The angle between the asymptotes of the hyperbola `x^(2) -3y^(2) =3 ` is

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To find the angle between the asymptotes of the hyperbola given by the equation \( x^2 - 3y^2 = 3 \), we can follow these steps:

### Step 1: Rewrite the equation of the hyperbola
The standard form of a hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
We can rewrite the given equation \( x^2 - 3y^2 = 3 \) in standard form by dividing through by 3:
\[ \frac{x^2}{3} - \frac{y^2}{1} = 1 \]
Here, we identify \( a^2 = 3 \) and \( b^2 = 1 \).

### Step 2: Identify the slopes of the asymptotes
The asymptotes of a hyperbola of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by the equations:
\[ y = \pm \frac{b}{a} x \]
In our case, \( a = \sqrt{3} \) and \( b = 1 \), so the slopes \( m_1 \) and \( m_2 \) of the asymptotes are:
\[ m_1 = \frac{b}{a} = \frac{1}{\sqrt{3}} \quad \text{and} \quad m_2 = -\frac{b}{a} = -\frac{1}{\sqrt{3}} \]

### Step 3: Calculate the angle between the asymptotes
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 values of \( m_1 \) and \( m_2 \):
\[ \tan(\theta) = \left| \frac{\frac{1}{\sqrt{3}} - \left(-\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)\left(-\frac{1}{\sqrt{3}}\right)} \right| \]
\[ = \left| \frac{\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} \right| \]
\[ = \left| \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \right| = \left| \frac{2}{\sqrt{3}} \cdot \frac{3}{2} \right| = \left| \frac{3}{\sqrt{3}} \right| = \sqrt{3} \]

### Step 4: Find the angle \( \theta \)
Since \( \tan(\theta) = \sqrt{3} \), we know that:
\[ \theta = \frac{\pi}{3} \]

### Conclusion
Thus, the angle between the asymptotes of the hyperbola \( x^2 - 3y^2 = 3 \) is:
\[ \theta = \frac{\pi}{3} \]