Question

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

• A. $2\sqrt{2}$
• B. $\sqrt{2}$
• C. $4\sqrt{2}$
• D. $8\sqrt{2}$

Answer 1

Answer: (B)

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 $=(\pm ae,0)$
$=(\pm 2\sqrt{2},0)=(2\sqrt{2},0)$
(taken positive sign)
The nearest distance of focus from directrix
$=\left|\frac{2\sqrt{2}-\sqrt{2}}{\sqrt{1}}\right|=\sqrt{2}$