Given an ellipse with the parametric definition:
$x = a \cos(t)$, $y = b \sin(t)$
Points on the ellipse $(a \cos(t),b \sin(t))$, have the line normal to the ellipse:
$a\sin\left(t\right)x-b\cos\left(t\right)y=\frac{1}{2}\left(a^2-b^2\right)\sin\left(2t\right)$
I am looking for values of $t$ where this normal line passes through a given point, $p$.
$p = (p_x, p_y)$
This problem can be reduced down to a singular equation:
$ap_x\sin\left(t\right)-bp_y\cos\left(t\right)=\frac{1}{2}\left(a^2-b^2\right)\sin\left(2t\right)$
I am not sure how to solve this equation to get valid values for $t$.
A solution for this equation would be helpful, or if this can be solved using a different method that would also help.
Given the ellipse
$$ E(x,y) = \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2-1=0 $$
and a point $p_0 = (x_0,y_0)$, the normal line to $E$ passing by $p_0$ is given by
$$ L\to p = p_0 + \lambda \nabla E $$
where $p = (x, y) \in E(x,y)$ and $\nabla E = (\frac{x}{a^2},\frac{y}{b^2})$ so we have to solve
$$ x = x_0 + \lambda \frac{x}{a^2}\\ y = y_0 + \lambda \frac{y}{b^2} $$
or
$$ x(1-\frac{\lambda}{a^2}) = x_0\\ y(1-\frac{\lambda}{b^2}) = y_0 $$
or
$$ a\cos(t)(1-\frac{\lambda}{a^2}) = x_0\\ b\sin(t)(1-\frac{\lambda}{b^2}) = y_0 $$
This system should be solved for $t,\lambda$ giving lenghtly expressions.
As an intermediary step we have
$$ \left(\frac{a x_0}{a^2-\lambda}\right)^2+\left(\frac{b y_0}{b^2-\lambda}\right)^2 = 1\\ \tan(t) = \frac{b(a^2-\lambda)y_0}{a(b^2-\lambda)y_0} $$
I hope this helps.