Let p be a positive real, let S be the parabola y^{2} = 4px, and let P be a point with coordinates (a, b). Show that there are 1, 2 or 3 normals from P to S according as 4(2p - a)^{2} + 27 pb^{2} >, = or < 0.

**Solution**

Straightforward.

The general point on the parabola is (pt^{2}, 2pt). The slope of the tangent is 2p/y = 1/t, and so the slope of the normal is -t. Hence the equation of the normal is (y - 2pt) = -t(x - pt^{2}). This passes through (a, b) iff pt^{3} + (2p - a)t - b = 0 (*).

This is a cubic, so it has 1, 2 or 3 real roots. We have to decide which. If (2p - a) > 0, then pt^{3} + (2p - a)t is strictly increasing and takes all values in (-∞, ∞), so (*) has just one real root. The same is true if (2p - a) = 0, unless b = 0, in which case there are three coincident roots.

If (2p - a) < 0, then pt^{3} + (2p - a)t has a maximum and a minimum. Differentiating, we find that these are at t = ± √( -(2p - a)/(3p) ), with values ± 2(2p - a)/3 √( -(2p - a)/(3p) ). So there are 1, 2 or 3 real roots according as b^{2} >, = or < 4(2p - a)^{2}/9 -(2p - a)/3p, which is the condition in the question. Note that if (2p - a) ≥ 0, then this expression is certainly > 0, so the same rule applies.

© John Scholes

jscholes@kalva.demon.co.uk

15 Sep 1999