Let's solve the given hyperbola equations step by step.
### (i) For the equation `16x² - 9y² = 144`:
1. **Convert to Standard Form**:
\[
\frac{16x^2}{144} - \frac{9y^2}{144} = 1
\]
Simplifying gives:
\[
\frac{x^2}{9} - \frac{y^2}{16} = 1
\]
2. **Identify \(a^2\) and \(b^2\)**:
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we have:
\[
a^2 = 9 \quad \Rightarrow \quad a = 3
\]
\[
b^2 = 16 \quad \Rightarrow \quad b = 4
\]
3. **Lengths of Transverse and Conjugate Axes**:
- Length of Transverse Axis = \(2a = 2 \times 3 = 6\)
- Length of Conjugate Axis = \(2b = 2 \times 4 = 8\)
4. **Coordinates of the Center**:
The center is at \((0, 0)\).
5. **Coordinates of the Vertices**:
Vertices are at \((\pm a, 0) = (\pm 3, 0)\).
6. **Eccentricity**:
\[
e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}
\]
7. **Coordinates of the Foci**:
\[
\text{Foci} = (\pm ae, 0) = \left(\pm 3 \cdot \frac{5}{3}, 0\right) = (\pm 5, 0)
\]
### Summary for (i):
- Length of Transverse Axis: 6
- Length of Conjugate Axis: 8
- Center: (0, 0)
- Vertices: (3, 0) and (-3, 0)
- Eccentricity: \(\frac{5}{3}\)
- Foci: (5, 0) and (-5, 0)
---
### (ii) For the equation `2x² - 3y² - 6 = 0`:
1. **Convert to Standard Form**:
\[
2x^2 - 3y^2 = 6 \quad \Rightarrow \quad \frac{x^2}{3} - \frac{y^2}{2} = 1
\]
2. **Identify \(a^2\) and \(b^2\)**:
\[
a^2 = 3 \quad \Rightarrow \quad a = \sqrt{3}
\]
\[
b^2 = 2 \quad \Rightarrow \quad b = \sqrt{2}
\]
3. **Lengths of Transverse and Conjugate Axes**:
- Length of Transverse Axis = \(2a = 2\sqrt{3}\)
- Length of Conjugate Axis = \(2b = 2\sqrt{2}\)
4. **Coordinates of the Center**:
The center is at \((0, 0)\).
5. **Coordinates of the Vertices**:
Vertices are at \((\pm a, 0) = (\pm \sqrt{3}, 0)\).
6. **Eccentricity**:
\[
e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{3}} = \sqrt{\frac{5}{3}}
\]
7. **Coordinates of the Foci**:
\[
\text{Foci} = (\pm ae, 0) = \left(\pm \sqrt{3} \cdot \sqrt{\frac{5}{3}}, 0\right) = \left(\pm \sqrt{5}, 0\right)
\]
### Summary for (ii):
- Length of Transverse Axis: \(2\sqrt{3}\)
- Length of Conjugate Axis: \(2\sqrt{2}\)
- Center: (0, 0)
- Vertices: \((\sqrt{3}, 0)\) and \((- \sqrt{3}, 0)\)
- Eccentricity: \(\sqrt{\frac{5}{3}}\)
- Foci: \((\sqrt{5}, 0)\) and \((- \sqrt{5}, 0)\)
---
### (iii) For the equation `3x² - 2y² = 1`:
1. **Convert to Standard Form**:
\[
\frac{x^2}{\frac{1}{3}} - \frac{y^2}{\frac{1}{2}} = 1
\]
2. **Identify \(a^2\) and \(b^2\)**:
\[
a^2 = \frac{1}{3} \quad \Rightarrow \quad a = \frac{1}{\sqrt{3}}
\]
\[
b^2 = \frac{1}{2} \quad \Rightarrow \quad b = \frac{1}{\sqrt{2}}
\]
3. **Lengths of Transverse and Conjugate Axes**:
- Length of Transverse Axis = \(2a = \frac{2}{\sqrt{3}}\)
- Length of Conjugate Axis = \(2b = \frac{2}{\sqrt{2}}\)
4. **Coordinates of the Center**:
The center is at \((0, 0)\).
5. **Coordinates of the Vertices**:
Vertices are at \((\pm a, 0) = \left(\pm \frac{1}{\sqrt{3}}, 0\right)\).
6. **Eccentricity**:
\[
e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\frac{1}{2}}{\frac{1}{3}}} = \sqrt{1 + \frac{3}{2}} = \sqrt{\frac{5}{2}}
\]
7. **Coordinates of the Foci**:
\[
\text{Foci} = (\pm ae, 0) = \left(\pm \frac{1}{\sqrt{3}} \cdot \sqrt{\frac{5}{2}}, 0\right) = \left(\pm \frac{\sqrt{5}}{\sqrt{6}}, 0\right)
\]
### Summary for (iii):
- Length of Transverse Axis: \(\frac{2}{\sqrt{3}}\)
- Length of Conjugate Axis: \(\frac{2}{\sqrt{2}}\)
- Center: (0, 0)
- Vertices: \(\left(\frac{1}{\sqrt{3}}, 0\right)\) and \(\left(-\frac{1}{\sqrt{3}}, 0\right)\)
- Eccentricity: \(\sqrt{\frac{5}{2}}\)
- Foci: \(\left(\frac{\sqrt{5}}{\sqrt{6}}, 0\right)\) and \(\left(-\frac{\sqrt{5}}{\sqrt{6}}, 0\right)\)
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