Polar Conic to Cartesian

  • Thread starter Knissp
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Homework Statement


The equation of a conic in polar coordinates is:
[tex] r = \frac{r_o}{1-\epsilon cos(\theta)}[/tex].

[tex]\epsilon [/tex] is the eccentricity, 0 for a circle, (0,1) for an ellipse, 1 for a parabola, and >1 for a hyperbola.

What is this equation expressed in Cartesian coordinates?


Homework Equations


[tex]r=(x^2+y^2)^{1/2}[/tex]
[tex]x=r cos(\theta)[/tex]


The Attempt at a Solution


[tex] r = \frac{r_o}{1-\epsilon cos(\theta)}[/tex]

[tex] r (1-\epsilon cos(\theta)) = r_o[/tex]

[tex] r - r cos(\theta) \epsilon = r_o[/tex]

[tex] r - x \epsilon = r_o[/tex]

[tex] r = r_o + x \epsilon[/tex]

[tex] (r)^2 = (r_o + x \epsilon)^2 [/tex]

[tex] x^2 + y^2 = r_o^2 + 2 (x \epsilon) r_o + (x \epsilon)^2 [/tex]

[tex] x^2 - (x \epsilon)^2 - 2 \epsilon r_o x + y^2 = r_o^2 [/tex]

[tex] (1 - \epsilon^2)x^2 - 2 \epsilon r_o x + y^2 = r_o^2 [/tex]

My textbook says this equation should be:

[tex] (1 - \epsilon^2)x^2 - 2 r_o x + y^2 = r_o^2 [/tex]

(Notice that we differ on the coefficient of x. Is the textbook missing an epsilon by a misprint? or did I mess up somewhere?

Thanks for the help.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi Knissp! :smile:

(have an epsilon: ε :wink:)

No, you're correct …

if we put ε = 0, for a circle, then your equation is obviously correct, but the book's equation is (x - r0)2 + y2 = 2r02, which is a circle, but not the right one! :wink:
 

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