Given the ellipse below, if I try to find a point that intercects the ellipse at an angle $\theta$ I need to get the coordinates $x = a cos(\theta)$ and $y = b sin(\theta)$. The issue is that this parametric form will give me the point at the black dot when the correct one should have been the yellow dot. Researching online I found this website https://www.petercollingridge.co.uk/tutorials/computational-geometry/finding-angle-around-ellipse/ that provides a formula to get the real angle $\phi$ for an ellipse and it works: $$ \phi = atan(a\sin(\theta)/b\cos(\theta))\;\; x = a \cos(\phi) $$ Notice that we're doing $a\sin(\theta)$ instead of $a\cos(\theta)$.
This angle $\phi$ is different from the angle $\theta$ and im trying to understand what exactly it represents and why this formula works. What's the intuition behind it?
I created this project to show the issue: https://www.khanacademy.org/computer-programming/angle-of-ellipse/6148205981515776
A point on the ellipse is given parametrically as $P = ( a \cos \theta, b \sin \theta)$. $\theta$ here is the parameter called the eccentric angle. To express $P$ in terms of the actual angle $\phi$ we have to write
$ P = r ( cos \phi, \sin \phi ) $
Here $\phi$ is the polar angle. Dividing $y$ by $x$ in both representations
$\dfrac{y}{x} = \dfrac{ b \sin \theta }{a \cos \theta} = \tan \phi $
Hence,
$ \phi = \tan^{-1} \left( \dfrac{ b \sin \theta }{a \cos \theta} \right) = \tan^{-1} \left( \dfrac{b}{a} \tan \theta \right) $
$\phi$ is the angle that the vector extending from the origin to the point $P$ makes with the positive $x$-axis.