Find the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the hyperbola.
`x^(2)-4y^(2)=4`

Answer

To solve the problem step-by-step, we will analyze the hyperbola given by the equation \( x^2 - 4y^2 = 4 \).

### Step 1: Rewrite the equation in standard form
We start with the equation of the hyperbola:
\[ x^2 - 4y^2 = 4 \]
To convert this into standard form, we divide the entire equation by 4:
\[ \frac{x^2}{4} - \frac{y^2}{1} = 1 \]
This is now in the standard form of a hyperbola, which is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( a^2 = 4 \) and \( b^2 = 1 \).

### Step 2: Identify \( a \) and \( b \)
From the standard form, we can find:
\[ a = \sqrt{4} = 2 \quad \text{and} \quad b = \sqrt{1} = 1 \]

### Step 3: Find the center of the hyperbola
The center of a hyperbola in standard form is at the origin (0, 0) when there are no shifts in the equation. Thus, the center is:
\[ \text{Center} = (0, 0) \]

### Step 4: 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 \( b \):
\[ e = \sqrt{1 + \frac{1}{4}} = \sqrt{1 + 0.25} = \sqrt{1.25} = \frac{\sqrt{5}}{2} \]

### Step 5: Find the foci
The foci of the hyperbola are located at:
\[ (\pm ae, 0) \]
Substituting the values of \( a \) and \( e \):
\[ \text{Foci} = \left( \pm 2 \cdot \frac{\sqrt{5}}{2}, 0 \right) = (\pm \sqrt{5}, 0) \]

### Step 6: Find the equations of the directrices
The equations of the directrices for a hyperbola are given by:
\[ x = \pm \frac{a}{e} \]
Calculating this:
\[ x = \pm \frac{2}{\frac{\sqrt{5}}{2}} = \pm \frac{2 \cdot 2}{\sqrt{5}} = \pm \frac{4}{\sqrt{5}} \]

### Step 7: Calculate the length of the latus rectum
The length of the latus rectum \( L \) is given by the formula:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \( b \) and \( a \):
\[ L = \frac{2 \cdot 1^2}{2} = \frac{2 \cdot 1}{2} = 1 \]

### Summary of Results
- Center: \( (0, 0) \)
- Foci: \( (\pm \sqrt{5}, 0) \)
- Eccentricity: \( \frac{\sqrt{5}}{2} \)
- Equations of the Directrices: \( x = \pm \frac{4}{\sqrt{5}} \)
- Length of the Latus Rectum: \( 1 \)