Another polar / rectangular simplification

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SUMMARY

The discussion focuses on converting the polar equation r = 1/(1 - cos(θ)) into rectangular coordinates. The user, Sparky, initially struggles with the algebraic manipulation but ultimately arrives at the correct rectangular form, y² = 1 + 2x, by squaring both sides after isolating the square root. The key steps involved using the relationships r = √(x² + y²) and cos(θ) = x/r to facilitate the conversion. This process highlights the importance of algebraic skills in transforming polar equations to rectangular coordinates.

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  • Understanding of polar coordinates and their relationship to rectangular coordinates.
  • Proficiency in algebraic manipulation, including squaring both sides of an equation.
  • Familiarity with trigonometric functions, specifically cosine.
  • Knowledge of the Pythagorean theorem as it applies to coordinate systems.
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Students studying mathematics, particularly those focusing on algebra and coordinate geometry, as well as educators looking for examples of polar to rectangular conversions.

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Homework Statement



Convert into rectangular coordinates:

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



Homework Equations





The Attempt at a Solution



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


[tex] r –r(cos) = 1[/tex]
(why can't I get a minus sign to display correctly? - I'm trying to show r - r*cos = 1)

I used
[tex] r = \sqrt {x^2+y^2}[/tex]
and

[tex] cos = \frac {x}{r}<br /> <br /> x = (r)(cos)[/tex]

resulting combinations gives:
[tex] \sqrt {x^2+y^2} – x = 1[/tex]
(I think) - and again - I'm having a problem (maybe just on my end) displaying a minus sign. I'm seeing "8211" on the screen for the minus sign.

I'm trying to display sqrt(x^2+y^2) - x = 1.

The book gets

[tex] y^2} = 1 + 2x[/tex]

I‘ve tried some various algebra stuff but am not getting close to the book’s answer.

Thanks
-Sparky
 
Last edited:
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Never mind, I got it - I don't know how to delete the post (if I'm allowed)

I moved x over and squared both sides.

[tex] \sqrt {x^2+y^2} = 1 + x[/tex]

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

[tex] {x^2+y^2} = 1 + 2x + x^2[/tex]

[tex] y^2} = 1 + 2x[/tex]
 

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