Question

The eccentricities of conic sections with one focus at the origin and the directrix corresponding and sketch a graph:
\(\displaystyle{e}={2},\)
\(directrix \displaystyle{r}=-{2} \sec{\theta}.\)

Answer 1

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