α and β are positive real numbers such that 1/α + 1/β = 1. Prove that the line mx + ny = 1 with m, n positive reals is tangent to the curve x^{α} + y^{α} = 1 in the first quadrant (x, y ≥ 0) iff m^{β} + n^{β} = 1.

**Solution**

Suppose mx + ny = 1 is tangent to the curve. Suppose it touches at (a, b). Differentiating, we see that the tangent at (a, b) is a^{α-1}x + b^{β-1} = 1, so m = a^{α-1}, n = b^{β-1}. Hence, using αβ - β = α, we have that m^{β} + n^{β} = a^{α} + b^{α} = 1.

Conversely, suppose that m^{β} + n^{β} = 1. Take a = m^{β/α}, b = n^{β/α}. Then a^{α} + b^{α} = 1, so (a, b) lies on the curve in the first quadrant. Its tangent is Mx + Ny = 1, where M = a^{α-1}, N = b^{β-1}. But a = m^{β/α} and β/α (α - 1)= 1, so M = m. Similarly, N = n. Thus we have established that mx + ny = 1 is tangent to the curve in the first quadrant as required.

© John Scholes

jscholes@kalva.demon.co.uk

25 Jan 2002