Here is John Hewson's solution:
Let (t , t^2) be a point on the parabola. The slope of the tangent at that point is dy/dx =2 x = 2 t. The slope of the line from (t , t^2) to a point (X , Y) is (t^2 - Y)/(t - X) = 2t if the line is a tangent. Thus t^2 - Y = 2 t^2 - 2 t X , i. e. t ^2 - 2 X t + Y = 0. The roots of this equation, u, v (say) give the two points where the tangents from (X, Y) meet the parabola, and these lines have slopes 2u and 2v. They are perpendicular to one another if the product of their slopes = -1. i.e. if (2u)(2v) = -1 ie if 4uv = -1. The product of the roots of the quadratic equation is Y and the locus of (X , Y) is: 4 Y = -1, a straight line parallel to the X - axis. The slopes 2u and 2v are the tangents of the angles that the two tangents make with the x - axis. The tangent of the angle A between the two tangents is given by: Tan A =( 2v - 2u)/ (1 + 4uv)) = 1 when A = 45 degrees Now (v - u)^2 =(u + v)^2 - 4uv Thus 4( (u + v)^2 - 4uv) = (1 + 4uv)^2 From the original quadratic we have: uv = Y, and u + v = 2X, and sustituting gives: 4(4X^2 - 4Y) = (1 + 4Y)^2 16X^2 - 16Y = 1 + 8Y +16 Y^2 Giving 16X^2 - 16Y^2 - 24Y -1 = 0. The lower branch of this hyperbola is the locus of the intersection of the two tangents when the angle between them is 45 degrees.