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Q. The distance of the focus of $x^2-y^2=4$ from the directrix which is nearer to it, is

KCETKCET 2010Application of Derivatives

Solution:

Given equation of hyperbola is
$x^{2}-y^{2}=4$
or $\frac{x^{2}}{4}-\frac{y^{2}}{4}=1$
$\left(\because a^{2}=b^{2}=4\right)$
Now, $b^{2}=a^{2}\left(e^{2}-1\right)$
$4=4\left(e^{2}-1\right)$
$\Rightarrow e=\pm \sqrt{2}$
Equation of directrix $x=\pm \frac{a}{e}$
$\Rightarrow x=\pm \frac{2}{\sqrt{2}}$
$\Rightarrow x=\pm \sqrt{2}$
$\Rightarrow x-\sqrt{2}=0$
and the focus $=(\pm a e, 0)$
$=(\pm 2 \sqrt{2}, 0)=(2 \sqrt{2}, 0)$
(taken positive sign)
The nearest distance of focus from directrix
$=\left|\frac{2 \sqrt{2}-\sqrt{2}}{\sqrt{1}}\right|=\sqrt{2}$