For the hyperbola `4x^(2) - 9y^(2) = 1 `, find the axes, centre, eccentricity, foci and equations of the directrices.

Answer

To solve the problem step by step, we will analyze the given hyperbola equation and extract the required information.

### Step 1: Write the equation of the hyperbola
The given equation of the hyperbola is:
\[ 4x^2 - 9y^2 = 1 \]

### Step 2: Rewrite in standard form
To convert this into standard form, we divide the entire equation by 1:
\[ \frac{4x^2}{1} - \frac{9y^2}{1} = 1 \]
Now, we can rewrite it as:
\[ \frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{1}{9}} = 1 \]

### Step 3: Identify \(a^2\) and \(b^2\)
From the standard form, we can identify:
- \(a^2 = \frac{1}{4}\) which gives \(a = \frac{1}{2}\)
- \(b^2 = \frac{1}{9}\) which gives \(b = \frac{1}{3}\)

### Step 4: Find the center
The center of the hyperbola is at the origin (0, 0) since there are no shifts in the equation. Therefore:
- Center = (0, 0)

### Step 5: Determine 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 \(a^2\) and \(b^2\):
\[ e = \sqrt{1 + \frac{\frac{1}{9}}{\frac{1}{4}}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Step 6: Find the foci
The foci of a hyperbola are located at \((\pm ae, 0)\). Thus:
\[ F = \left(\pm \frac{1}{2} \cdot \frac{\sqrt{13}}{3}, 0\right) = \left(\pm \frac{\sqrt{13}}{6}, 0\right) \]

### Step 7: Find the equations of the directrices
The equations of the directrices for a hyperbola are given by:
\[ x = \pm \frac{a}{e} \]
Substituting the values of \(a\) and \(e\):
\[ x = \pm \frac{\frac{1}{2}}{\frac{\sqrt{13}}{3}} = \pm \frac{3}{2\sqrt{13}} \]

### Summary of Results
- **Axes**: Transverse axis along the x-axis
- **Center**: (0, 0)
- **Eccentricity**: \( \frac{\sqrt{13}}{3} \)
- **Foci**: \( \left(\frac{\sqrt{13}}{6}, 0\right) \) and \( \left(-\frac{\sqrt{13}}{6}, 0\right) \)
- **Equations of the Directrices**: \( x = \frac{3}{2\sqrt{13}} \) and \( x = -\frac{3}{2\sqrt{13}} \)