The angle between the asymptotes of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` is

Options

A. `cos^(-1)(1/e)`
B. `2cos^(-1)(1/e)`
C. `sin^(-1)(1/e)`
D. None of these

Answer

To find the angle between the asymptotes of the hyperbola given by the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can follow these steps:

### Step 1: Identify the asymptotes of the hyperbola
The asymptotes of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by the equations:
\[ y = \pm \frac{b}{a} x \]

### Step 2: Determine the slopes of the asymptotes
From the equations of the asymptotes, we can see that the slopes are:
\[ m_1 = \frac{b}{a} \quad \text{and} \quad m_2 = -\frac{b}{a} \]

### Step 3: Use the formula for the angle between two lines
The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) can be calculated using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the slopes:
\[ \tan \theta = \left| \frac{\frac{b}{a} - \left(-\frac{b}{a}\right)}{1 + \frac{b}{a} \cdot \left(-\frac{b}{a}\right)} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{\frac{b}{a} + \frac{b}{a}}{1 - \frac{b^2}{a^2}} \right| = \left| \frac{\frac{2b}{a}}{\frac{a^2 - b^2}{a^2}} \right| = \left| \frac{2b a^2}{a(a^2 - b^2)} \right| = \frac{2b}{a^2 - b^2} \]

### Step 4: Find the angle \(\theta\)
To find the angle \(\theta\), we can use the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{2b}{a^2 - b^2}\right) \]

### Step 5: Conclusion
Thus, the angle between the asymptotes of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is given by:
\[ \theta = 2 \tan^{-1}\left(\frac{b}{a}\right) \]