Find the co-ordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola `y^(2) - 25x^(2) = 25`.

Explanation

To solve the problem of finding the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola given by the equation \( y^2 - 25x^2 = 25 \), we will follow these steps:

### Step 1: Rewrite the Hyperbola in Standard Form
The given equation is:
\[ y^2 - 25x^2 = 25 \]
To convert this into standard form, we will first divide the entire equation by 25:
\[ \frac{y^2}{25} - \frac{x^2}{1} = 1 \]
This can be rewritten as:
\[ \frac{y^2}{5^2} - \frac{x^2}{1^2} = 1 \]
This is now in the standard form of a hyperbola:
\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]
where \( a^2 = 1 \) and \( b^2 = 25 \).

### Step 2: Identify Values of a and b
From the standard form, we can identify:
\[ a = 1 \quad \text{and} \quad b = 5 \]

### Step 3: Calculate the Eccentricity (e)
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 \( b \):
\[ e = \sqrt{1 + \frac{25}{1}} = \sqrt{26} \]

### Step 4: Find the Coordinates of the Foci
For a hyperbola of the form \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the foci are located at:
\[ (0, \pm be) \]
Calculating the coordinates:
\[ be = 5\sqrt{26} \]
Thus, the coordinates of the foci are:
\[ (0, \pm 5\sqrt{26}) \]

### Step 5: Find the Coordinates of the Vertices
The vertices for this hyperbola are located at:
\[ (0, \pm b) \]
Thus, the coordinates of the vertices are:
\[ (0, \pm 5) \]

### Step 6: Calculate the Length of the Latus Rectum
The length of the latus rectum \( L \) for a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \( a \) and \( b \):
\[ L = \frac{2 \cdot 25}{1} = 50 \]

### Summary of Results
- **Coordinates of the Foci**: \( (0, \pm 5\sqrt{26}) \)
- **Coordinates of the Vertices**: \( (0, \pm 5) \)
- **Eccentricity**: \( \sqrt{26} \)
- **Length of the Latus Rectum**: \( 50 \)