Determining the angle degree of an arc in ellipse?

Question

Is it possible to determine the angle in degree of an arc in ellipse by knowing the arc length, ellipse semi-major and semi-minor axis ? If I have an arc length at the first quarter of an ellipse and I want to know the angle of it, what is the data that I will need it to use it and what is the exact method to use it?

Please take a look at this picture:

Actually I know how to determine the arc length of a ellipse here. but I want to do the obverse.

Answer 1

Parameterization of an ellipse by angle from the center is $$ \gamma(\phi)=(\cos(\phi),\sin(\phi))\frac{ab}{\sqrt{a^2\sin^2(\phi)+b^2\cos^2(\phi)}} $$ and $$ \left|\gamma'(\phi)\right|=ab\sqrt{\frac{a^4\sin^2(\phi)+b^4\cos^2(\phi)}{\left(a^2\sin^2(\phi)+b^2\cos^2(\phi)\right)^3}} $$ and integrating $\left|\gamma'(\phi)\right|$ gets extremely messy.

So instead, we use $\theta$, where $$ b\tan(\theta)=a\tan(\phi) $$ Then $$ \gamma(\theta)=(a\cos(\theta),b\sin(\theta)) $$ and $$ \left|\gamma'(\theta)\right|=\sqrt{a^2\sin^2(\theta)+b^2\cos^2(\theta)} $$ Now, integrating $\left|\gamma'(\theta)\right|$ is a lot simpler. $$ \int\left|\gamma'(\theta)\right|\,\mathrm{d}\theta =b\,\mathrm{EllipticE}\left(\theta,\frac{b^2-a^2}{b^2}\right) $$ However, to go from from arc length to angle, we still need to invert the Elliptic integral.

---

Solution of the problem given using Mathematica

Here we get the solution to 30 places.

In[]=  Phi[a_,b_,s_,opts:OptionsPattern[]] := Block[ {t}, t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)] /. FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]]

In[]=  Phi[4, 2.9`32, 3.31`32, WorkingPrecision->30]

Out[]= 0.87052028743193111752524449959

Thus, $\phi\doteq0.87052028743193111752524449959\text{ radians}$.

Let me describe the algorithm a bit.

FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]

inverts the elliptic integral to get $\theta$ from $a$, $b$, and $s$.

Now, we want to find $\phi$ so that $b\tan(\theta)=a\tan(\phi)$. However, simply using $\tan^{-1}\left(\frac ba\tan(\theta)\right)$ will only return a result in $\left(-\frac\pi2,\frac\pi2\right)$. To get the correct value, we use the relation $$ \begin{align} \tan(\phi-\theta) &=\frac{\tan(\phi)-\tan(\theta)}{1+\tan(\phi)\tan(\theta)}\\ &=\frac{\frac ba\tan(\theta)-\tan(\theta)}{1+\frac ba\tan(\theta)\tan(\theta)}\\ &=\frac{b\tan(\theta)-a\tan(\theta)}{a+b\tan(\theta)\tan(\theta)}\\ \phi &=\theta+\tan^{-1}\left(\frac{(b-a)\tan(\theta)}{a+b\tan^2(\theta)}\right) \end{align} $$ This is why we have

t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)]