If $\mathrm{\alpha }$ is a variable complex number such that |$\mathrm{\alpha }$|>1(|$\mathrm{\alpha }$|  is a constant) and $\mathrm{z}=\mathrm{\alpha }+\frac{1}{\mathrm{\alpha }}$ lies on a conic then

$\text{ Let }\left|\mathrm{\alpha }\right|=\mathrm{r}>1\text{ and }\mathrm{\alpha }=\mathrm{rcis}\mathrm{\theta }\text{ then }$
$\mathrm{z}=\mathrm{x}+\mathrm{iy}=\mathrm{\alpha }+\frac{1}{\mathrm{\alpha }}=\mathrm{rcis}\mathrm{\theta }+\frac{\mathrm{cis}(-\mathrm{\theta })}{\mathrm{r}}$
$\Rightarrow \mathrm{x}=\left(\mathrm{r}+\frac{1}{\mathrm{r}}\right)\cos \mathrm{\theta }\text{ and }\mathrm{y}=\left(\mathrm{r}-\frac{1}{\mathrm{r}}\right)\sin \mathrm{\theta }$
$\text{ Eliminating }\mathrm{\theta }\text{ gives }\frac{{\mathrm{x}}^2}{{\left(\mathrm{r}+\frac{1}{\mathrm{r}}\right)}^2}+\frac{{\mathrm{y}}^2}{{\left(\mathrm{r}-\frac{1}{\mathrm{r}}\right)}^2}=1$
Which is an ellipse . $\mathrm{a}=\mathrm{r}+\frac{1}{\mathrm{r}},\mathrm{b}=\mathrm{r}-\frac{1}{\mathrm{r}}$
$(\mathrm{r}=|\mathrm{\alpha }|>1\Rightarrow \mathrm{a}>\mathrm{b});\therefore \mathrm{e}=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\frac{2}{\mathrm{r}+\frac{1}{\mathrm{r}}},$
distance between foci = 2ae = 4
$\Rightarrow \mathrm{distance} \mathrm{between} \mathrm{directrices} = \frac{2\mathrm{a}}{\mathrm{e}}$