Find the eccentricity, foci, vertices, length of latus rectum and equations of directrices of hyperbola.
(i) `x^(2)-4y^(2)=4`
(ii) `4x^(2)-9y^(2)=27`

Answer

To solve the given hyperbola equations step by step, we will analyze each hyperbola separately.

### (i) For the hyperbola \( x^2 - 4y^2 = 4 \):

1. **Rewrite the equation in standard form**:
\[ \frac{x^2}{4} - \frac{y^2}{1} = 1 \]
Here, \( a^2 = 4 \) and \( b^2 = 1 \).

2. **Find \( a \) and \( b \)**:
\[ a = \sqrt{4} = 2, \quad b = \sqrt{1} = 1 \]

3. **Calculate \( c \)** using the formula \( c^2 = a^2 + b^2 \):
\[ c^2 = 4 + 1 = 5 \implies c = \sqrt{5} \]

4. **Find the eccentricity \( e \)**:
\[ e = \frac{c}{a} = \frac{\sqrt{5}}{2} \]

5. **Determine the foci**:
\[ \text{Foci} = (\pm c, 0) = (\pm \sqrt{5}, 0) \]

6. **Determine the vertices**:
\[ \text{Vertices} = (\pm a, 0) = (\pm 2, 0) \]

7. **Calculate the length of the latus rectum**:
\[ \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \cdot 1}{2} = 1 \]

8. **Find the equations of the directrices**:
\[ x = \frac{a^2}{c} = \frac{4}{\sqrt{5}} \quad \text{and} \quad x = -\frac{4}{\sqrt{5}} \]

### Summary for (i):
- Eccentricity \( e = \frac{\sqrt{5}}{2} \)
- Foci \( (\pm \sqrt{5}, 0) \)
- Vertices \( (\pm 2, 0) \)
- Length of latus rectum \( 1 \)
- Equations of directrices \( x = \frac{4}{\sqrt{5}} \) and \( x = -\frac{4}{\sqrt{5}} \)

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### (ii) For the hyperbola \( 4x^2 - 9y^2 = 27 \):

1. **Rewrite the equation in standard form**:
\[ \frac{x^2}{\frac{27}{4}} - \frac{y^2}{3} = 1 \]
Here, \( a^2 = \frac{27}{4} \) and \( b^2 = 3 \).

2. **Find \( a \) and \( b \)**:
\[ a = \sqrt{\frac{27}{4}} = \frac{3\sqrt{3}}{2}, \quad b = \sqrt{3} \]

3. **Calculate \( c \)** using the formula \( c^2 = a^2 + b^2 \):
\[ c^2 = \frac{27}{4} + 3 = \frac{27}{4} + \frac{12}{4} = \frac{39}{4} \implies c = \frac{\sqrt{39}}{2} \]

4. **Find the eccentricity \( e \)**:
\[ e = \frac{c}{a} = \frac{\frac{\sqrt{39}}{2}}{\frac{3\sqrt{3}}{2}} = \frac{\sqrt{39}}{3\sqrt{3}} = \frac{2\sqrt{39}}{3\sqrt{3}} \]

5. **Determine the foci**:
\[ \text{Foci} = \left( \pm \frac{\sqrt{39}}{2}, 0 \right) \]

6. **Determine the vertices**:
\[ \text{Vertices} = \left( \pm \frac{3\sqrt{3}}{2}, 0 \right) \]

7. **Calculate the length of the latus rectum**:
\[ \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \cdot 3}{\frac{3\sqrt{3}}{2}} = \frac{12}{3\sqrt{3}} = \frac{4\sqrt{3}}{3} \]

8. **Find the equations of the directrices**:
\[ x = \frac{a^2}{c} = \frac{\frac{27}{4}}{\frac{\sqrt{39}}{2}} = \frac{27}{2\sqrt{39}} \quad \text{and} \quad x = -\frac{27}{2\sqrt{39}} \]

### Summary for (ii):
- Eccentricity \( e = \frac{2\sqrt{39}}{3\sqrt{3}} \)
- Foci \( \left( \pm \frac{\sqrt{39}}{2}, 0 \right) \)
- Vertices \( \left( \pm \frac{3\sqrt{3}}{2}, 0 \right) \)
- Length of latus rectum \( \frac{4\sqrt{3}}{3} \)
- Equations of directrices \( x = \frac{27}{2\sqrt{39}} \) and \( x = -\frac{27}{2\sqrt{39}} \)

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