If the pair of lines `b^2x^2-a^2y^2=0` are inclined at an angle `theta`, then the eccentricity of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` is

Options

A. `sec theta`
B. `cos theta`
C. `tan""(theta)/(2)`
D. `sec""(theta)/(2)`

Answer

To solve the problem step by step, we will analyze the given information and derive the eccentricity of the hyperbola based on the angle of inclination of the pair of lines.

### Step-by-Step Solution:

1. **Identify the Given Equation of the Pair of Lines:**
The equation of the pair of lines is given as:
\[ b^2x^2 - a^2y^2 = 0 \]
This can be rewritten as:
\[ \frac{b^2}{a^2}x^2 = y^2 \]

2. **Express the Lines in Slope Form:**
Taking the square root of both sides, we get:
\[ y = \pm \frac{b}{a} x \]
This indicates that the slopes of the two lines are:
\[ m_1 = \frac{b}{a} \quad \text{and} \quad m_2 = -\frac{b}{a} \]

3. **Calculate the Angle of Inclination (θ):**
The angle θ between the two lines can be calculated using the formula for the tangent of the angle between two lines:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting the values of \(m_1\) and \(m_2\):
\[ \tan \theta = \frac{\frac{b}{a} - \left(-\frac{b}{a}\right)}{1 + \left(\frac{b}{a}\right)\left(-\frac{b}{a}\right)} = \frac{\frac{2b}{a}}{1 - \frac{b^2}{a^2}} \]
Simplifying this gives:
\[ \tan \theta = \frac{2ab}{a^2 - b^2} \]

4. **Relate the Angle to the Eccentricity:**
We know that:
\[ \tan \frac{\theta}{2} = \frac{b}{a} \]
From the double angle identity, we can express \( \tan \theta \) in terms of \( \tan \frac{\theta}{2} \):
\[ \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} \]
Substituting \( \tan \frac{\theta}{2} = \frac{b}{a} \):
\[ \tan \theta = \frac{2 \cdot \frac{b}{a}}{1 - \left(\frac{b}{a}\right)^2} \]

5. **Use the Eccentricity Formula:**
The eccentricity \( e \) of the hyperbola is given by the formula:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
We can express \( \frac{b^2}{a^2} \) in terms of \( \tan \frac{\theta}{2} \):
\[ e^2 - 1 = \tan^2 \frac{\theta}{2} \]
Therefore:
\[ e^2 = 1 + \tan^2 \frac{\theta}{2} \]

6. **Final Expression for Eccentricity:**
Using the identity \( 1 + \tan^2 x = \sec^2 x \), we can write:
\[ e^2 = \sec^2 \left(\frac{\theta}{2}\right) \]
Thus, the eccentricity \( e \) is:
\[ e = \sec \left(\frac{\theta}{2}\right) \]

### Conclusion:
The eccentricity of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is:
\[ e = \sec \left(\frac{\theta}{2}\right) \]