The angle of the tangent on the surface of an ellipse 
My question is, given $a, b$ and $h$, whilst $c$ is unknown, is it possible to find the angle $β$ of an ellipsoid. Ive tried to illustrate my problem in the image above.

The ellipse has equation $(x/b)^2+(y/a)^2=1$. You can plug $y=h$ into this to solve for $c$.
Suppose the tangent line intersects the $y$ and $x$-axes at $(0,u)$ and $(v,0)$ respectively. We may shrink the $y$-axis by a factor of $a$, and the $x$-axis by a factor of $b$, to turn the ellipse into a unit circle. The point $(c,h)$ becomes the point $(\frac{c}{b},\frac{h}{a})$, and the tangent line now intersects $(0,\frac{u}{a})$ and $(\frac{v}{b},0)$. The line segment joining the origin $(0,0)$ to $(\frac{c}{b},\frac{h}{a})$ and the tangent line respectively have slopes
$$ m=\frac{(h/a)}{(c/b)}, \qquad m'=-\frac{(u/a)}{(v/b)}=-\frac{1}{m}. $$
Note $m$ and $m'$ are opposite reciprocals, because the segment and tangent are perpendicular.
Substitute the formulas for $m$ and $m'$ into $m'=-1/m$ to solve for $\tan\beta=u/v$ and get
$$ \beta=\tan^{-1}\left(\frac{ca^2}{hb^2}\right). $$
