Trace of the cissoid of Diocles

Question

I want to show that you can parameterized the cissoid of Diocles using $$ \alpha(t) = \left(\frac{2at^2}{1+t^2}, \frac{2at^3}{1+t^2}\right),\quad\text{where }t=\tan\theta. $$

Here is an image of the cissoid of Diocles, along with its construction. Each point along the trace is the point given by the vector $p$, which is the vector with origin at $0$, given by $CB$.

Now, by drawing a line segment from $C$ to the line $AV$, parallel to the $x$-axis, I quite readily get that $p_1t = p_2$, just using that $\tan\theta = \frac{p_2}{p_1}$. However, after this I'm stuck. I've tried using the Pythagorean theorem on the triangle inside the circle, made up of points $0$, $C$, and the straight line parallel to the $y$-axis that descends down from $C$ and intersect the $x$-axis. However, this is just a messy computation that yields the already known relation above.

My geometry is very rusty, and there might be a theorem or two that relates to circles cut in this way, but I'd prefer if there was a more simple proof, that just relied on basic geometry and trigonometry.

Answer 1

In $\triangle OAB, \frac{OA}{OB} = \cos\theta \implies OB = 2a \sec\theta$

As $C$ is on the circumference of the circle, $\angle OCA = 90^\circ$ and so in $\triangle OCA, OC = OA \cos\theta = 2a \cos\theta$

$CB = OB - OC = 2a (\sec\theta - \cos\theta) = 2a \sin\theta \tan\theta$

Now as per definition of Cissoid of Diocles, it is traced by point $P$ such that $OP = CB = 2a \sin\theta \tan\theta$

Now using $x = OP \cos\theta, y = OP \sin\theta$, can you get to the parametrization asked in the question?