P0 is the parabola y2 = mx, vertex K (0, 0). If A and B points on P0 whose tangents are at right angles, let C be the centroid of the triangle ABK. Show that the locus of C is a parabola P1. Repeat the process to define Pn. Find the equation of Pn.
Solution
The gradient at the point (x1 , y1) is 1/2 m/y1. So the tangents at (x1 , y1), (x2 , y2) are perpendicular iff y1y2 = -m2/4. So we may take one point as (x, y) = (t2/m, t) and the other as (1/16 m3/t2, -1/4 m2/t). Hence the centroid is (x, y), where x = 1/3 (t2/m + 1/16 m3/t2), y = 1/3 (t - 1/4 m2/t). But this lies on the parabola y2 = m/3 (x - m/6). But since any value of t was possible and hence any value of y, any point of this parabola is a centroid from some pair of points A, B.
Repeating, we get that the equation for Pn is y2 = m/3n(x - m/6(1 + 1/3 + ... + 1/3n-1) ) = m/3n(x - m/4 (1 - 1/3n) ).
© John Scholes
jscholes@kalva.demon.co.uk
14 Dec 1999