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:

enter image description here

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

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    $\begingroup$ Just as to get the arc length from the angle you need to use an elliptic integral, to get the angle from the arc length you need to invert the elliptic integral. $\endgroup$ – robjohn Jul 4 '13 at 13:14

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)]

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    $\begingroup$ Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework $\endgroup$ – Mohammad Fakhrey Jul 5 '13 at 7:32
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    $\begingroup$ @MohammadFakhrey: I've added the example, but that will only show you how to solve. $\endgroup$ – robjohn Jul 5 '13 at 14:53

The nice answer by robjohn demands a comment. His result for the integral of the arc-length of the elipse (with major axis $2a$ and minor axis $2b$) is $b\,E(\theta,1-a^2/b^2)$, but this is only a quarter of the complete elliptic circumference. Also, for those more interested readers, it is easy to show that $b\,E(\theta,1-a^2/b^2) = a\,E(\theta,1-b^2/a^2)$, as found at Wolfram World webpage http://mathworld.wolfram.com/Ellipse.html. This change rule is useful for avoiding negative values of parameters. F. M. S. Lima, University of Brasilia (fabio-at-fis.unb.br)

  • $\begingroup$ Between $\theta=0$ and $\theta=2\pi$, $b\,\text{EllipticE}\left(\theta,\frac{b^2-a^2}{b^2}\right)$ gives the entire circumference. Perhaps I am missing something. $\endgroup$ – robjohn Feb 6 '16 at 0:45

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