The parabola P has focus a distance m from the directrix. The chord AB is normal to P at A. What is the minimum length for AB?

**Solution**

Answer: 3√3 m.

We may take the equation of P as 2my = x^{2}. The gradient at the point A (a, a^{2}/2m) is a/m, so the normal at (a, b) is (y - a^{2}/2m) = -m/a (x - a). Substituting in 2my = x^{2}, it meets P at (x, y) where x^{2} + 2m^{2}/a x - (2m^{2} + a^{2}) = 0, so the other point B has x = -(2m^{2}/a + a).

Thus AB^{2} = (2a + 2m^{2}/a)^{2} + 4m^{2}(1 + m^{2}/a^{2})^{2} = 4a^{2}(1 + m^{2}/a^{2})^{3}. Differentiating, we find the minimum is at a^{2} = 2m^{2} and is AB^{2} = 27m^{2}.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002