\(\displaystyle{e}={2},{r}=-{2} \sec{\theta}\)

Since \(\displaystyle{e}>{1}\), therefore the conic is a hyperbola.

Now consider the directrix,

\(\displaystyle{r}=-{2} \sec{\theta}\)

\(\displaystyle\Rightarrow{r}=-\frac{2}{{ \cos{\theta}}}\)

\(\displaystyle\Rightarrow{r} \cos{\theta}=-{2}\)

\(\displaystyle\Rightarrow{x}=-{2}\)

Now comparing whith \(\displaystyle{x}=-{p},\) we get

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

Therefore equation is,

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

\(\displaystyle\Rightarrow{r}=\frac{{{2}\cdot{2}}}{{{1}-{2} \cos{\theta}}}\)

\(\displaystyle\Rightarrow{r}=\frac{4}{{{1}-{2} \cos{\theta}}}\)

Now the graph is,

Since \(\displaystyle{e}>{1}\), therefore the conic is a hyperbola.

Now consider the directrix,

\(\displaystyle{r}=-{2} \sec{\theta}\)

\(\displaystyle\Rightarrow{r}=-\frac{2}{{ \cos{\theta}}}\)

\(\displaystyle\Rightarrow{r} \cos{\theta}=-{2}\)

\(\displaystyle\Rightarrow{x}=-{2}\)

Now comparing whith \(\displaystyle{x}=-{p},\) we get

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

Therefore equation is,

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

\(\displaystyle\Rightarrow{r}=\frac{{{2}\cdot{2}}}{{{1}-{2} \cos{\theta}}}\)

\(\displaystyle\Rightarrow{r}=\frac{4}{{{1}-{2} \cos{\theta}}}\)

Now the graph is,