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

$8\sqrt{2}$

Explanation for the correct option:

Determining the distance:

Given: $x=8\sec \theta ,y=8\tan \theta$

Firstly, we need to eliminate $\theta$:

$\therefore \sec \theta =\frac{x}{8}\text{and}\tan \theta =\frac{y}{8}$

We know that: ${\sec }^2\theta -{\tan }^2\theta =1$, therefore putting the values we have:

$$\begin{gathered} \Rightarrow \frac{x^2}{8^2}-\frac{y^2}{8^2}=1 \\ \Rightarrow \frac{x^2}{64}-\frac{y^2}{64}=1...\left(1\right) \end{gathered}$$

On Comparing the equation with standard hyperbola:

$$\begin{gathered} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ a=8 \\ b=8 \end{gathered}$$

Calculating the Eccentricity :

$$\begin{gathered} e=\sqrt{1+\frac{b^2}{a^2}} \\ =\sqrt{1+\frac{8^2}{8^2}} \\ =\sqrt{2} \end{gathered}$$

The distance between the directrices of the hyperbola is give as:

$$\begin{gathered} =\frac{a}{e}-\left(\frac{-a}{e}\right) \\ =\frac{2a}{e} \\ =\frac{2\times 8}{\sqrt{2}}[\because a=8,e=\sqrt{2}] \\ =8\sqrt{2} \end{gathered}$$

Hence, the correct answer is option (A)