# How to parameterize the elliptic equation with length of arc $t$

As we know, the equation of elliptic is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=r^2$$ But I want to use its parametric form,the $t$ is representative of the arc length.So how to parameterize the elliptic equation?

• The usual analog for the $(r\cos t, r\sin(t)$ parameterizatoin of a circle is $(ar\cos t, br\sin(t)$ where $t$ again runs from $0$ to $2\pi$. But this is a parameterization in terms of angle from the origin, not arc length. I would not be shocked to learn that an ellipse can be parameterized in terms of its arc length using Jacobi elliptic functions sn and cn, with $k$ for those functions depending on the eccentricity of the ellipse. I would be shocked to learn that there is any more elementary parameterization in terms of arc length. Jun 12, 2017 at 4:25
• @MarkFischler I want to partition a elliptic into some isometric arc length.For the sake of it,I want to use a parameter of arc length to parameterize the elliptic equation. :)
– mayi
Jun 12, 2017 at 5:01
• Is there a way to prove that a function does not have an anti derivative or is it simply that no one has found one yet. For example $$\int_{}^{} \sqrt{1 +a\cdot cos^{2}(x) }\hspace{3mm} dx$$ Is there a way to prove that this integral cannot be expressed in terms of a finite number of elementary functions? Jan 4, 2020 at 18:58
• See a newer post here. Oct 26, 2021 at 6:36

As Mark Fischler commented, I do not see how this could be done.

Using $$x=a\, r\cos(t)$$, $$y=a \,r\sin(t)$$, the arc length is given by $$L=r\int_{0}^{u} \sqrt{a^2 \cos^2(t)+b^2 \sin^2(t)}\,dt=br\, E\left(u\left|1-\frac{a^2}{b^2}\right. \right)$$ You cannot extract explicitely $$u$$ from this equation except writing $$u=\text{InverseFunction}[\text{EllipticE},1,2]\left[\frac{L}{b r},1-\frac{a^2}{b^2}\right]$$ which is not the most convenient.

However, for cases, what you could do is to use parametric splines $$x=f(L)$$ and $$y=g(L)$$

Edit (six years later)

If you are optimistic and want to solve for $$u$$ the equation $$L=b\,r\, E\left(u\left|1-\frac{a^2}{b^2}\right. \right)$$ let $$A=\frac L{b\,r} \qquad \text{and}\qquad k=1-\frac{a^2}{b^2}$$ and use the series expansion $$E(u|k)=u-\frac{k}{6}u^3+\frac{(4-3 k) k }{120} u^5+O\left(u^7\right)$$ and power series reversion would give $$u=A+\frac{ k}{6}A^3+\frac{k (13 k-4)}{120} A^5 +O\left(A^7\right)$$

• @ mayi Attempting this parametrically as $( x(s),y(s))= \int ( \cos \phi(s), \sin \phi (s)) \,ds$ bbgodfrey's answer to my question.[ArcSoln][1] [1]: mathematica.stackexchange.com/questions/147688/… Jun 12, 2017 at 8:22
• @Narasimham. Thanks for the link ! It is very interesting. Jun 12, 2017 at 8:28
• Glad you liked it, am still looking at further generalization including rotations in the plane.. Jun 12, 2017 at 11:28