Differential Geometry - rational parameterization of the Cissoid of Diocles (2)

Problem:

Find a formula with only $ x $ and $ y $ describing the Cissoid of Diocles as defined in the last post.

Solution:

The key idea is that we should phase shift the parametrization of the circle. I learn this trick when I worked on this problem.

In particular, we could let $ \phi $ = $ \theta - \frac{\pi}{2} $, that allow us to write the circle as $ (r \cos \theta, r \sin \theta) = (r \cos (\phi + \frac{\pi}{2}), r \sin(\phi + \frac{\pi}{2}) = (-r\sin \phi, r\cos\phi) $.

The circle looked pretty similar, but magic happen when we find the parametrization of the cissoid. The code is basically the same as in the last post, of course, except the phase shift

syms t;
cx = -t/(1 + t*t);
cy = 0.5*(1 - t*t)/(1 + t*t) + 0.5;
m  = cy/cx;
lx = 1/m;
dx = lx - cx;
dy = m * dx;

This time, we get the parametrization as $ x = \frac{-t^3}{t^2 + 1} $, $ y = \frac{t^2}{t^2 + 1} $.

The magic of the phase shift originate from the geometric insight. If we start the circle at the y-axis, we obtain symmetry!

Having the simplified parametrization, now we can simply obtain $ t = \frac{-x}{y} $. Substituting that back to the $ y $ formula, we get:

$ y = \frac{(\frac{-x}{y})^2}{(\frac{-x}{y})^2 + 1} $

That simplifies to $ (x^2 + y^2)y = x^2 $, and this is the formula we sought for.

So long since the last post, sometimes, ideas just kick in.