distance is