The eccentricity of the hyperbola canjugate to `(x-1)^(2)-3(y-1)^(2)=1` is

Options

A. 3
B. 2
C. `2/sqrt3`
D. `3/2`

Answer

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

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

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

### Step 3: Identify the conjugate hyperbola
The conjugate hyperbola of the given hyperbola has the form:
\[ \frac{(y-1)^2}{(1/\sqrt{3})^2} - \frac{(x-1)^2}{1^2} = 1 \]
This means the roles of \(a\) and \(b\) are switched:
- For the conjugate hyperbola, \(a' = \frac{1}{\sqrt{3}}\) and \(b' = 1\).

### Step 4: Calculate the eccentricity of the conjugate hyperbola
The eccentricity \(e'\) of the conjugate hyperbola is given by:
\[ e' = \sqrt{1 + \frac{a'^2}{b'^2}} \]
Substituting the values:
\[ e' = \sqrt{1 + \frac{\left(\frac{1}{\sqrt{3}}\right)^2}{1^2}} = \sqrt{1 + \frac{1/3}{1}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Conclusion
The eccentricity of the hyperbola conjugate to \((x-1)^2 - 3(y-1)^2 = 1\) is \(\frac{2}{\sqrt{3}}\).

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