Step 1

We convert the equation into rectangular form first

\(\displaystyle{r}=\frac{4}{{{1}- \cos{\theta}}}\)

\(\displaystyle{r}{\left({1}- \cos{\theta}\right)}={4}\)

\(\displaystyle{r}-{r} \cos{\theta}={4}\)

\(\displaystyle\sqrt{{{x}^{2}+{y}^{2}}}-{x}={4}\)

\(\displaystyle\sqrt{{{x}^{2}+{y}^{2}}}={x}+{4}\)

\(\displaystyle{x}^{2}+{y}^{2}={\left({x}+{4}\right)}^{2}\)

\(\displaystyle{x}^{2}+{y}^{2}={x}^{2}+{8}{x}+{16}\)

\(\displaystyle{y}^{2}={8}{x}+{16}\)

\(\displaystyle{y}^{2}={8}{\left({x}+{2}\right)}\)

This is a parabola.

Step 2

Compare the equation with the standard form

\(\displaystyle{\left({y}-{k}\right)}^{2}={4}{p}{\left({x}-{h}\right)}\)

\(\displaystyle{h}=-{2},{k}={0}\)

\(\displaystyle{8}{p}={2}\)

\(\displaystyle{\quad\text{or}\quad},{p}={2}\)

Directrix \(\displaystyle={x}={h}-{p}{\quad\text{or}\quad}{x}=-{2}-{2}{\quad\text{or}\quad}{x}=-{4}\)

Answer: Parabola, \(\displaystyle{x}=-{4}\)

We convert the equation into rectangular form first

\(\displaystyle{r}=\frac{4}{{{1}- \cos{\theta}}}\)

\(\displaystyle{r}{\left({1}- \cos{\theta}\right)}={4}\)

\(\displaystyle{r}-{r} \cos{\theta}={4}\)

\(\displaystyle\sqrt{{{x}^{2}+{y}^{2}}}-{x}={4}\)

\(\displaystyle\sqrt{{{x}^{2}+{y}^{2}}}={x}+{4}\)

\(\displaystyle{x}^{2}+{y}^{2}={\left({x}+{4}\right)}^{2}\)

\(\displaystyle{x}^{2}+{y}^{2}={x}^{2}+{8}{x}+{16}\)

\(\displaystyle{y}^{2}={8}{x}+{16}\)

\(\displaystyle{y}^{2}={8}{\left({x}+{2}\right)}\)

This is a parabola.

Step 2

Compare the equation with the standard form

\(\displaystyle{\left({y}-{k}\right)}^{2}={4}{p}{\left({x}-{h}\right)}\)

\(\displaystyle{h}=-{2},{k}={0}\)

\(\displaystyle{8}{p}={2}\)

\(\displaystyle{\quad\text{or}\quad},{p}={2}\)

Directrix \(\displaystyle={x}={h}-{p}{\quad\text{or}\quad}{x}=-{2}-{2}{\quad\text{or}\quad}{x}=-{4}\)

Answer: Parabola, \(\displaystyle{x}=-{4}\)