Find the angle between the asymptotes of the hyperbola `(x^2)/(16)-(y^2)/9=1` .

Explanation

To find the angle between the asymptotes of the hyperbola given by the equation

\[ \frac{x^2}{16} - \frac{y^2}{9} = 1, \]

we can follow these steps:

### Step 1: Identify the values of \(a\) and \(b\)

The standard form of the hyperbola is

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \]

From the given equation, we can identify:

\[ a^2 = 16 \quad \text{and} \quad b^2 = 9. \]

Thus, we find:

\[ a = \sqrt{16} = 4 \quad \text{and} \quad b = \sqrt{9} = 3. \]

### Step 2: Determine the slopes of the asymptotes

For a hyperbola of the form

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \]

the slopes of the asymptotes are given by:

\[ m_1 = \frac{b}{a} \quad \text{and} \quad m_2 = -\frac{b}{a}. \]

Substituting the values of \(b\) and \(a\):

\[ m_1 = \frac{3}{4} \quad \text{and} \quad m_2 = -\frac{3}{4}. \]

### Step 3: Use the formula for the angle between the asymptotes

The angle \(\theta\) between the asymptotes can be found using the formula:

\[ \tan(\theta) = \frac{m_1 - m_2}{1 + m_1 m_2}. \]

Substituting the values of \(m_1\) and \(m_2\):

\[ \tan(\theta) = \frac{\frac{3}{4} - \left(-\frac{3}{4}\right)}{1 + \left(\frac{3}{4}\right)\left(-\frac{3}{4}\right)}. \]

### Step 4: Simplify the expression

Calculating the numerator:

\[ m_1 - m_2 = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2}. \]

Calculating the denominator:

\[ 1 + m_1 m_2 = 1 + \left(\frac{3}{4}\right)\left(-\frac{3}{4}\right) = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16}. \]

Now substituting back into the tangent formula:

\[ \tan(\theta) = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \cdot \frac{16}{7} = \frac{48}{14} = \frac{24}{7}. \]

### Step 5: Find the angle \(\theta\)

To find \(\theta\), we take the inverse tangent:

\[ \theta = \tan^{-1}\left(\frac{24}{7}\right). \]

This gives us the angle between the asymptotes of the hyperbola.

### Final Answer

The angle between the asymptotes of the hyperbola is

\[ \theta = \tan^{-1}\left(\frac{24}{7}\right). \]

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