1. **State the problem:** Rewrite the hyperbola equation $$9x^2 - 4y^2 - 36x - 40y - 388 = 0$$ in standard form and find its center, foci, vertices, and asymptotes. 2. **Rewrite the equation grouping x and y terms:** $$9x^2 - 36x - 4y^2 - 40y = 388$$ 3. **Complete the square for x and y terms:** For x: $$9(x^2 - 4x)$$ For y: $$-4(y^2 + 10y)$$ 4. **Complete the square inside parentheses:** $$x^2 - 4x = (x^2 - 4x + 4) - 4 = (x - 2)^2 - 4$$ $$y^2 + 10y = (y^2 + 10y + 25) - 25 = (y + 5)^2 - 25$$ 5. **Substitute back:** $$9[(x - 2)^2 - 4] - 4[(y + 5)^2 - 25] = 388$$ 6. **Distribute:** $$9(x - 2)^2 - 36 - 4(y + 5)^2 + 100 = 388$$ 7. **Combine constants:** $$9(x - 2)^2 - 4(y + 5)^2 + 64 = 388$$ 8. **Isolate terms:** $$9(x - 2)^2 - 4(y + 5)^2 = 388 - 64$$ $$9(x - 2)^2 - 4(y + 5)^2 = 324$$ 9. **Divide both sides by 324:** $$\frac{9(x - 2)^2}{324} - \frac{4(y + 5)^2}{324} = 1$$ 10. **Simplify fractions:** $$\frac{(x - 2)^2}{36} - \frac{(y + 5)^2}{81} = 1$$ **Standard form:** $$\frac{(x - 2)^2}{6^2} - \frac{(y + 5)^2}{9^2} = 1$$ --- **(a) Center:** The center is at $$(h, k) = (2, -5)$$ **(b) Foci:** For hyperbola $$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$, foci are at $$(h \pm c, k)$$ where $$c = \sqrt{a^2 + b^2}$$. Calculate $$c$$: $$c = \sqrt{36 + 81} = \sqrt{117} = 3\sqrt{13}$$ Foci: $$(2 \pm 3\sqrt{13}, -5)$$ **(c) Vertices:** Vertices are at $$(h \pm a, k)$$: $$(2 \pm 6, -5)$$ So vertices are at $$(8, -5)$$ and $$(-4, -5)$$ **(d) Equation of asymptotes:** Asymptotes for horizontal hyperbola: $$y - k = \pm \frac{b}{a}(x - h)$$ Substitute values: $$y + 5 = \pm \frac{9}{6}(x - 2) = \pm \frac{3}{2}(x - 2)$$ So asymptotes are: $$y = -5 + \frac{3}{2}(x - 2)$$ and $$y = -5 - \frac{3}{2}(x - 2)$$ **(e) Sketch:** The hyperbola opens left and right centered at $$(2, -5)$$ with vertices at $$(8, -5)$$ and $$(-4, -5)$$, foci further out at $$(2 \pm 3\sqrt{13}, -5)$$, and asymptotes as above.