Angle between the asymptotes of a hyperbola is `x^(2)-3y^(2)=1` is a) `15^(@)` b) `45^(@)` c) `60^(@)` d) `30^(@)`

Options

A. `15^(@)`
B. `45^(@)`
C. `60^(@)`
D. `30^(@)`

Answer

To find the angle between the asymptotes of the hyperbola given by the equation \( x^2 - 3y^2 = 1 \), we can follow these steps:

### Step 1: Identify the standard form of the hyperbola
The given equation is \( x^2 - 3y^2 = 1 \). This can be rewritten in the standard form of a hyperbola:
\[ \frac{x^2}{1} - \frac{y^2}{\frac{1}{3}} = 1 \]
This indicates that the hyperbola is centered at the origin and opens along the x-axis.

### Step 2: Determine 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:
\[ y = \pm \frac{b}{a} x \]
For our hyperbola:
- \( a^2 = 1 \) so \( a = 1 \)
- \( b^2 = \frac{1}{3} \) so \( b = \frac{1}{\sqrt{3}} \)

Thus, the equations of the asymptotes are:
\[ y = \pm \frac{1/\sqrt{3}}{1} x = \pm \frac{x}{\sqrt{3}} \]

### Step 3: Find the slopes of the asymptotes
From the equations of the asymptotes, we can see that:
- The slope of the first asymptote \( m_1 = \frac{1}{\sqrt{3}} \)
- The slope of the second asymptote \( m_2 = -\frac{1}{\sqrt{3}} \)

### Step 4: Calculate the angle between the asymptotes
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{\frac{1}{\sqrt{3}} - \left(-\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)\left(-\frac{1}{\sqrt{3}}\right)} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \right| = \left| \frac{2}{\sqrt{3}} \cdot \frac{3}{2} \right| = \left| \frac{3}{\sqrt{3}} \right| = \sqrt{3} \]

### Step 5: Determine the angle
Now we find \( \theta \):
\[ \theta = \tan^{-1}(\sqrt{3}) = 60^\circ \]

Thus, the angle between the asymptotes of the hyperbola \( x^2 - 3y^2 = 1 \) is \( 60^\circ \).

### Final Answer
The correct option is (c) \( 60^\circ \).

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