The eccentricity of the hyperbola `y^(2) - 25x^(2) = 25` is

Options

A. `sqrt(26)`
B. `(1)/(sqrt(26))`
C. `(sqrt(26))/(5)`
D. `(1)/(5)`

Answer

To find the eccentricity of the hyperbola given by the equation \( y^2 - 25x^2 = 25 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation of the hyperbola:
\[ y^2 - 25x^2 = 25 \]
To rewrite it in standard form, we divide every term by 25:
\[ \frac{y^2}{25} - \frac{25x^2}{25} = \frac{25}{25} \]
This simplifies to:
\[ \frac{y^2}{25} - \frac{x^2}{1} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form of the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), we can identify:
- \(a^2 = 25\) which gives \(a = \sqrt{25} = 5\)
- \(b^2 = 1\) which gives \(b = \sqrt{1} = 1\)

### Step 3: 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 \(b^2\) and \(a^2\):
\[ e = \sqrt{1 + \frac{1}{25}} \]
Calculating the fraction:
\[ e = \sqrt{1 + 0.04} = \sqrt{1.04} \]

### Step 4: Simplify the expression
To simplify \(\sqrt{1.04}\):
\[ e = \sqrt{\frac{26}{25}} = \frac{\sqrt{26}}{\sqrt{25}} = \frac{\sqrt{26}}{5} \]

### Final Answer
Thus, the eccentricity of the hyperbola is:
\[ e = \frac{\sqrt{26}}{5} \]

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