If asymptotes of hyperbola bisect the angles between the transverse axis and conjugate axis of hyperbola, then what is eccentricity of hyperbola?

Explanation

To find the eccentricity of the hyperbola given that its asymptotes bisect the angles between the transverse and conjugate axes, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Hyperbola**:
The standard form of a hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(a\) is the distance from the center to the vertices along the transverse axis, and \(b\) is the distance from the center to the vertices along the conjugate axis.

2. **Equations of the Asymptotes**:
The equations of the asymptotes for the hyperbola are:
\[ y = \pm \frac{b}{a} x \]

3. **Angle Bisecting Condition**:
According to the problem, the asymptotes bisect the angles between the transverse axis (x-axis) and the conjugate axis (y-axis). This means that the angles formed by the asymptotes with the axes are equal.

4. **Angles Formed by the Asymptotes**:
The angles formed by the asymptotes with the axes can be determined using the slopes of the lines. The slope of the asymptotes is given by:
\[ m = \frac{b}{a} \quad \text{and} \quad -\frac{b}{a} \]
The angles formed with the x-axis are:
\[ \theta_1 = \tan^{-1}\left(\frac{b}{a}\right) \quad \text{and} \quad \theta_2 = \tan^{-1}\left(-\frac{b}{a}\right) \]

5. **Equal Angles**:
Since the asymptotes bisect the angles, we have:
\[ \theta_1 + \theta_2 = 90^\circ \]
This implies:
\[ 2\theta_1 = 90^\circ \quad \Rightarrow \quad \theta_1 = 45^\circ \]

6. **Finding the Relationship Between a and b**:
From the angle \( \theta_1 = 45^\circ \):
\[ \tan(45^\circ) = 1 \quad \Rightarrow \quad \frac{b}{a} = 1 \quad \Rightarrow \quad b = a \]

7. **Finding the Eccentricity**:
The eccentricity \(e\) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \(b = a\):
\[ e = \sqrt{1 + \frac{a^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2} \]

### Final Answer:
The eccentricity of the hyperbola is:
\[ e = \sqrt{2} \]