Problem 46 - Precalculus: Mathematics for Calculus

Question

Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Opens upward with focus 5 units away from the vertex

Step-by-Step Solution

Answer

The equation is \(y = \frac{1}{20}x^2\).

Step 1: Understand the Parabola's Vertex Form

Since the vertex is at the origin (0, 0) and the general equation of a parabola that opens upward is \(y = ax^2\).

Step 2: Use the Distance to the Focus

For a parabola that opens upwards and has its vertex at the origin, the distance from the vertex to the focus is \(\frac{1}{4a}\). Given that the focus is 5 units away from the vertex, we have:\(\frac{1}{4a} = 5.\)

Step 3: Solve for 'a'

Solve the equation \(\frac{1}{4a} = 5\) for \(a\).1. Multiply both sides by \(4a\) to get 1 = 20a.2. Divide both sides by 20 to find \(a\).Thus, \(a = \frac{1}{20}\).

Step 4: Write the Equation of the Parabola

Substitute \(a = \frac{1}{20}\) into the vertex form equation from Step 1 \(y = ax^2\).Therefore, the parabola's equation is:\[y = \frac{1}{20}x^2.\]

Key Concepts

Vertex Form

In parabolas, the vertex form of an equation is crucial because it reveals key features like the vertex and the way the parabola opens. When a parabola's vertex is located at the origin (0, 0), the equation simplifies to a very basic form compared to when the vertex is elsewhere. The standard vertex form of a parabola when it opens upwards or downwards is given by:\[y = a(x-h)^2 + k\]Here,

• \(h\) and \(k\) are the horizontal and vertical coordinates of the vertex, respectively.
• When the vertex is at the origin, \(h\) and \(k\) are both 0, making the equation: \(y = ax^2\).
• The parameter \(a\) determines the opening direction and the width of the parabola.

For positive \(a\), the parabola opens upward, while a negative \(a\) means it opens downward. The larger the absolute value of \(a\), the 'narrower' the parabola. Always remember: changing \(a\) affects how 'stretched' or 'compressed' the parabola appears on a graph.

Focus and Directrix

To fully understand a parabola, one must consider both its focus and directrix. These elements are essential parts of a parabola's definition.The focus is a point from which distances are measured to construct the parabola. For a parabola that opens upward and has its vertex on the origin:

• The focus is at the point \((0, p)\), where \(p\) is a positive number.
• In our case, the focus is 5 units from the vertex, making the focus at \((0, 5)\).

The directrix, meanwhile, is a line that is equidistant from the parabola as the focus. For an upward opening parabola with the vertex at the origin,

• The directrix is the horizontal line \(y = -p\).
• Thus, our directrix is located at \(y = -5\).

The set of points (each, equidistant to this focus and directrix) outlines the parabola's curve on the graph. Together with the vertex, these components help define the parabola's precise shape and location.

Distance to Focus

The distance from the vertex to the focus of a parabola is a critical piece of information. This distance determines key variables in the equation of the parabola. For a parabola that opens upwards with the vertex at the origin, the relationship between the distance to the focus \(p\) and the parameter \(a\) is given by:\[p = \frac{1}{4a}\]Here,

• \(p\) is the distance from the vertex to the focus.
• \(a\) dictates how quickly the parabola opens.

In the problem, the focus is 5 units from the vertex point. Substituting into the formula, we have:\[\frac{1}{4a} = 5\]Solving for \(a\), we multiply both sides by \(4a\) to isolate \(1\):1 = 20aDividing by 20, we find\[a = \frac{1}{20}\].This value allows us to complete our equation, ensuring the parabola opens at the correct rate with the focus precisely 5 units from the origin. Thus, our final equation becomes \(y = \frac{1}{20}x^2\), perfectly matching the given conditions.