# How do you find the circumference of the ellipse x^2+4y^2=1?

Feb 15, 2017

Using numerical techniques, we can get a approximation for this as:

$C = 4.8442$

#### Explanation:

Although this seems like quite a simple question, the answer is actually ridiculous complicated.

We need to first put the ellipse equation in standard form:

${x}^{2} + 4 {y}^{2} = 1$
$\therefore {\left(\frac{x}{1}\right)}^{2} + {\left(\frac{y}{\frac{1}{2}}\right)}^{2} = 1$

Comparing with the standard equation;

${\left(\frac{x}{a}\right)}^{2} + {\left(\frac{y}{b}\right)}^{2} = 1$

We can identify this as an ellipse with semi-major axis $b = \frac{1}{2}$ and semi-minor axis $a = 1$, and the eccentricity of the ellipse is given by:

 e=sqrt(1-(b/a)^2))
$\setminus = \sqrt{1 - {\left(\frac{\frac{1}{2}}{1}\right)}^{2}}$
$\setminus = \sqrt{\frac{3}{4}}$
$\setminus = \frac{1}{2} \sqrt{3}$

Then the exact circumference is given by:

$C = 4 a E \left(e\right)$

where $E \left(e\right)$ is a complete elliptic integral of the second kind,

$E \left(e\right) = {\int}_{0}^{\frac{\pi}{2}} \setminus \sqrt{1 - {e}^{2} {\sin}^{2} \theta} \setminus d \theta$

Using numerical techniques, we can get a approximation for this as:

$C = 4.8442$