Question about finding a third point on an ellipse given angle

Question

If I have a known point $Y$ on an ellipse in the first quadrant, and known point $X$ on the $x$-axis, and some angle $\theta$ between $XY$ and $YZ$ with $Z$ being some mystery third point on the ellipse, what are the coordinates of $Z$?

Answer 1

Let's choose some coordinates:

\begin{align*} X&=\begin{pmatrix}x_X\\y_X\end{pmatrix}& Y&=\begin{pmatrix}x_Y\\y_Y\end{pmatrix} \end{align*}

Start with a line through the origin with angle $\theta$ against the $x$ axis. A point on that line has the form

$$\lambda\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix} \qquad\lambda\in\mathbb R$$

Now rotate your setup such that the original $x$ axis becomes a line through the origin but parallel to the line $XY$. You can scale your coordinate system while you rotate it, which makes for easy numbers:

$$ \lambda \begin{pmatrix} x_X-x_Y & y_Y-y_X \\ y_X-y_Y & x_X-x_Y \end{pmatrix} \begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix} $$

Now move the origin of the coordinate system to point $Y$:

$$ \lambda \begin{pmatrix} x_X-x_Y & y_Y-y_X \\ y_X-y_Y & x_X-x_Y \end{pmatrix} \begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}+ \begin{pmatrix}x_Y\\y_Y\end{pmatrix} $$

Plug the resulting coordinates into the equation of the ellipse, and you obtain a quadratic equation in $\lambda$. But one solution will be $\lambda=0$ for the point $Y$ itself. So you can divide the equation by $\lambda$ to obtain the other solution, which is your pint $Z$.