The eccentricity of the conjugate hyperbola of the hyperbola `x^2-3y^2=1` is (a) 2 (b) `2sqrt(3)` (c) 4 (d) `4/5`

Options

A. 2
B. `2//sqrt3`
C. 4
D. `4//5`

Explanation

To find the eccentricity of the conjugate hyperbola of the given hyperbola \( x^2 - 3y^2 = 1 \), we can follow these steps:

### Step 1: Identify the standard form of the hyperbola
The given equation of the hyperbola is:
\[ x^2 - 3y^2 = 1 \]
This can be rewritten in the standard form:
\[ \frac{x^2}{1^2} - \frac{y^2}{\left(\frac{1}{\sqrt{3}}\right)^2} = 1 \]
From this, we can identify \( a^2 = 1 \) and \( b^2 = \frac{1}{3} \).

### Step 2: Calculate the eccentricity of the hyperbola
The eccentricity \( E_1 \) of a hyperbola is given by the formula:
\[ E_1 = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ E_1 = \sqrt{1 + \frac{\frac{1}{3}}{1}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Step 3: Find the eccentricity of the conjugate hyperbola
The relationship between the eccentricity of the hyperbola \( E_1 \) and the eccentricity of the conjugate hyperbola \( E_2 \) is given by:
\[ \frac{1}{E_1^2} + \frac{1}{E_2^2} = 1 \]
First, we calculate \( E_1^2 \):
\[ E_1^2 = \frac{4}{3} \]
Now substituting into the equation:
\[ \frac{1}{\frac{4}{3}} + \frac{1}{E_2^2} = 1 \]
This simplifies to:
\[ \frac{3}{4} + \frac{1}{E_2^2} = 1 \]
Rearranging gives:
\[ \frac{1}{E_2^2} = 1 - \frac{3}{4} = \frac{1}{4} \]
Taking the reciprocal:
\[ E_2^2 = 4 \]
Thus, taking the square root gives:
\[ E_2 = 2 \]

### Conclusion
The eccentricity of the conjugate hyperbola is \( 2 \).

### Final Answer
The correct option is (a) 2.