# Perimeter of an ellipse from radii, and vice-versa

### Perimeter of an ellipse from radii, and vice-versa

Hi there,

I'm new here. Greetings.

I need to figure out how to calculate the perimeter, or circumference, of an ellipse using the radii, and vice-versa: how to figure out the radii given the perimeter and the ratio of the two radii to each other.

I found an equation for the perimeter, or circumference here: http://www.numericana.com/answer/ellipse.htm#elliptic
given the radii x and y, where x>y:

P=3.1415*(((2*((x^2)+(y^2)))-(((x-y)^2)/2))^.5)

The result will be an approximation but it will be close enough.

Any chance you could help me solve this equation for x in terms of C and y, and then also for y in terms of C and x?
I’m pretty rusty on my equations!

Many Thanks,

rgesh
San Francisco, CA
Guest

### Re: Perimeter of an ellipse from radii, and vice-versa

You have $$P= \pi (2(x^2- y^2)- (x^2- y^2)/2 )^{0,5}$$
(You have a lot more parentheses than you need but too many is better than not enough!)

The first thing you will have to do is get rid of that square root. And that means you will have to square both sides of the equation: $$P^2= \pi^2(2(x^2- y^2)- (x^2- y^2)/2)$$.

Divide both sides of the equation by $$\pi^2$$ and multiply by 2:
$$\frac{2P^2}{\pi^2}= 4(x^2- y^2)- (x^2- y^2)= 4x^2- 4y^2- x^2+ y^2= 3x^2- 3y^2$$.

Add $$3y^2$$ to both sides of the equation:
$$\frac{2P^2}{\pi^2}+ 3y^2= 3x^2$$.

Divide both sides of the equation by 3:
$$\frac{2P^2}{3\pi^2}+ y^2= x^2$$.

Finally, take the square root of both sides:
$$x= \pm\sqrt{\frac{2P^2}{3\pi^2}+ y^2}$$.
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