1. **Problem Statement:** Given the conic section equation $$9x^2 - 4y^2 + 18x - 8y + 6 = 0,$$ we need to: (a) Classify the conic. (b) Find its center, vertices, and foci. (c) Sketch its graph. 2. **Classifying the Conic:** The general form is $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.$$ Here, $$A=9,$$ $$B=0,$$ $$C=-4.$$ Calculate the discriminant $$\Delta = B^2 - 4AC = 0^2 - 4(9)(-4) = 144 > 0.$$ Since $$\Delta > 0,$$ the conic is a **hyperbola**. 3. **Rewrite the equation by completing the square:** Group $$x$$ and $$y$$ terms: $$9x^2 + 18x - 4y^2 - 8y + 6 = 0.$$ Divide terms to complete the square: $$9(x^2 + 2x) - 4(y^2 + 2y) = -6.$$ Complete the square inside each parenthesis: - For $$x^2 + 2x$$, add and subtract $$1$$: $$x^2 + 2x + 1 - 1 = (x+1)^2 - 1.$$ - For $$y^2 + 2y$$, add and subtract $$1$$: $$y^2 + 2y + 1 - 1 = (y+1)^2 - 1.$$ Substitute back: $$9[(x+1)^2 - 1] - 4[(y+1)^2 - 1] = -6.$$ Expand: $$9(x+1)^2 - 9 - 4(y+1)^2 + 4 = -6.$$ Simplify constants: $$9(x+1)^2 - 4(y+1)^2 - 5 = -6.$$ Add 5 to both sides: $$9(x+1)^2 - 4(y+1)^2 = -1.$$ Multiply both sides by $$-1$$ to get standard form: $$-9(x+1)^2 + 4(y+1)^2 = 1.$$ Rewrite as: $$4(y+1)^2 - 9(x+1)^2 = 1.$$ Divide both sides by 1: $$\frac{(y+1)^2}{\frac{1}{4}} - \frac{(x+1)^2}{\frac{1}{9}} = 1.$$ 4. **Identify center, vertices, and foci:** - Center is at $$(-1, -1).$$ - Since the $$y$$ term is positive and comes first, the transverse axis is vertical. - $$a^2 = \frac{1}{4}$$ so $$a = \frac{1}{2}.$$ - $$b^2 = \frac{1}{9}$$ so $$b = \frac{1}{3}.$$ - Vertices are $$a$$ units above and below the center along the $$y$$-axis: $$(-1, -1 + \frac{1}{2}) = (-1, -\frac{1}{2}),$$ $$(-1, -1 - \frac{1}{2}) = (-1, -\frac{3}{2}).$$ - Calculate $$c$$ for foci using $$c^2 = a^2 + b^2 = \frac{1}{4} + \frac{1}{9} = \frac{9}{36} + \frac{4}{36} = \frac{13}{36}.$$ So $$c = \frac{\sqrt{13}}{6}.$$ - Foci are $$c$$ units above and below the center: $$(-1, -1 + \frac{\sqrt{13}}{6}),$$ $$(-1, -1 - \frac{\sqrt{13}}{6}).$$ 5. **Summary:** - Type: Hyperbola - Center: $$(-1, -1)$$ - Vertices: $$(-1, -\frac{1}{2})$$ and $$(-1, -\frac{3}{2})$$ - Foci: $$(-1, -1 + \frac{\sqrt{13}}{6})$$ and $$(-1, -1 - \frac{\sqrt{13}}{6})$$ 6. **Graph Sketch:** The hyperbola opens vertically with center at $$(-1,-1)$$.