Question: Find the center of the hyperbola $ 9x^2 - 18x - 16y^2 - 64y = 144 $.
["Find the Center of the Hyperbola: Understanding the Hyperbola Equation $ 9x^2 - 18x - 16y^2 - 64y = 144 $", "When working with conic sections, identifying key features like the center is essential for graphing and analyzing the shape. A common question in algebra and precalculus is: Find the center of the hyperbola defined by $ 9x^2 - 18x - 16y^2 - 64y = 144 $. Let’s break down the process step-by-step to locate the center accurately.", "---", "### Step 1: Rewrite the Equation in Standard Form", "The standard form of a hyperbola reveals its center, vertices, and orientation. To convert the given quadratic equation into standard form, we use completing the square for both $ x $ and $ y $ terms.", "Start with the original equation:", "$$\n9x^2 - 18x - 16y^2 - 64y = 144\n$$", "Group the $ x $ and $ y $ terms:", "$$\n(9x^2 - 18x) + (-16y^2 - 64y) = 144\n$$", "Factor out the coefficients of $ x^2 $ and $ y^2 $:", "$$\n9(x^2 - 2x) - 16(y^2 + 4y) = 144\n$$", "---", "### Step 2: Complete the Square", "Complete the square for $ x^2 - 2x $:", "- Take half of $-2$: $-1$, square it: $(-1)^2 = 1$\n- Add and subtract 1 inside the parentheses:\n$$\nx^2 - 2x + 1 - 1 = (x - 1)^2 - 1\n$$", "Complete the square for $ y^2 + 4y $:", "- Half of 4 is 2, square is 4\n- Add and subtract 4 inside the group:\n$$\ny^2 + 4y + 4 - 4 = (y + 2)^2 - 4\n$$", "Substitute back into the equation:", "$$\n9\left[(x - 1)^2 - 1\right] - 16\left[(y + 2)^2 - 4\right] = 144\n$$", "---", "### Step 3: Distribute and Simplify", "Distribute the constants:", "$$\n9(x - 1)^2 - 9 - 16(y + 2)^2 + 64 = 144\n$$", "Combine constants on the left:", "$$\n9(x - 1)^2 - 16(y + 2)^2 + 55 = 144\n$$", "Move constant to the right:", "$$\n9(x - 1)^2 - 16(y + 2)^2 = 89\n$$", "---", "### Step 4: Write in Standard Hyperbola Form", "Divide both sides by 89 to get 1 on the right:", "$$\n\frac{9(x - 1)^2}{89} - \frac{16(y + 2)^2}{89} = 1\n$$", "Rewriting:", "$$\n\frac{(x - 1)^2}{\frac{89}{9}} - \frac{(y + 2)^2}{\frac{89}{16}} = 1\n$$", "This is the standard form of a horizontal-axis hyperbola:", "$$\n\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\n$$", "where the center is at $ (h, k) $.", "---", "### Step 5: Identify the Center", "From the equation:", "- $ h = 1 $\n- $ k = -2 $", "Thus, the center of the hyperbola is at:", "$$\n\boxed{(1, -2)}\n$$", "---", "### Why Find the Center Important?", "Locating the center helps in several ways:\n- Determines the hyperbola’s location on the coordinate plane\n- Guides graphing by establishing the hyperbola’s crossing point\n- Is crucial when writing equations of tangents, asymptotes, or performing translations\n- Enhances conceptual understanding of conic sections and symmetry", "---", "### Final Thoughts", "Understanding how to transform a general quadratic equation into standard conic form enables you to identify vital features like the center of any hyperbola. In this case, applying completing the square carefully reveals the hyperbola centered at $ (1, -2) $. Mastering this process strengthens your ability to analyze and graph hyperbolas accurately.", "---", "Keywords: hyperbola center, hyperbola equation, completing the square, standard form conic sections, how to find center hyperbola, algebra tutoring, coordinate geometry, conic sections tutorial.\nMeta Description: Learn how to find the center of the hyperbola $ 9x^2 - 18x - 16y^2 - 64y = 144 $ by completing the square and rewriting in standard form."]
