Differential Geometry - rational parameterization of the Cissoid of Diocles (1)

Question:

Find a rational parameterization of the Cissoid of Diocles.

The curve is defined as follow.

Draw a circle of radius 0.5 centered at (0, 0.5).
Draw a line y = 1
Shoot a ray from the origin, measure the distance between the intersection point with the circle and the intersection point with the line.
Compute the point with that length from the origin lying of the same ray.
The locus is the Cissoid of Diocles.

Solution:

We know a rational parameterization of the circle through the t-substitution.

$\begin{eqnarray*} \tan^2 \theta &=& \frac{\sin^2 \theta}{\cos^2 \theta} \\ 1 + \tan^2 \theta &=& 1 + \frac{\sin^2 \theta}{\cos^2 \theta} \\ &=& \frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} \\ &=& \frac{1}{\cos^2 \theta} \\ \cos^2 \theta &=& \frac{1}{1 + \tan^2\theta} \\ 2\cos^2 \theta - 1 &=& 2 \frac{1}{1 + \tan^2\theta} - 1 \\ \cos 2\theta &=& 2 \frac{1}{1 + \tan^2\theta} - \frac{1 + \tan^2\theta}{1 + \tan^2\theta} \\ &=& \frac{1 - \tan^2 \theta}{1 + \tan^2\theta} \\ \sqrt{1 - \cos^2 2\theta} &=& \sqrt{1 - (\frac{1 - \tan^2 \theta}{1 + \tan^2\theta})^2} \\ \sin 2\theta &=& \sqrt{(\frac{1 + \tan^2 \theta}{1 + \tan^2\theta})^2 - (\frac{1 - \tan^2 \theta}{1 + \tan^2\theta})^2} \\ &=& \sqrt{(\frac{2 \tan \theta}{1 + \tan^2\theta})^2} \\ &=& \frac{2 \tan \theta}{1 + \tan^2\theta} \\ \end{eqnarray*}$

The key above is that we have a square root of a perfect square above, so we can get to the rational parameterization of the circle. With careful scaling and shifting, we have $c_x$ and $c_y$ in the below code represents the circle. To compute the intersection of the ray with the line, we compute the slope $m$ of the ray, and then since the line has $y$ coordinate 1, it must have $x$ coordinate $l_x = \frac{1}{m}$. Finally we construct $d_x = l_x - c_x$, that will create the $x$ coordinate of the point on the Cissoid of Diocles, $d_y$ is simply $m d_x$.

Last but not least, the key is to simplify it not manually, but automatically using MATLAB, here is the code for the purpose.

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

Now it gives our answer: $x = \frac{(1-t)^3}{2(1+t^2)(1+t)}$ and $y = \frac{(1-t)^2}{2(1+t^2)}$