Answer
a) The equation $r=\frac{1}{3-3\cos \theta }$ can be written in the form of $9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}}$
b) The equation in the rectangular coordinate form is $9{{y}^{2}}=1+6x$ and the curve is a parabola.
Work Step by Step
(a)
Let us consider the equation:
$r=\frac{1}{3-3\cos \theta }$
We cross multiply the terms:
$\begin{align}
& r\left( 3-3\cos \theta \right)=1 \\
& 3r-3r\cos \theta =1 \\
& 3r=1+3r\cos \theta
\end{align}$
We square both sides of this equation:
$\begin{align}
& 3r=1+3r\cos \theta \\
& 9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}}
\end{align}$
Thus, the equation $r=\frac{1}{3-3\cos \theta }$ can be written in the form of $9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}}$.
(b)
Let us consider the equation:
$9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}}$
In polar form, $x=r\cos \theta $ and $y=r\sin \theta $
Therefore, ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$
Then, put these values in the equation
$\begin{align}
& 9\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( 1+3x \right)}^{2}} \\
& 9{{x}^{2}}+9{{y}^{2}}=1+9{{x}^{2}}+6x \\
& 9{{y}^{2}}=1+6x
\end{align}$
Thus, the provided equation is a parabola opening towards the right.
