Find the coordinate of the foci, vertice eccentricity and the length of the latus rectum of the hyperbola
`49y ^(2) - 16x ^(2) = 784`

Answer

To solve the problem, we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation of the hyperbola given as:
\[ 49y^2 - 16x^2 = 784 \]

To convert this into standard form, we divide the entire equation by 784:
\[ \frac{49y^2}{784} - \frac{16x^2}{784} = 1 \]
This simplifies to:
\[ \frac{y^2}{16} - \frac{x^2}{49} = 1 \]

### Step 2: Identify the values of \(a^2\) and \(b^2\)
From the standard form of the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), we can identify:
- \(a^2 = 16\) → \(a = 4\)
- \(b^2 = 49\) → \(b = 7\)

### Step 3: Calculate the value of \(c\)
The relationship between \(a\), \(b\), and \(c\) (the distance from the center to the foci) is given by:
\[ c = \sqrt{a^2 + b^2} \]
Substituting the values we found:
\[ c = \sqrt{16 + 49} = \sqrt{65} \]

### Step 4: Find the coordinates of the foci
For a hyperbola that opens upwards and downwards, the foci are located at \((0, \pm c)\):
\[ \text{Foci: } (0, \pm \sqrt{65}) \]

### Step 5: Find the coordinates of the vertices
The vertices are located at \((0, \pm a)\):
\[ \text{Vertices: } (0, \pm 4) \]

### Step 6: Calculate the eccentricity \(e\)
The eccentricity \(e\) is given by:
\[ e = \frac{c}{a} \]
Substituting the values:
\[ e = \frac{\sqrt{65}}{4} \]

### Step 7: Calculate the length of the latus rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values:
\[ L = \frac{2 \times 49}{4} = \frac{98}{4} = 24.5 \]

### Final Results
- **Foci:** \((0, \sqrt{65})\) and \((0, -\sqrt{65})\)
- **Vertices:** \((0, 4)\) and \((0, -4)\)
- **Eccentricity:** \(\frac{\sqrt{65}}{4}\)
- **Length of the Latus Rectum:** \(24.5\)