Question

The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is
(a) $8\sqrt{2}$
(b) $16\sqrt{2}$
(c) $4\sqrt{2}$
(d) $6\sqrt{2}$

Answer 1

(a) $8\sqrt{2}$

We have:
$x=8\sec \theta ,y=8\tan \theta$

$$\begin{gathered} \mathrm{On}\mathrm{squaring}\mathrm{and}\mathrm{subtracting}: \\ {x}^2-y^2=8{\sec }^2\theta -8{\tan }^2\theta \\ \Rightarrow x^2-y^2=8 \\ \Rightarrow \frac{x^2}{8}-\frac{y^2}{8}=1 \end{gathered}$$

∴ a = b = 8
Distance between the directrices of the hyperbola = $\frac{2a^2}{\sqrt{a^2+b^2}}$
$$\begin{gathered} \mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{directrices}=\frac{2\times 64}{\sqrt{64+64}} \\ =\frac{128}{8\sqrt{2}} \\ =\frac{16}{\sqrt{2}} \\ =8\sqrt{2} \end{gathered}$$