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Problem: Transform the following equation from polar to rectangular coordinates. \begin{eqnarray*} \rho &=& \frac{2}{1 - \cos \theta} \\ \end{eqnarray*}

Answer:

Recall that: \begin{eqnarray*} \rho &=& {(x^2+y^2)} ^ {\frac{1}{2}} \\ \cos \theta &=& \frac{x}{\rho} =\frac{x}{(x^2+y^2)^{\frac{1}{2}}} \\ \end{eqnarray*} This gives us: \begin{eqnarray*} {(x^2+y^2)}^{\frac{1}{2}} &=& \frac{2}{1 - \frac{x}{(x^2+y^2)^{\frac{1}{2}}}} \\ 2 &=& {(x^2+y^2)^\frac{1}{2}} - x \\ x + 2 &=& (x^2+y^2)^\frac{1}{2} \\ (x + 2)^2 &=& x^2+y^2 \\ x^2 + 4x + 4 &=& x^2+y^2 \\ 4x + 4 &=& y^2 \\ y^2 &=& 4x + 4 \end{eqnarray*} However, the book gets: \begin{eqnarray*} y^2 &=& 4(x + 2) \end{eqnarray*} What am I missing?

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    $\begingroup$ A typo in the book? $\endgroup$ – Aretino Sep 5 '15 at 22:16
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Let's check with $\theta=60^\circ$ so that $\rho=4$. Then: $x=2$, $y=2\sqrt3$, $y^2=12$. But $4(x+2)=16$ and $4x+4=12$. So you are right and the book is wrong.

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    $\begingroup$ Perhaps even easier, check at $\theta = \frac\pi2$ ($90$ degrees). Then $\cos\theta = 0$, $x=0$, and $\rho = 2$, hence $y = 2$ and $y^2 = 4$. The book says $y^2$ would be $8$ when $x = 0$. $\endgroup$ – David K Sep 5 '15 at 22:46

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