If we consider the ellipse in the picture here
How do we determine the angle $\lambda$ of the vector v (tangent at point x = 2 ,y = 3) with the line joining the center (10,0)and the point (2,3)?
Edit:
a = 10, b = 5
it seems impossible it is 80°, it looks more like 56-57°
$$ \dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} =1 $$
Differentiate to find slope
$$ y^{'}= \dfrac{x\, b^2 }{y\, a^2 }= \dfrac{8\, 5^2 }{3 \,10^2} = \dfrac23 $$
Find arctan of above and add to $ {20.55}^0 $ to get correct angle.
The angle $ \lambda $ is: 90°- 9.46° = 80.54°.
If you look closely you can see that the angle $ \lambda $ is not measured from the line joining the center of the ellipse to the point (the green line), but from the line joining the apopsis to the point (the red line).
The angle of the velocity vector is: $$ \arctan\left( \frac{ e + \cos(\nu)}{ \sin(\nu)} \right) = 39.44\!° $$
where e is the eccentricity, and $\nu$ is the true anomaly $( e = \frac{ \sqrt{3} }{2} ; \ \ \nu = 92.52\!° )$.
The angle of the the line joining the center [10,0] and the point [2,3] ([10,0]-[2,3]=[8,-3]) is: $$ \arctan \left( -\frac38 \right) = -20.56 \!° $$ So the angle between that line and the tangent vector is: 39.44° + 20.56° = 60°.
