Question

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. $6\sqrt{2}$

Answer 1

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

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

The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by,

$e=\sqrt{1+\frac{b^2}{a^2}}$ and the distance between the directrices is

given by $\frac{2a}{e}.$

Here, $a=8$ and $b=8$ since ${\sec }^2\theta -{\tan }^2\theta =1$

Therefore, the distance becomes $\frac{2\times 8}{\sqrt{1+1}}=8\sqrt{2}$.

Hence, option A is correct.