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

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Answer 1 · 2017-02-15T00:41:20

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

$C = 4.8442$ $C = 4.8442$

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:

$x^{2} + 4 y^{2} = 1$ $x^2+4y^2=1$
$:. (x/1)^2+(y/(1/2))^2=1$

Comparing with the standard equation;

$(x/a)^2+(y/b)^2=1$

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

$e=sqrt(1-(b/a)^2))$
$\ = sqrt(1-((1/2)/1)^2)$
$\ = sqrt(3/4)$
$\ = 1/2sqrt(3)$

Then the exact circumference is given by:

$C=4aE(e)$

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

$E(e) = int_0^(pi/2) \ sqrt(1-e^2sin^2 theta) \ d theta$

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

$C = 4.8442$

How do you find the circumference of the ellipse x^2+4y^2=1x2+4y2=1x^2+4y^2=1?