Q. The distance between the directrices of the hyperbola $x = 8\sec\theta,\ y = 8\ tan\ \theta$ is
A) $8\sqrt{2}$
B) $16\sqrt{2}$
C) $4\sqrt{2}$
D) $6\sqrt{2}$
Explanation
(a)
Given, $x^{2} = 64\sec^{2}{\theta,y^{2} = 64\tan^{2}\theta}$
$\therefore\ x^{2} - y^{2} = 64\ (\sec^{2}{\theta - \tan^{2}{\theta)}}$
$\Rightarrow \ x^{2} - y^{2} = 64$
$\therefore$ It is a rectangular hyperbola whose eccentricity is$\ \sqrt{2}$
The distance between directrices $= \frac{2a}{e} = \frac{2 \times 8}{\sqrt{2}} = 8\sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.