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**Finding focus with Archimedes**

Hard as it is to imagine now, for most of the history of mathematics there was no coordinate geometry system and therefore graphs were not drawn using algebraic equations but instead were constructed. The ancient Greeks such as Archimedes made detailed studies of conic sections (parabolas, ellipses and hyperbola) using ideas of relationships in constructions. The nice approach to this method is that it makes clear the link between conic sections and their properties in reflecting light – a property which can then be utilized when making lenses. A parabolic telescope for example uses the property that all light collected through the scope will pass through a single focus point.

Let’s see how we can construct a parabola without any algebra – simply using the constructions of the Greeks. We start with a line and a focus point F not on the line. This now defines a **unique parabola**.

This unique parabola is defined as all the points A such that the distance from A to F is equal as the perpendicular distance from A to the line.

We can see above that point A must be on our parabola because the distance AB is the same as the distance AF.

We can also see that point C must be on our parabola because the length CD is the same as CF. Following this same method we could eventually construct every point on our parabola. This would finally create the following parabola:

**Focus point of a parabolic **mirror

We can now see how this parabola construction gives us an intrinsic understanding of reflective properties. If we have a light source entering parallel to the perpendicular though the focus then we can use the fact that this light will pass through the focus to find the path the light traces before it is reflected out.

Newton made use of this property when designing his parabolic telescope. It’s interesting to note how a different method leads to a completely different appreciation of the properties of a curve.

**Finding the area under a quadratic curve without calculus**

Amazingly a method for finding the area under a quadratic curve was also discovered by the Greek scientist and mathematician Archimedes around 2200 years ago – and nearly 2000 years before calculus. Archimedes’ method was as follows.

Choose 2 points on the curve, join them to make 2 sides of a triangle. Choose the 3rd point of the triangle as the point on the quadratic with the same gradient as the chord. This is best illustrated as below. Here I generated a parabola with focus at (0,1) and line with the x axis.

Here I chose points B and C, joined these with a line and then looked for the point on the triangle with the same gradient. This then gives a triangle with area 4. Archimedes then discovered that the area of the parabolic segment (i.e the total area enclosed by the line BC and the parabola) is 4/3 the area of the triangle. This gives 4/3 of 4 which is 5 1/3. Once we have this we can find the area under the curve (i.e the integral) using simple areas of geometric shapes.

**Using calculus**

We can check that Archimedes’ method does indeed work. We want to find the area enclosed by the 2 following equations:

This is given by:

It works! Now we can try a slightly more difficult example. This time I won’t choose 2 points parallel to the x-axis.

This time I find the gradient of the line joining B and C and then find the point on the parabola with the same gradient. This forms my 3rd point of the triangle. The area of this triangle is approximately 1.68. Therefore Archimedes’ method tells us the area enclosed between the line and the curve will be approximately 4/3 (1.68) = 2.24. Let’s check this with calculus:

Again we can see that this method works – our only error was in calculating an approximate area for the triangle rather than a more precise answer.

So, nearly 2000 years before the invention of calculus the ancient Greeks were already able to find areas bounded by line and parabolic curves – and indeed Archimedes was already exploring the ideas of the limit of sums of areas upon which calculus in based.