I have drawn this image to explain the problem. Consider the below situation:
Information:
We have an ellipse of size $~a~ (X$ axis$)$ and $~b~ (Y$ Axis$)$. We know these values.
Given the angle $~α~$, we know that point $~P~$ is $~(a\cos α, b\sin α)~$. $($we also know the value of $~α)$
Problem:
What I really want to determine is the coordinates of $~P'~$, which is the point where the line $~OA~$ intersects the ellipse. I am not interested in $~P~$. As such, the difficulty I am having is that I want to determine the coordinates of point $~P'~$. $~P'~$'s coordinates will be $~(a\cos(α+β), b\sin(α+β))~$, so the real question becomes: How do I find $~β~$ ?
Thanks a lot for the help, it really sucks not remembering high school geometry.
Kind regards
The equation of the ellipse is $\quad\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1$
Let $\rho=|OP'|=\sqrt{x^2+y^2} \quad\begin{cases} x=\rho\cos(\alpha) \\ y=\rho\sin(\alpha) \end{cases}$
$\left(\frac{\rho\cos(\alpha)}{a}\right)^2+\left(\frac{\rho\sin(\alpha)}{b}\right)^2=1$
$|OP'|=\rho=\frac{1}{\sqrt{\left(\frac{\cos(\alpha)}{a}\right)^2+\left(\frac{\sin(\alpha)}{b}\right)^2}}=\frac{ab}{\sqrt{b^2\cos^2(\alpha)+a^2\sin^2(\alpha)}}$
The coordinates of point $P'$ are : $\begin{cases} x=\frac{ab\cos(\alpha)}{\sqrt{b^2\cos^2(\alpha)+a^2\sin^2(\alpha)}} \\ y=\frac{ab\sin(\alpha)}{\sqrt{b^2\cos^2(\alpha)+a^2\sin^2(\alpha)}} \end{cases}$
The abscisses of points $A'$ and $P'$ are the same : $$a\cos(\alpha+\beta)=\frac{ab\cos(\alpha)}{\sqrt{b^2\cos^2(\alpha)+a^2\sin^2(\alpha)}}$$ from which $\beta$ is derived : $$\beta = -\alpha+\cos^{-1}\left(\frac{b\cos(\alpha)}{\sqrt{b^2\cos^2(\alpha)+a^2\sin^2(\alpha)}} \right)$$ Or, on an equivalent form : $$\beta = -\alpha+\cos^{-1}\left(\frac{1}{\sqrt{1+\frac{a^2}{b^2}\tan^2(\alpha)}} \right)$$