QUESTION FOUR ( 20 MARKS) a) A hyperbola has the equation; \( 9 x^{2}-16 y^{2}-18 x-64 y-199=0 \) Find its center; vertices, foci, eccentricity, length of latus rectum and equatio of the asymptotes ( 10 Marks)

To find the various components of the hyperbola defined by the equation \( 9 x^{2}-16 y^{2}-18 x-64 y-199=0 \), we first need to convert it into standard form. ### Step 1: Rearranging and Completing the Square 1. Rearrange the equation: \( 9x^{2} - 18x - 16y^{2} - 64y - 199 = 0 \) 2. Group \( x \) terms and \( y \) terms: \( 9(x^{2} - 2x) - 16(y^{2} + 4y) = 199 \) 3. Complete the square: - For \( x \): \( x^{2} - 2x = (x-1)^{2} - 1 \) - For \( y \): \( y^{2} + 4y = (y+2)^{2} - 4 \) Insert into the equation: \( 9((x-1)^{2}-1) - 16((y+2)^{2}-4) = 199 \) This simplifies to: \( 9(x-1)^{2} - 16(y+2)^{2} - 9 + 64 = 199 \) Combine constants: \( 9(x-1)^{2} - 16(y+2)^{2} + 55 = 199 \) Now isolate: \( 9(x-1)^{2} - 16(y+2)^{2} = 144 \) Dividing everything by 144 gives: \(\frac{(x-1)^{2}}{16} - \frac{(y+2)^{2}}{9} = 1 \) ### Step 2: Identifying Elements from the Standard Form From the standard form \(\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1\): - The center is at \((h, k) = (1, -2)\). - \(a^{2} = 16 \Rightarrow a = 4\) (transverse axis) - \(b^{2} = 9 \Rightarrow b = 3\) ### Step 3: Finding Vertices, Foci, Eccentricity, Length of Latus Rectum, and Equations of Asymptotes 1. **Vertices**: These are \((h \pm a, k) = (1 \pm 4, -2)\) → Vertices at \((-3, -2)\) and \((5, -2)\). 2. **Foci**: The distance \(c\) is given by \(c^2 = a^2 + b^2 = 16 + 9 = 25\) → \(c = 5\). - Foci at \((h \pm c, k) = (1 \pm 5, -2)\) → Foci at \((-4, -2)\) and \((6, -2)\). 3. **Eccentricity**: \(e = \frac{c}{a} = \frac{5}{4}\). 4. **Length of Latus Rectum**: Given by \(\frac{2b^2}{a} = \frac{2 \cdot 9}{4} = 4.5\). 5. **Asymptotes**: The equations are \(y - k = \pm \frac{b}{a}(x - h)\) → - \(y + 2 = \pm \frac{3}{4}(x - 1)\), leading to \(y = \frac{3}{4}x - \frac{11}{4}\) and \(y = -\frac{3}{4}x - \frac{5}{4}\). ### Summary - **Center**: \((1, -2)\) - **Vertices**: \((-3, -2)\) and \((5, -2)\) - **Foci**: \((-4, -2)\) and \((6, -2)\) - **Eccentricity**: \(\frac{5}{4}\) - **Length of Latus Rectum**: \(4.5\) - **Asymptotes**: \(y = \frac{3}{4}x - \frac{11}{4}\) and \(y = -\frac{3}{4}x - \frac{5}{4}\)