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. Solution
: r 2 = 2k 2 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 = k 2 . Then the tangent at (a, k 2 /a) is (y - k 2 /a) = - k 2 /a 2 (x - a). The perpendicular line through the origin is y = a 2 /k 2 x. They intersect at x = 2k 2 /(a (a 2 /k 2 + k 2 /a 2 ) ), y = 2a/(a 2 /k 2 + k 2 /a 2 ). So the polar coordinates r, θ satisfy tan θ = y/x = a 2 /k 2 , r 2 = x 2 + y 2 = 4a 2 (k 4 /a 4 + 1)/(k 2 /a 2 + a 2 /k 2 ) = 4a 2 (cot 2 θ + 1)/(tan θ + cot θ) 2 = 4 a 2 cos 2 θ/(sin 2 θ + cos 2 θ) 2 = 4a 2 cos 2 θ = 4k 2 sin θ cos θ = 2k 2 sin 2θ. Thus the polar equation of the locus is r 2 = 2k 2 sin 2θ. 1st Putnam 1938 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002