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Find the eccentricity, foci, vertices, l...

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`

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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 \( x = -\frac{4}{\sqrt{5}} \) --- ### (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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