Question

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

• A. $8\sqrt{2}$
• B. $6\sqrt{2}$
• C. $\sqrt{2}$
• D. $7$

Answer 1

The correct option is A $8\sqrt{2}$

Given,

$x=8\sec \varphi$

$\Rightarrow \sec \varphi =\frac{x}{8}$.....(1)

$y=8\tan \varphi$

$\Rightarrow \tan \varphi =\frac{y}{8}$......(2)

wkt,

${\sec }^2\varphi -{\tan }^2\varphi =1$

using (1) and (2), we get,

$\frac{x^2}{64}-\frac{y^2}{64}=1$

equation of hyperbola,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

comparing the equations, we get,

$a=8=b$

Eccentricity,

$b^2=a^2(e^2-1)$

$\Rightarrow 8^2=8^2(e^2-1)$

$1=e^2-1$

$e=\sqrt{2}$

The distance between their directrix

$=\frac{a}{e}-(-\frac{a}{e})$

$=2\frac{a}{e}$

$=2\frac{8}{\sqrt{2}}$

$=8\sqrt{2}$