$ 4(x+3)^2 - 9(y-1)^2 = 18 $.

["Understanding the Equation: $ 4(x+3)^2 - 9(y-1)^2 = 18 $", "The equation $ 4(x+3)^2 - 9(y-1)^2 = 18 $ represents a conic section, specifically a hyperbola, and understanding its structure provides insight into its geometric properties and solutions. This article explains how to rewrite, analyze, and interpret this equation step-by-step, enhancing clarity for students, educators, and math enthusiasts.", "---", "### What Type of Conic Section is This?", "Given the form $ A(x-h)^2 - B(y-k)^2 = C $, this equation describes a hyperbola centered at $(h, k)$ with axes aligned to the coordinate axes. The positive coefficient on the $x$-term indicates the hyperbola opens left and right, along the horizontal axis.", "---", "### Step 1: Normalize the Equation to Standard Form", "To identify key features such as center, vertices, and asymptotes, we rewrite the equation in standard form:", "[\n4(x+3)^2 - 9(y-1)^2 = 18\n]", "Divide both sides by 18:", "[\n\frac{4(x+3)^2}{18} - \frac{9(y-1)^2}{18} = 1\n]", "Simplify the fractions:", "[\n\frac{(x+3)^2}{\frac{18}{4}} - \frac{(y-1)^2}{2} = 1 \quad \Rightarrow \quad \frac{(x+3)^2}{4.5} - \frac{(y-1)^2}{2} = 1\n]", "We can write it as:", "[\n\frac{(x+3)^2}{\left(\frac{3\sqrt{2}}{2}\right)^2} - \frac{(y-1)^2}{(\sqrt{2})^2} = 1\n]", "This confirms standard hyperbolic form:", "[\n\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\n]", "where:\n- Center: $ (h, k) = (-3, 1) $\n- $ a^2 = 4.5 \Rightarrow a = \frac{3}{\sqrt{2}} $\n- $ b^2 = 2 \Rightarrow b = \sqrt{2} $", "---", "### Step 2: Key Features of the Hyperbola", "- Center: $(-3, 1)$ — the midpoint of the hyperbola’s conjugate and transverse axes.\n- Vertices: Since the transverse axis is horizontal, the vertices are located at\n $$\n (h \pm a, k) = \left(-3 \pm \frac{3}{\sqrt{2}},\ 1\right)\n $$\n Thus, two vertices:\n $$\n \left(-3 + \frac{3}{\sqrt{2}},, 1\right) \quad \ ext{and} \quad \left(-3 - \frac{3}{\sqrt{2}},, 1\right)\n $$", "- Asymptotes: The slope of the asymptotes for this hyperbola is $ \pm \frac{b}{a} = \pm \frac{\sqrt{2}}{3/\sqrt{2}} = \pm \frac{2}{3} $.\n Using center point $(-3, 1)$, the asymptotes are:", "$$\n y - 1 = \pm \frac{2}{3}(x + 3)\n $$", "---", "### Step 3: Solving for $ y $ to Graph or Analyze", "To express $ y $ explicitly, rearrange the original equation:", "[\n4(x+3)^2 - 9(y-1)^2 = 18\n]", "Solve for $ (y-1)^2 $:", "[\n9(y-1)^2 = 4(x+3)^2 - 18\n]", "[\n(y-1)^2 = \frac{4(x+3)^2 - 18}{9}\n]", "Then,", "[\ny - 1 = \pm \sqrt{ \frac{4(x+3)^2 - 18}{9} } = \pm \frac{ \sqrt{4(x+3)^2 - 18} }{3}\n]", "For real solutions, the expression under the square root must be non-negative:", "[\n4(x+3)^2 - 18 \geq 0 \quad \Rightarrow \quad (x+3)^2 \geq \frac{18}{4} = 4.5\n]", "So, the domain is:\n$$\nx \leq -3 - \frac{3}{\sqrt{2}} \quad \ ext{or} \quad x \geq -3 + \frac{3}{\sqrt{2}}\n$$", "---", "### Step 4: Real-World Applications & Interpretation", "Hyperbolas like this appear in physics (e.g., satellite tracking), architecture (reflectors), navigation (GPS signals), and economics (cost functions). The center $(-3, 1)$ provides a reference point, while vertices help locate focus-related properties.", "---", "### Summary", "The equation $ 4(x+3)^2 - 9(y-1)^2 = 18 $ is a standard horizontal hyperbola centered at $(-3, 1)$ with distinct asymptotes and defined only in two asymptotic regions. Understanding its algebraic form reveals critical geometric elements vital for graphing, solving, and application.", "---", "### SEO Keywords\nhyperbola equation, standard form hyperbola, center and asymptotes hyperbola, solving $ 4(x+3)^2 - 9(y-1)^2 = 18 $, hyperbolic functions, conic section analysis", "---", "Start modeling conic sections with precision — understanding equations unlocks powerful geometry insights!"]