Question

The distance between the directrices of the hyperbola $x = 8$ $\sec\theta$, $y = 8\tan\theta$is

• A. $16\sqrt{2}$
• B. $\sqrt{2}$
• C. $h^{2} > ab$
• D. $4,\sqrt{2}$

Answer 1

Answer: (C)

$h^{2} > ab$

Answer 2

Equation of hyperbola is $x = 8\sec\theta,y = 8\tan\theta$ ⇒ $\frac{x}{8} = \sec\theta$, $\frac{y}{8} = \tan\theta$

$\therefore\sec^{2}\theta - \tan^{2}\theta = 1$ ⇒ $\frac{x^{2}}{8^{2}} - \frac{y^{2}}{8^{2}} = 1$

Here $a = 8,b = 8$. Now $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{8^{2}}{8^{2}}} = \sqrt{2}$

$\therefore$ Distance between directrices = $\frac{2a}{e}$ = $\frac{2 \times 8}{\sqrt{2}} = 8\sqrt{2}$