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 = x2. The gradient at the point A (a, a2/2m) is a/m, so the normal at (a, b) is (y - a2/2m) = -m/a (x - a). Substituting in 2my = x2, it meets P at (x, y) where x2 + 2m2/a x - (2m2 + a2) = 0, so the other point B has x = -(2m2/a + a).
Thus AB2 = (2a + 2m2/a)2 + 4m2(1 + m2/a2)2 = 4a2(1 + m2/a2)3. Differentiating, we find the minimum is at a2 = 2m2 and is AB2 = 27m2.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002