<s>[INST] Polar to Rectangular Coordinates 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?
 [/INST] 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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