The distance between the directrices of the hyperabola `x=8 sec theta, y =8 tan theta` is-

Options

A. `4sqrt(2)`
B. `6sqrt(2)`
C. `8sqrt(2)`
D. `10sqrt(2)`

Answer

To find the distance between the directrices of the hyperbola given by the parametric equations \( x = 8 \sec \theta \) and \( y = 8 \tan \theta \), we can follow these steps:

### Step 1: Convert the Parametric Equations
The given parametric equations are:
\[ x = 8 \sec \theta \]
\[ y = 8 \tan \theta \]
We can express \( \sec \theta \) and \( \tan \theta \) in terms of \( x \) and \( y \):
\[ \sec \theta = \frac{x}{8}, \quad \tan \theta = \frac{y}{8} \]

### Step 2: Use the Identity
We know the identity:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
Substituting the expressions for \( \sec \theta \) and \( \tan \theta \):
\[ \left(\frac{x}{8}\right)^2 - \left(\frac{y}{8}\right)^2 = 1 \]

### Step 3: Simplify the Equation
Multiplying through by \( 64 \) (which is \( 8^2 \)):
\[ x^2 - y^2 = 64 \]
This can be rearranged to fit the standard form of a hyperbola:
\[ \frac{x^2}{64} - \frac{y^2}{64} = 1 \]

### Step 4: Identify \( a \) and \( b \)
From the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we can identify:
\[ a^2 = 64 \quad \Rightarrow \quad a = 8 \]
\[ b^2 = 64 \quad \Rightarrow \quad b = 8 \]

### Step 5: Calculate the Eccentricity \( e \)
The eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{64}{64}} = \sqrt{2} \]

### Step 6: Find the Distance Between the Directrices
The distance between the directrices of a hyperbola is given by the formula:
\[ \text{Distance} = \frac{2a}{e} \]
Substituting the values of \( a \) and \( e \):
\[ \text{Distance} = \frac{2 \times 8}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 16 \cdot \frac{\sqrt{2}}{2} = 8\sqrt{2} \]

### Final Answer
Thus, the distance between the directrices of the hyperbola is:
\[ \boxed{8\sqrt{2}} \]