# Constant Product Curve

Problem: Find a locus of all points such that the product of the distances from a point in the locus to two fixed points is constant.

The problem is made simple if we find the locus in terms of polar coordinate system and assign the two points the coordinates A1(–a, 0) and A2(a, 0) as done in Figure 1. We let the product equal P and the distance between A1 and P be d1 and A2 and P be d2 so that d1d2 = P. Therefore, we have to express r in terms of θ. Using the Law of Cosines, we have the following equalities:

(i) $d_{1}^{2} = a^2 + r^2 - 2 \left|a\right|r\cos(\pi-\theta) =$ $a^2+r^2+2\left|a\right|r\cos\theta$

(ii) $d_{2}^{2} = a^2 + r^2 - 2 \left|a\right|r\cos(\theta) =$ $a^2+r^2-2\left|a\right|r\cos\theta$

Multiplying (i) and (ii), we obtain:

(iii) $d_{1}^{2}d_{2}^{2} = P^2 =$ $(a^2+r^2+2\left|a\right|r\cos\theta)(a^2+r^2-2\left|a\right|r\cos\theta)=$ $(a^2+r^2)^2-4a^2r^2\cos^2\theta$

(iv) $P^2 = r^4+a^4-2a^2r^2(2\cos^2\theta-1) =$ $r^4+a^4-2a^2r^2\cos(2\theta)$

Solving for r2, we obtain the polar form as:

(v) $r^2 = a^2\cos(2\theta) \pm \sqrt{P^2-a^4\sin^2(2\theta)}$

A locus of all points P such that the product of the distances between the points in the locus and two fixed points A1(–a, 0) and A2(a, 0) is equal to P is satisfied by the polar equation:

$r^2 = a^2\cos(2\theta) \pm \sqrt{P^2-a^4\sin^2(2\theta)}$

## Lemniscate

An interesting case occurs when the product P equals a2. The equation reduces to that of a Lemniscate.

(v) $r = \pm \sqrt{a^2\cos(2\theta) \pm a^2\sqrt{1-\sin^2(2\theta)}} =$ $\pm a\sqrt{2\cos(2\theta)}$

The rectangular form of the lemniscate is $(x^2+y^2)^2 = 2a^2(x^2-y^2)$. If we solve for a, we get $a = \frac{x^2+y^2}{\sqrt{2(x^2-y^2)}}$.

The Geogebra activity below shows a lemniscate for a = 2. You can drag the point around to see the lengths change and the product remains constant.