1st Putnam 1938

Problem B5

Find the locus of the foot of the perpendicular from the center of a rectangular hyperbola to a tangent. Obtain its equation in polar coordinates and sketch it.



Answer: r2 = 2k2 sin 2θ. It is a figure of 8 with its axis along the line y = x and touching the x-axis and y-axis at the origin.

Take the hyperbola as xy = k2. Then the tangent at (a, k2/a) is (y - k2/a) = - k2/a2 (x - a). The perpendicular line through the origin is y = a2/k2 x. They intersect at x = 2k2/(a (a2/k2 + k2/a2) ), y = 2a/(a2/k2 + k2/a2). So the polar coordinates r, θ satisfy tan θ = y/x = a2/k2, r2 = x2 + y2 = 4a2(k4/a4 + 1)/(k2/a2 + a2/k2) = 4a2(cot2θ + 1)/(tan θ + cot θ)2 = 4 a2 cos2θ/(sin2θ + cos2θ)2 = 4a2cos2θ = 4k2sin θ cos θ = 2k2 sin 2θ. Thus the polar equation of the locus is r2 = 2k2 sin 2θ.



1st Putnam 1938

© John Scholes
5 Mar 2002