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Ellipse and Kepler's Law in Polar Coordinates 
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#1
Apr1312, 04:02 PM

P: 121

Greetings everyone,
I am having difficulties grasping the polar form of the ellipse equation, and there seems to be more than one way to express an ellipse in this form, if I am not mistaken. For example on the following webpage http://farside.ph.utexas.edu/teachin...s/node155.html the ellipse is represented in a different way than I am accustomed. How can I convert this into other forms? Thanks 


#2
Apr1412, 02:12 AM

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One option with the equation for an ellipse is whether to set the origin at a focus or at the centre. The link you provided gives the polar equation with a focus as origin. Do you have another link for contrast?



#3
Apr1412, 12:07 PM

P: 121

I do not have at the moment, I remember coming across one a year ago in a text I read. Do you have any site that I can learn conics and their equations in polar coordinates ?



#4
Apr1412, 06:08 PM

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Ellipse and Kepler's Law in Polar Coordinates
Consider a string length 2L with endpoints fixed at (A, 0), (+A, 0) (XY coords).
With polar coordinates at the same origin, I get r^{2}(L^{2}A^{2}.cos^{2}(θ)) = L^{2}(L^{2}A^{2}) Does that look familiar? Converting back to XY: (x^{2}+y^{2})L^{2}  x^{2}.A^{2} = L^{2}(L^{2}A^{2}) or x^{2}/L^{2} + y^{2}/(L^{2}A^{2}) = 1 Which does indeed appear to be an ellipse centred at the origin. 


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