# Why does the Ellipsograph/Trammel of Archimedes draw an ellipse, really?

Here's a diagram of the device I mean, hard at work drawing an ellipse. I find this quite surprising, and would like to get to the bottom of things.

Essentially, a rod (black line in animation) is anchored to two sliders (in blue): One at an end, the other somewhere in the middle. The non-anchored end (black square) can move freely, tracing an ellipse, as the sliders move along perpendicular axes -- the end anchor along the minor axis, the interior anchor the major axis.

If you're willing to set up shop and start giving coordinates to the end that draws the ellipse (letting $a$ be the length of the rod, $b$ the distance from interior anchor to free end), it's not too hard to see that

\begin{align*} x &= a \cos \theta \\ y &= b\sin \theta, \end{align*}

and so in turn $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$

verifying the motion is indeed elliptical.

But I find this pretty unsatisfactory; until introducing coordinates, I had no reason to believe that an ellipse should be created. Even having introduced coordinates and verifying, I still find it very mysterious. What property of an ellipse am I missing that makes this device "obviously" draw an ellipse?

To put it another way: How could someone knowing only classical geometry (i.e., someone without coordinate geometry) have designed this machine to draw an ellipse? I am assuming that, being named after Archimedes, the device at least predates Descartes and thus coordinate geometry. Although, the runners do form something of a coordinate axis...

Compare this mysterious motion with the much more widely known method of putting pushpins down and tying one end of a string to each, pulling the string taut. This method is more apparent in how it generates an ellipse: the string has a fixed length which is the sum of distances from the pencil to each pushpin. This obeys the key and classically-known property that the sum of distances from foci to a point on the ellipse is a fixed distance.

• Join the center of the figure to the midpoint between the two sliders to get a vector $u$ of fixed length rotating at a constant speed. You can then observe that the two sliders are at $u \pm v$, where $v$ is a vector of the same length rotating in the opposite direction. So the endpoint tracing out the curve is $u + \alpha v$, and then... something something Lissajous figures? I haven't figured it out to the end, but maybe someone else can take it from here. – Rahul May 18 '16 at 5:42
• Is it clear why it produces a circle when b=a? – user_of_math May 18 '16 at 8:59
• @Rahul I forgot how neat those Lissajous curves are! It'd be quite interesting if that was a fruitful approach (although I know even less about Lissajous curves, I certainly wouldn't mind the motivation to learn something new). – pjs36 May 18 '16 at 15:37
• @user_of_math Aside from the fact that the sliders wouldn't move (and the device would be significantly less cool in that case), yes, I think that's clear enough (although I hadn't thought about it until now, good question!) – pjs36 May 18 '16 at 15:39
• According to your last paragraph, is it that you accept the "two-foci-fixed-total-length" construct as an intuitive definition for ellipse and reject the "two-circles" construct? I personally find the two-circles parametric construct ($x = a\cos\theta,\, y=b\sin\theta$) also very nice and geometric, as it says that the ellipse is a stretched (linearly transformed) circle. Now, how to make a intuitive connection (proof) between the two-foci construct and the two-circles construct is indeed worth pursuing. Meanwhile, there's also the "conic-intersection" construct. – Lee David Chung Lin Dec 9 '16 at 0:17