The length of conjugate axis of a hyperbola is greater than the length of transverse axis. Then the eccentricity e is

Options

A. `=sqrt(2)`
B. `gtsqrt(2)`
C. `ltsqrt2`
D. `lt(1)/(sqrt(2))`

Answer

To solve the problem, we need to analyze the relationship between the lengths of the conjugate and transverse axes of a hyperbola and its eccentricity.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The length of the transverse axis of a hyperbola is given by \(2a\).
- The length of the conjugate axis of a hyperbola is given by \(2b\).
- For a hyperbola, the relationship between \(a\), \(b\), and the eccentricity \(e\) is given by the formula:
\[ e^2 = 1 + \frac{b^2}{a^2} \]

2. **Set Up the Inequality**:
- According to the problem, the length of the conjugate axis is greater than the length of the transverse axis:
\[ 2b > 2a \]
- Dividing both sides by 2, we get:
\[ b > a \]

3. **Express the Relationship**:
- Since \(b > a\), we can express this in terms of ratios:
\[ \frac{b}{a} > 1 \]
- Squaring both sides gives:
\[ \left(\frac{b}{a}\right)^2 > 1 \quad \Rightarrow \quad \frac{b^2}{a^2} > 1 \]

4. **Substituting into the Eccentricity Formula**:
- From the eccentricity formula, we substitute \(\frac{b^2}{a^2}\):
\[ e^2 = 1 + \frac{b^2}{a^2} \]
- Since \(\frac{b^2}{a^2} > 1\), we can say:
\[ e^2 > 1 + 1 = 2 \]

5. **Conclusion about Eccentricity**:
- Taking the square root of both sides, we find:
\[ e > \sqrt{2} \]

### Final Answer:
Thus, the eccentricity \(e\) is greater than \(\sqrt{2}\).

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