Problem 19
Question
An equation of a parabola is given. (a) Find the focus, directrix, and focal diameter of the parabola. (b) Sketch a graph of the parabola and its directrix. $$5 y=x^{2}$$
Step-by-Step Solution
Verified Answer
Focus: \((0, \frac{5}{4})\), Directrix: \(y = -\frac{5}{4}\), Focal Diameter: 5.
1Step 1: Identify the Parabola's Direction
The given equation is \(5y = x^2\). By rearranging it, we get \(y = \frac{1}{5}x^2\). This equation represents a parabola that opens upwards because the \(x^2\) term is positive.
2Step 2: Determine Standard Form
The standard form for a parabola that opens upwards is \((x-h)^2 = 4p(y-k)\). Comparing \(x^2 = 5y\) to \(x^2 = 4py\), we find \(4p = 5\), so \(p = \frac{5}{4}\). The vertex \((h, k)\) is \((0,0)\).
3Step 3: Find the Focus
The focus of the parabola \((x-h)^2 = 4p(y-k)\) is located at \((h, k+p)\). Since \(h=0\) and \(k=0\), the focus is at \((0, \frac{5}{4})\).
4Step 4: Determine the Directrix
The directrix of a parabola \((x-h)^2 = 4p(y-k)\) is the line \(y = k-p\). Here, the directrix is \(y = 0 - \frac{5}{4} = -\frac{5}{4}\).
5Step 5: Calculate the Focal Diameter
The focal diameter of a parabola is the absolute value of \(4p\). We have \(4p = 5\), so the focal diameter is 5.
6Step 6: Sketch the Parabola and Directrix
Draw the parabola with its vertex at the origin, opening upward. Mark the focus at \((0, \frac{5}{4})\) and draw the directrix as the horizontal line \(y = -\frac{5}{4}\). The parabola is symmetrically centered between the focus and the directrix.
Key Concepts
Focus of a ParabolaDirectrix of a ParabolaFocal Diameter of a Parabola
Focus of a Parabola
The concept of the focus is integral to understanding parabolas. In any parabola, the focus is a special point that helps define the curve. For the given parabola defined by the equation \(5y = x^2\), after rearranging, it becomes \(y = \frac{1}{5}x^2\), indicating it opens upwards. The focus is located at a position that ensures every point on the parabola is equidistant from both the focus and the directrix, a line we'll discuss shortly. For a parabola in the form \((x-h)^2 = 4p(y-k)\), the focus is found at \((h, k+p)\). In our equation, \(h = 0\) and \(k = 0\), so with \(p = \frac{5}{4}\), the focus is at \((0, \frac{5}{4})\). The presence of the focus highlights the parabolic shape and is essential for graphing and understanding the curve's behavior. Such relations you might notice in real-world applications include satellite dishes or lights reflecting off a parabolic mirror, where rays passing through the focus reach a stage of concentric symmetry.
Directrix of a Parabola
The directrix is another critical component in understanding a parabola. It represents a fixed straight line used in the geometric definition of the parabola, ensuring that each point on the curve is equidistant from both the focus and the directrix. This distance property is crucial in applications ranging from physics to architecture.
The given parabola \(y = \frac{1}{5}x^2\) is structured in its standard form, allowing us to derive the directrix. For a parabola given as \((x-h)^2 = 4p(y-k)\), the directrix is defined by \(y = k-p\). In our case with \(h=0, k=0\) and \(p=\frac{5}{4}\), this provides us with the line \(y = -\frac{5}{4}\). The directrix is a mathematical mediator providing balance and symmetry to the parabola, which, like the focus, is pivotal in accurately sketching the graph.
The given parabola \(y = \frac{1}{5}x^2\) is structured in its standard form, allowing us to derive the directrix. For a parabola given as \((x-h)^2 = 4p(y-k)\), the directrix is defined by \(y = k-p\). In our case with \(h=0, k=0\) and \(p=\frac{5}{4}\), this provides us with the line \(y = -\frac{5}{4}\). The directrix is a mathematical mediator providing balance and symmetry to the parabola, which, like the focus, is pivotal in accurately sketching the graph.
Focal Diameter of a Parabola
Understanding the focal diameter is key to visualizing the width of a parabola at the focus. Also known as the latus rectum, it measures the distance across the parabola, through its focus, parallel to the directrix.
For parabolas of the form \((x-h)^2 = 4p(y-k)\), the focal diameter is simply the absolute value of \(4p\). In our given equation, we have \(4p = 5\), thus making the focal diameter 5. This parameter provides insight into the spread of the parabola and how open or "narrow" the curve is. The focal diameter is significant as it assists in determining how light or signals might be focused by a parabolic reflector, in real-world situations.
For parabolas of the form \((x-h)^2 = 4p(y-k)\), the focal diameter is simply the absolute value of \(4p\). In our given equation, we have \(4p = 5\), thus making the focal diameter 5. This parameter provides insight into the spread of the parabola and how open or "narrow" the curve is. The focal diameter is significant as it assists in determining how light or signals might be focused by a parabolic reflector, in real-world situations.
Other exercises in this chapter
Problem 18
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