The distance of the focus of \(x^{2}-y^{2}=4\), form the directrix which is nearer to it, is

Given equation of hyperbola is
\(\mathrm{x}^{2}-\mathrm{y}^{2}=4\)
or \(\quad \frac{\mathrm{x}^{2}}{4}-\frac{\mathrm{y}^{2}}{4}=1 \quad\left(\because \mathrm{a}^{2}=\mathrm{b}^{2}=4\right)\)
Now, \(\quad \mathrm{b}^{2}=\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)\) \[ \begin{aligned} 4 &=4\left(\mathrm{e}^{2}-1\right) \Rightarrow \mathrm{e}=\pm \sqrt{2} \end{aligned} \]
Equation of directrix \(\mathrm{x}=\pm \frac{\mathrm{a}}{\mathrm{e}}\)
\(\Rightarrow \quad \mathrm{x}=\pm \frac{2}{\sqrt{2}}\)
\(\Rightarrow \quad \quad \mathrm{x}=\pm \sqrt{2} \Rightarrow \mathrm{x}-\sqrt{2}=0\)
and the focus \(=(\pm \mathrm{ae}, 0)\)
\(=(\pm 2 \sqrt{2}, 0)=(2 \sqrt{2}, 0)\)
The nearest distance of focus from directrix \(\quad=\left|\frac{2 \sqrt{2}-\sqrt{2}}{\sqrt{1}}\right|=\sqrt{2}\)