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Find the Cartesian equation corresponding to $r = \frac{5}{3-2\cos(\theta)}$

I got it into the form:

$(x^2 + y^2)(3-2x)^2 = 25$

and can see that maybe the equation of a circle will appear, but I can't seem to get much further without complicating things. I have a note saying that completing the square was used but I can't see how or where.

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Marvis has given a correct solution. Looking at your modified form, it appears to me that you made the mistake of substituting $x$ for $2\cos\theta$. Remember, the correspondence is $x=r\cos\theta$. –  Brian M. Scott May 18 '12 at 5:58
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1 Answer 1

up vote 4 down vote accepted

Rearranging we get that $3r - 2r\cos(\theta) = 5$. Note that $r = \sqrt{x^2 + y^2}$ and $r \cos(\theta) = x$.

This gives us $3 \sqrt{x^2+y^2} -2x = 5$.

Hence, \begin{align} 9(x^2+y^2) & = (2x+5)^2\\ 9x^2+9y^2 & = 4x^2 + 20x + 25\\ 5x^2 -20x + 9y^2 & = 25\\ 5(x-2)^2 + 9y^2 & = 45\\ \frac{(x-2)^2}{3^2} + \frac{y^2}{\left(\sqrt{5} \right)^2} & = 1 \end{align} The above is an ellipse with center $(2,0)$ with semi-major axis along $x$ of length $3$ and semi-minor axis along $y$ of length $\sqrt{5}$.

EDIT

In general, $$r(\theta) = \dfrac{a(1-e^2)}{1 \pm e \cos(\theta)}$$ represents an ellipse whose semi-major axis is $a$ and eccentricity $e$.

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Based on the problem statement, I don't see why you're not done at $3 \sqrt{x^2+y^2} -2x = 5$. It may be worth noting that the squaring is reversible, so equivalence to the original equation is not lost. –  Jonas Meyer May 18 '12 at 5:57
    
@JonasMeyer True. But I wanted to get it in the canonical ellipse form. –  user17762 May 18 '12 at 5:59
    
Yes, it is nice. One way to see that the squaring doesn't affect the solution set is to note that $3\sqrt{x^2+y^2}=-2x-5$ has no solutions, which follows from $3\sqrt{x^2+y^2}\geq 3|x|\geq 2|x|\geq -2x>-2x-5$. –  Jonas Meyer May 18 '12 at 6:12
    
@JonasMeyer Nice. When I actually did the squaring initially, I did not consider the fact that this could introduce artificial solutions, till you pointed it out in your earlier comment. Thanks. A good thing to remember. –  user17762 May 18 '12 at 6:20
    
You're welcome. I wondered if I was being hypervigilant, and we're safe here anyway. A worse case would be if we weren't careful in trying to simplify instead the equation $3\sqrt{x^2+y^2}=-2x-5$ without first noticing that it has no solutions (with $x,y\in\mathbb R$). –  Jonas Meyer May 18 '12 at 6:27
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