9(x - 3)^2 - 16(y - 2)^2 = 88

["Understanding the Equation 9(x – 3)² – 16(y – 2)² = 88: A Comprehensive Guide to Its Graph and Properties", "When studying conic sections, the equation 9(x – 3)² – 16(y – 2)² = 88 presents an intriguing hyperbola with a centered vertex and defined axes. Whether you’re a high school student diving into algebra or a lifelong learner exploring analytic geometry, understanding this equation unlocks deeper insights into its shape, orientation, and key graphical features.", "---", "### What Is the Equation?", "The given equation is:", "[\n9(x – 3)^2 - 16(y – 2)^2 = 88\n]", "This is a hyperbola in standard form, centered away from the origin, with its center at the point (x₀, y₀) = (3, 2). The subtraction of the squared terms indicates a hyperbola, not an ellipse.", "---", "### Rewriting in Standard Form", "To better understand the geometric features, we convert the equation into standard form by dividing both sides by 88:", "[\n\frac{9(x - 3)^2}{88} - \frac{16(y - 2)^2}{88} = 1\n]", "Simplify the fractions:", "[\n\frac{(x - 3)^2}{\frac{88}{9}} - \frac{(y - 2)^2}{\frac{88}{16}} = 1\n]", "Calculating the denominators:", "- (\frac{88}{9} \approx 9.78)\n- (\frac{88}{16} = 5.5)", "So the standard form becomes:", "[\n\frac{(x - 3)^2}{\frac{88}{9}} - \frac{(y - 2)^2}{5.5} = 1\n]", "This matches the standard hyperbola form:", "[\n\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\n]", "where:\n- Center: ((h, k) = (3, 2))\n- (a^2 = \frac{88}{9}) → (a = \sqrt{\frac{88}{9}} = \frac{\sqrt{88}}{3} = \frac{2\sqrt{22}}{3})\n- (b^2 = 5.5 = \frac{11}{2}) → (b = \sqrt{\frac{11}{2}} = \frac{\sqrt{22}}{2})", "---", "### Axis Orientations", "Because the x-term is positive and the y-term is negative, the hyperbola opens left and right (horizontal transverse axis), not vertically.", "---", "### Key Graph Features", "#### Center\nAt ((3, 2)). This is where the hyperbola’s two branches symmetrically branch off.", "#### Vertices\nSince the transverse axis is horizontal, vertices lie a units to the left and right of the center:", "- Vertex 1: ((3 + a, 2) = \left(3 + \frac{2\sqrt{22}}{3},\ 2\right))\n- Vertex 2: ((3 - a, 2) = \left(3 - \frac{2\sqrt{22}}{3},\ 2\right))", "#### Asymptotes", "For hyperbolas of the form (\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1), the asymptotes are straight lines passing through the center and with slopes:", "[\nm = \pm \frac{b}{a}\n]", "Substituting (a = \frac{2\sqrt{22}}{3}), (b = \frac{\sqrt{22}}{2}):", "[\n\frac{b}{a} = \frac{\frac{\sqrt{22}}{2}}{\frac{2\sqrt{22}}{3}} = \frac{\sqrt{22}}{2} \cdot \frac{3}{2\sqrt{22}} = \frac{3}{4}\n]", "Thus, the asymptotes are:", "- (y - 2 = \frac{3}{4}(x - 3))\n- (y - 2 = -\frac{3}{4}(x - 3))", "These lines guide the drawing of the hyperbola branches.", "---", "### Plotting the Hyperbola", "1. Plot the center at ((3, 2)).\n2. Mark vertices using (a = \frac{2\sqrt{22}}{3} \approx 3.03):\n - Right vertex ≈ ( (6.03, 2) )\n - Left vertex ≈ ( (-0.03, 2) )\n3. Draw asymptotes with slope (\pm 0.75) through ((3, 2)).\n4. Sketch two open branches extending left and right from the center, asymptotically approaching the lines.", "---", "### Significance and Applications", "This hyperbola appears in physics (e.g., trajectory modeling), engineering (antenna radiation patterns), and calculus (as functions of raising the conic to analyze symmetry and asymptotes). Understanding its structure aids in sketching, solving equations, and interpreting real-world phenomena tied to these curves.", "---", "### Final Thoughts", "The equation (9(x – 3)^2 – 16(y – 2)^2 = 88) reveals a centrally located hyperbola opening horizontally, centered at ((3, 2)), with a fixed relationship between its axes derived from (a) and (b). Mastery of such equations builds a powerful foundation in analytic geometry and prepares learners for advanced mathematical modeling.", "---", "SEO Keywords: \nHyperbolaEquation #9(x–3)²–16(y–2)²=88 #ConicSections #GraphofHyperbola #CenterOfHyperbola #Asymptotes #SkipToContent #AlgebraGeometry", "---", "Summary:\nThis hyperbola’s center at (3,2), horizontal transverse axis, vertices, and asymptotes follow clearly from its standard form. Simplifying and recognizing standard properties enables accurate visualization and deeper analytical understanding—key skills for excelling in mathematics and related disciplines."]