To prove it, find B on the graph at x=1, f(1) = p. Draw the tangent (DB) at B and take the symmetric of line AB with respect to this tangent. This is line BE' intersecting the y-axis on the focus of the parabola. The tangent at B and the parallel to it from D form a rhombus ABFD, whose diagonals intersect orthogonally at K(x = 1/2). Then (KE)

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