A point A(h, 0) , with h 0 , is chosen on the axis of the parabola y^2 = 4x such that the shortest distance from A to the parabola is 2 5 . Let P and Q be the points on the parabola that are closest to A . If the tangents to the parabola at P and Q intersect at a point T , then the distance between T and the orthocentre of the triangle PTQ is :

Step-by-step solution

Let the parametric coordinates of a point on the parabola y^2 = 4x be (t^2, 2t) . The equation of the normal to the parabola at this point is: y + tx = 2t + t^3 Since the shortest distance is measured along the normal, it must pass through A(h, 0) : 0 + th = 2t + t^3 h = 2 + t^2 (for t 0 ) The square of the distance from A(h, 0) to (t^2, 2t) is: D^2 = (t^2 - h)^2 + (2t - 0)^2 Substitute h = 2 + t^2 into the distance formula: D^2 = (t^2 - (2 + t^2))^2 + 4t^2 = 4 + 4t^2 Given that the shortest distance is 2 5 , we ha