Distance between the directrices of the hyperbola `(x^(2))/(49)-(y^(2))/(16)` =1 is

Options

A. `sqrt(65)/(7)`
B. `(49)/sqrt(65)`
C. `sqrt(33)/(4)`
D. `(98)/sqrt(65)`

Answer

To find the distance between the directrices of the hyperbola given by the equation \(\frac{x^2}{49} - \frac{y^2}{16} = 1\), we can follow these steps:

### Step 1: Identify \(a^2\) and \(b^2\)
From the hyperbola equation, we can identify:
- \(a^2 = 49\)
- \(b^2 = 16\)

### Step 2: Calculate \(a\)
To find \(a\), we take the square root of \(a^2\):
\[ a = \sqrt{49} = 7 \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(b^2\) and \(a^2\):
\[ e = \sqrt{1 + \frac{16}{49}} = \sqrt{1 + \frac{16}{49}} = \sqrt{\frac{49 + 16}{49}} = \sqrt{\frac{65}{49}} = \frac{\sqrt{65}}{7} \]

### Step 4: Calculate the distance between the directrices
The distance \(D\) between the directrices of the hyperbola is given by the formula:
\[ D = \frac{2a}{e} \]
Substituting the values of \(a\) and \(e\):
\[ D = \frac{2 \times 7}{\frac{\sqrt{65}}{7}} = \frac{14}{\frac{\sqrt{65}}{7}} = 14 \times \frac{7}{\sqrt{65}} = \frac{98}{\sqrt{65}} \]

### Final Answer
Thus, the distance between the directrices of the hyperbola is:
\[ \frac{98}{\sqrt{65}} \]

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