$ 4[(x+3)^2 - 9] - 9[(y-1)^2 - 1] = -9 $.

["Optimize Your Understanding: Solving and Analyzing the Equation — $ 4[(x+3)^2 - 9] - 9[(y-1)^2 - 1] = -9 $", "Ec-qua-ti-on-solve: mastering a specific algebraic equation is essential for students, math enthusiasts, and professionals who rely on precision in problem-solving. In this article, we’ll deep-dive into understanding and solving the equation", "$$\n4[(x+3)^2 - 9] - 9[(y-1)^2 - 1] = -9\n$$", "and explore its geometric and algebraic significance. If you're searching for efficient ways to simplify, solve, or analyze this equation, you’re in the right place.", "---", "## Step 1: Expand and Simplify the Equation", "To solve this equation, start by expanding the squared terms:", "$$\n4[(x+3)^2 - 9] = 4(x^2 + 6x + 9 - 9) = 4(x^2 + 6x)\n$$\n$$\n= 4x^2 + 24x\n$$", "Similarly for the $y$-terms:", "$$\n-9[(y-1)^2 - 1] = -9[(y^2 - 2y + 1) - 1] = -9(y^2 - 2y + 1 - 1) = -9(y^2 - 2y)\n$$\n$$\n= -9y^2 + 18y\n$$", "Now substitute these back:", "$$\n4x^2 + 24x - 9y^2 + 18y = -9\n$$", "---", "## Step 2: Rearrange into Standard Form", "Bring all terms to one side:", "$$\n4x^2 + 24x - 9y^2 + 18y + 9 = 0\n$$", "Group $x$ and $y$ terms:", "$$\n(4x^2 + 24x) + (-9y^2 + 18y) + 9 = 0\n$$", "We complete the square for both $x$ and $y$ terms.", "### Squaring and Completing the Square:", "For $x$:\nFactor out 4:\n$$\n4(x^2 + 6x)\n$$\nComplete the square:\n$$\nx^2 + 6x = (x+3)^2 - 9\n$$\nSo:\n$$\n4[(x+3)^2 - 9] = 4(x+3)^2 - 36\n$$", "For $y$:\nFactor out -9:\n$$\n-9[(y^2 - 2y)]\n$$\nComplete the square inside:\n$$\ny^2 - 2y = (y-1)^2 - 1\n$$\nSo:\n$$\n-9[(y-1)^2 - 1] = -9(y-1)^2 + 9\n$$", "Now substitute back:", "$$\n4(x+3)^2 - 36 - 9(y-1)^2 + 9 + 9 = 0\n$$", "Simplify constants:\n$$\n4(x+3)^2 - 9(y-1)^2 - 18 = 0\n$$", "Move constant to the right:", "$$\n4(x+3)^2 - 9(y-1)^2 = 18\n$$", "Divide entire equation by 18 to normalize:", "$$\n\frac{4(x+3)^2}{18} - \frac{9(y-1)^2}{18} = 1\n\Rightarrow \frac{(x+3)^2}{4.5} - \frac{(y-1)^2}{2} = 1\n$$", "This is the standard form of a hyperbola centered at $(-3, 1)$, opening horizontally.", "---", "## Step 3: Key Features of the Hyperbola", "- Center: $(-3, 1)$\n- Vertices: $ \pm \sqrt{4.5} $ along the $x$-axis → $ \pm \frac{3\sqrt{2}}{2} \approx \pm 2.12 $ from $(-3, 1)$\n- Asymptotes: Lines $ y - 1 = \pm \frac{\sqrt{2}}{3}(x + 3) $\n- Transverse axis length: $ 2a = 2\sqrt{4.5} = 3\sqrt{2} $\n- Conjugate axis length: $ 2b = 2\sqrt{2} $", "---", "## Step 4: Why This Equation Matters (Practical Applications)", "Equations of this form model precise real-world phenomena:\n- Satellite dish geometry relies on hyperbolic surfaces for signal focus.\n- Optics and laser reflectors use hyperbolic paths.\n- Economics and physics employ conic sections for optimization and motion analysis.", "Solving such algebraic expressions is more than academic—it’s a gateway to applying analytical thinking in engineering, design, and scientific modeling.", "---", "## Step 5: How to Solve for $x$ or $y$?", "Solving for $x$:\nFrom simplified step:\n$$\n4(x+3)^2 = 9(y-1)^2 + 18\n\Rightarrow (x+3)^2 = \frac{9(y-1)^2 + 18}{4}\n\Rightarrow x+3 = \pm \sqrt{ \frac{9(y-1)^2 + 18}{4} }\n\Rightarrow x = -3 \pm \frac{1}{2} \sqrt{9(y-1)^2 + 18}\n$$", "Solving for $y$:\n$$\n-9(y-1)^2 = -4(x+3)^2 + 18\n\Rightarrow (y-1)^2 = \frac{4(x+3)^2 - 18}{9}\n\Rightarrow y = 1 \pm \sqrt{ \frac{4(x+3)^2 - 18}{9} }\n$$", "---", "## Final Thoughts", "The equation $ 4[(x+3)^2 - 9] - 9[(y-1)^2 - 1] = -9 $ elegantly reveals a hyperbolic curve with defined symmetry and algebraic properties. Mastering its solution enhances algebraic fluency and paves the way for advanced geometry and calculus applications.", "Whether you're a student tackling quadratic equations, a teacher illustrating conic sections, or a professional using mathematical modeling, this equation demonstrates how simplification and completing the square unlock deeper mathematical insight.", "---", "Keywords:\n$ 4[(x+3)^2 - 9] - 9[(y-1)^2 - 1] = -9 $, hyperbola equation, algebraic manipulation, completing the square, conic sections, center $(-3, 1)$, vertex calculation, educational algebra.", "Optimize your algebra — understand, solve, and apply!"]