Find the eccentricity of hyperbola `x^(2)-9y^(2)=1`.

Answer

To find the eccentricity of the hyperbola given by the equation \( x^2 - 9y^2 = 1 \), we can follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation is:
\[ x^2 - 9y^2 = 1 \]
We can rewrite it in the standard form of a hyperbola, which is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
To do this, we divide the entire equation by 1:
\[ \frac{x^2}{1} - \frac{9y^2}{1} = 1 \]
This can be rewritten as:
\[ \frac{x^2}{1} - \frac{y^2}{\frac{1}{9}} = 1 \]
From this, we can identify:
\[ a^2 = 1 \quad \text{and} \quad b^2 = \frac{1}{9} \]

### Step 2: Calculate the eccentricity
The formula for the eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ e = \sqrt{1 + \frac{\frac{1}{9}}{1}} = \sqrt{1 + \frac{1}{9}} \]
This simplifies to:
\[ e = \sqrt{1 + 0.1111} = \sqrt{\frac{10}{9}} \]

### Step 3: Simplify the expression
We can simplify \( \sqrt{\frac{10}{9}} \):
\[ e = \frac{\sqrt{10}}{3} \]

### Final Answer
Thus, the eccentricity of the hyperbola \( x^2 - 9y^2 = 1 \) is:
\[ e = \frac{\sqrt{10}}{3} \]

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