# The radius of an ellipse from the origin.

1. Mar 21, 2008

### Erez

Hello,
given (x^2)/(a^2) + (y^2)/(b^2) = 1.
and using polar coordinates x=rcos(phi) , y=rsin(phi),
equating gives r^2 = 1/[(cos^2(phi)/a^2) + (sin^2(phi)/b^2)].
or if we leave b in the nominator :
r= b/[(sin^2(phi)+(b^2/a^2)cos^2(phi)]^1/2.

-could someone give a hint as to how the demoninator of the last expression can be turned into [1 - (e^2)cos^2(phi)]^1/2 where e is the eccentricity of the ellipse?
and what is the value of e?

thank you.

Last edited: Mar 21, 2008
2. Mar 21, 2008

$$e^2=1-\frac{b^2}{a^2}$$

And by the way this is introductory physics section.

3. Mar 21, 2008

### tiny-tim

Welcome to PF!

Hi Erez! Welcome to PF!

Hint: use cos^2(phi) + sin^(phi) = 1.

Then … ?

4. Mar 21, 2008

### Erez

Hello,
I am surprised I recived a reply so quick,
thank you.
p.s. in what category/section would this post belong ?

5. Mar 21, 2008

Pre calculus mathematics, perhaps

6. Mar 21, 2008

### tiny-tim

Hi Erez

Well, it was a short question, clearly stated, without loads of irrelevant gumph to read through … and some of us give questions like that priority!

Well, this is just geometry, so it should really have gone into "Precalculus Mathematics", which is defined as "All math courses prior to calculus"