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**

Answer: 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.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002