Step 1
GIven polar equation is:
\(\displaystyle{r}=\ {\frac{{{10}}}{{{5}\ +\ {2}\ {\cos{\theta}}}}}\)
The graph of the given polar equation is:
Step 2
Therefore the given polar equation represents an ellipse with major axis as the x-axis, center at \(\displaystyle{\left(-{1},\ {0}\right)}.\)

\(\displaystyle{r}=\ {\frac{{{10}}}{{{\left({5}\ +\ {2}\ {\cos{\theta}}\right\rbrace}}}}\)

\(\displaystyle{r}=\ {\frac{{{2}}}{{{\left[{1}\ +\ {\left({\frac{{{2}}}{{{5}}}}\right)}\ {\cos{\theta}}\right]}}}}\)

Eccentricity \(\displaystyle{e}=\ {\frac{{{2}}}{{{5}}}}\ {<}\ {1}\)</span> Therefore the given equation represents an ellipse. Now we get, \(\displaystyle{2}={e}{p}\)

\(\displaystyle{2}=\ {\left({\frac{{{2}}}{{{5}}}}\right)}{p}\)

\(\displaystyle{p}={5}\) The directrix is \(\displaystyle{y}={5}.\)

\(\displaystyle{r}=\ {\frac{{{10}}}{{{\left({5}\ +\ {2}\ {\cos{\theta}}\right\rbrace}}}}\)

\(\displaystyle{r}=\ {\frac{{{2}}}{{{\left[{1}\ +\ {\left({\frac{{{2}}}{{{5}}}}\right)}\ {\cos{\theta}}\right]}}}}\)

Eccentricity \(\displaystyle{e}=\ {\frac{{{2}}}{{{5}}}}\ {<}\ {1}\)</span> Therefore the given equation represents an ellipse. Now we get, \(\displaystyle{2}={e}{p}\)

\(\displaystyle{2}=\ {\left({\frac{{{2}}}{{{5}}}}\right)}{p}\)

\(\displaystyle{p}={5}\) The directrix is \(\displaystyle{y}={5}.\)