Solution: The standard form of a parabola with vertical axis is $ y = \frac{1}{4p}x^2 $. The focus is at $ (0, p) $, so $ p = 5 $. The width of 20 units implies the parabola passes through $ (10, 5) $. Substituting: $ 5 = \frac{1}{20}(100) \Rightarrow 5 = 5 $, which holds. Thus, the equation is $ y = \frac{1}{20}x^2 $. \boxed{y = \dfrac{1}{20}x^2}

Solution: The standard form of a parabola with vertical axis is $ y = \frac{1}{4p}x^2 $. The focus is at $ (0, p) $, so $ p = 5 $. The width of 20 units implies the parabola passes through $ (10, 5) $. Substituting: $ 5 = \frac{1}{20}(100) \Rightarrow 5 = 5 $, which holds. Thus, the equation is $ y = \frac{1}{20}x^2 $. \boxed{y = \dfrac{1}{20}x^2}

["Solution: The Standard Form of a Parabola with Vertical Axis", "Understanding the standard form of a parabola with a vertical axis is essential in both algebra and geometry. This elegant equation reveals key features such as the vertex, focus, and axis of symmetry. For a parabola defined as\n$$\ny = \frac{1}{4p}x^2,\n$$\nthe focus is located at $ (0, p) $, and the directrix lies at $ y = -p $. The parameter $ p $ directly determines the parabola’s width and orientation.", "---", "### Using the Given Value of $ p $", "In this problem, we are told that $ p = 5 $. Substituting into the standard form gives:\n$$\ny = \frac{1}{4 \cdot 5}x^2 = \frac{1}{20}x^2.\n$$\nThis concise expression captures the unique shape of the parabola based on the focus position.", "---", "### Verifying the Equation with a Known Point", "To ensure accuracy, we verify that the parabola passes through the point $ (10, 5) $, a point on the curve since its width spans 20 units (from $ x = -10 $ to $ x = 10 $).", "Plug $ x = 10 $ into the equation:\n$$\ny = \frac{1}{20}(10)^2 = \frac{1}{20} \cdot 100 = 5.\n$$\nThis confirms that $ (10, 5) $ lies on the parabola — satisfying the 20-unit width condition.", "---", "### Deeper Insight: What Does $ y = \frac{1}{20}x^2 $ Represent?", "The equation $ y = \frac{1}{20}x^2 $ defines a vertically oriented parabola opening upwards with vertex at the origin $ (0, 0) $. Given $ p = 5 $, the focus is indeed at $ (0, 5) $, distance $ p $ above the vertex — perfectly aligned with the geometric definition.", "This form simplifies calculations in applications ranging from projectile motion modeling to optics and architectural design.", "---", "### Final Equation", "Thus, the standard form of the parabola with vertical axis and $ p = 5 $, validated by known geometry and algebra, is:\n$$\n\boxed{y = \dfrac{1}{20}x^2}\n$$\nThis equation is both mathematically sound and practically useful for analyzing parabolic curves in science and engineering."]

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