The Witch of Agnesi and its cartesian equation. I want to learn to simply take an ancient geometer's description of an interesting curve and try to find its cartesian equation. I'm struggling to find the equation of the locus of all points described by the 'Witch of Agnesi' The description is as follows:
"Starting with a fixed circle, a point O on the circle is chosen. For any other point A on the circle, the secant line OA is drawn. The point M is diametrically opposite to O. The line OA intersects the tangent of M at the point N. The line parallel to OM through N, and the line perpendicular to OM through A intersect at P. As the point A is varied, the path of P is the Witch of Agnesi."
I'd really appreciate insight and not plug and chug kind of answers if any. I'm looking for your ideas and tips. It would be very helpful if you could explain the answer step by step.
(I know the answer thanks to Wikipedia but I'd like to know how to arrive at the result)
 A: A picture is worth a thousand words. 

Ideas: 


*

*What are the parametric equations of the circle and the line $\overline{OA}$? 

*What is the x coordinate of $A$? What is the y coordinate of the intersection of the circle and the line, $Q$? 

*How can these be combined to get the coordinates of $P$ and then a parametric equation?

*How can this parametric equation be turned into a Cartesian equation?

A: Note that the coordinates are formed as projections of known lengths.
The line through origin cuts the circle and line $y= 2a$ at $ K,N $ respectively.
The curve is found by drawing lines parallel to $x-,y-$ axes through $K,N$ to get cutting point at $P$.
$$x= 2a \tan \theta$$
$$ OK=OM \cos \theta= 2a \cos \theta$$
$$ y= OQ=PL=OK \cos \theta = 2a \cos^2 \theta $$

To get this into Cartesian form, eliminate $\theta$ using trig relation $ \sec^2(...)= 1 + \tan^2(...) $
$$ \dfrac{2a}{y}=1+\dfrac{x^2}{4a^2} \to y=\dfrac{8a^3}{x^2+4a^2}$$
Notice that the Cartesian equation form is obtained only through this trig identity after the construction procedure is over.
