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

Answer

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

### Step 1: Write the equation in standard form
We start with the given hyperbola equation:
\[ x^2 - 3y^2 = 3 \]
To convert it into standard form, we divide both sides by 3:
\[ \frac{x^2}{3} - \frac{y^2}{1} = 1 \]

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

### Step 3: Calculate the angles corresponding to the slopes
The angles \(\theta_1\) and \(\theta_2\) corresponding to the slopes can be found using the tangent function:
\[ \tan \theta_1 = \frac{1}{\sqrt{3}} \implies \theta_1 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
\[ \tan \theta_2 = -\frac{1}{\sqrt{3}} \implies \theta_2 = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \]

### Step 4: Find the angle between the asymptotes
The angle \(\theta\) between the two asymptotes is given by the difference between these angles:
\[ \theta = \theta_1 - \theta_2 = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) \]
\[ \theta = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

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

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