How to find the polar coordinate angle of the tangent of any point on an ellipse? I have an ellipse centered at 0,0 with a height of 75 and a width of 150. Now let say I know both the x and y and also the angle of a point on the circumference of the ellipse. I want to know what the angle of the point tangent to ellipse is relative to polar coordinates.

 A: The parametric equations for a central ellipse (aligned with the X & Y axes) are
$$x = a\cos(\theta)$$
$$y = b\sin(\theta)$$
where $a$ and $b$ are the "radii" of the ellipse on the X & Y axes, respectively; in other words, $a$ and $b$ are the lengths of the semimajor and semiminor axes of the ellipse.
However, $\theta$ is not the central angle of the ellipse, it's the central angle of the auxiliary circles associated with the ellipse. In astronomy, it's known as the eccentric anomaly. Here's a diagram, courtesy of Wikipedia (click the image for an SVG version):


The eccentric anomaly of point P is the angle E. The center of the ellipse is point C, and the focus is point F.

The blue circle provides the X coordinate of P, and the green circle provides its Y coordinate.
As Intelligenti pauca mentions in the comments, the polar angle $\phi$ can be calculated using:
$$\tan\phi={y\over x}={b\over a}\tan\theta$$
Differentiating the previous equations for $x$ & $y$,
$$\begin{align}
dx & = -a\sin(\theta)\,d\theta\\
dy & = b\cos(\theta)\,d\theta
\end{align}$$
Therefore,
$$\frac{dy}{dx} = \frac{b\cos(\theta)}{-a\sin(\theta)}$$
and that's the slope of the required tangent. So the angle $\alpha$ that tangent makes with the (positive) X axis is
$$\alpha = \tan^{-1}\left(\frac{b\cos(\theta)}{-a\sin(\theta)}\right)$$
