Find the coordinate of the foci, vertice eccentricity and the length of the latus rectum of the hyperbola
`(y ^(2))/(9) - (x ^(2))/(27) =1 `

Answer

To solve the problem, we need to find the coordinates of the foci, vertices, eccentricity, and the length of the latus rectum of the hyperbola given by the equation:

\[ \frac{y^2}{9} - \frac{x^2}{27} = 1 \]

### Step 1: Identify the standard form of the hyperbola
The given equation is already in the standard form of a hyperbola, which is:

\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the equation, we can identify:
- \(a^2 = 9\) → \(a = \sqrt{9} = 3\)
- \(b^2 = 27\) → \(b = \sqrt{27} = 3\sqrt{3}\)

### Step 3: Calculate \(c\) (the distance from the center to the foci)
The relationship between \(a\), \(b\), and \(c\) for hyperbolas is given by:

\[ c^2 = a^2 + b^2 \]

Substituting the values of \(a^2\) and \(b^2\):

\[ c^2 = 9 + 27 = 36 \]
\[ c = \sqrt{36} = 6 \]

### Step 4: Find the coordinates of the foci
For hyperbolas of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the foci are located at \((0, \pm c)\).

Thus, the coordinates of the foci are:

\[ (0, 6) \text{ and } (0, -6) \]

### Step 5: Find the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by:

\[ e = \frac{c}{a} \]

Substituting the values of \(c\) and \(a\):

\[ e = \frac{6}{3} = 2 \]

### Step 6: Calculate the length of the latus rectum
The length of the latus rectum \(L\) for hyperbolas is given by:

\[ L = \frac{2b^2}{a} \]

Substituting the values of \(b^2\) and \(a\):

\[ L = \frac{2 \times 27}{3} = \frac{54}{3} = 18 \]

### Summary of Results
- Coordinates of the foci: \((0, 6)\) and \((0, -6)\)
- Eccentricity: \(2\)
- Length of the latus rectum: \(18\)