The radius of an ellipse from the origin.

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.

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and what is the value of e
$$e^2=1-\frac{b^2}{a^2}$$

And by the way this is introductory physics section.

tiny-tim
Homework Helper
Welcome to PF!

Hi Erez! Welcome to PF!

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

Then … ?

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

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