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The radius of an ellipse from the origin.

  • Thread starter Erez
  • Start date
2
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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:

Answers and Replies

and what is the value of e
[tex]e^2=1-\frac{b^2}{a^2}[/tex]

And by the way this is introductory physics section.
 
tiny-tim
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Homework Helper
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Welcome to PF!

Hi Erez! Welcome to PF! :smile:

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

Then … ? :smile:
 
2
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Hello,
I am surprised I recived a reply so quick,
thank you.
p.s. in what category/section would this post belong ?
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
Hello,
I am surprised I recived a reply so quick,
thank you.
p.s. in what category/section would this post belong ?
Hi Erez :smile:

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! :wink:

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

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