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Chapter 10: Problem 11

Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$y=5 x^{2}$$

Short Answer

Expert verified
Focus: \((0, \frac{1}{20})\), Directrix: \( y = -\frac{1}{20} \), Focal diameter: \( \frac{1}{5} \).

Step by step solution

01

Standard Form Comparison

Recognize the given equation of the parabola: \( y = 5x^2 \). The standard form for a parabola opening upwards or downwards is \( y = a(x-h)^2 + k \). Therefore, this equation resembles \( y = 5(x-0)^2 + 0 \), indicating a vertex form at \((h, k) = (0, 0)\) and \( a = 5 \).
02

Determine Parabola Orientation and Focus Distance

Since the parabola's equation is \( y = 5x^2 \), it opens upwards with the vertex at \((0, 0)\). The formula for the distance 'p' from the vertex to the focus or directrix is \( \frac{1}{4a} \). So, \( p = \frac{1}{4 \times 5} = \frac{1}{20} \).
03

Find the Focus

Given that the parabola opens upwards from the vertex \((0, 0)\), the focus is \( p \) units above the vertex. Therefore, the focus is located at \( (0, \frac{1}{20}) \).
04

Locate the Directrix

The directrix is a line parallel to the opening direction of the parabola but on the opposite side of the vertex, \( p \) units below it. Therefore, the directrix is the line \( y = -\frac{1}{20} \).
05

Calculate the Focal Diameter

The focal diameter, also known as the latus rectum, is \(|4p|\). With \( p = \frac{1}{20} \), the focal diameter is \( |4 \times \frac{1}{20}| = \frac{1}{5} \).
06

Sketch the Graph

To sketch the graph, place the vertex at the origin \((0,0)\), plot the focus at \((0, \frac{1}{20})\), and draw the line \( y = -\frac{1}{20} \) as the directrix. The graph should be a parabola opening upwards with symmetry along the y-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabolas
In mathematics, a parabola is a specific type of curve that results from slicing a cone vertically. This shape is seen in various natural and engineered contexts, from satellite dishes to paths of thrown baseballs. The standard form of a parabola's equation can offer insights into its geometric properties and orientation.
  • A parabola's equation typically appears like this: \( y = a(x-h)^2 + k \), where \((h, k)\) represents its vertex, the peak or bottom point.
  • "\(a\)" affects how tight or wide the parabola is; positive values create upward-opening parabolas, while negatives point downwards.
For example, in our exercise with \( y = 5x^2 \), the equation simplifies to show a parabola opening upwards, with a vertex at the origin \((0, 0)\). Understanding these basics is essential for identifying the parabola's other components, such as its focus and directrix.
Focus and Directrix
The focus and directrix are fundamental in defining a parabola. Imagine a parabolic mirror: light rays entering it parallel to the axis will reflect and travel through a single point, the focus.
  • The focus is a notable point inside the parabola. It helps shape the curve's path by ensuring consistency in how distances are measured across the parabola.
  • The directrix is a line outside the parabola. Contrary to the focus, it doesn't cut through the curve but helps define it spatially.
  • For any point on the parabola, the distance to the focus equals its distance to the directrix.
In our given equation \( y = 5x^2 \), the focus lies just above the vertex at \((0, \frac{1}{20})\), and the directrix sits opposite at \( y = -\frac{1}{20} \). Knowing these ensures one can precisely locate the shape in a coordinate plane, showing its relation to both its vertex and these defining features.
Graphing Parabolas
Graphing a parabola helps visualize its properties, especially when there are components like the focus and directrix. A step-by-step approach is vital.
  • Start by pinpointing the vertex \((0, 0)\). It's your base for constructing the parabola.
  • Then, locate the focus at \((0, \frac{1}{20})\). This point not only tells you the direction in which to open the curve but also how focused or stretched it might appear.
  • Next is the directrix, a line \( y = -\frac{1}{20} \) beneath the vertex. Think of it as the guideline ensuring your parabola arcs correctly.
  • Draw the parabola. Ensure it's equidistant in its span from both the focus and directrix, reflecting symmetry along the y-axis.
In practice, these steps create a harmonious representation of the parabola on graph paper or screen, demonstrating its equilibrium balance between focus and directrix. Understanding graphics in this way not only supports comprehension but bolsters your abilities to engage with applied mathematics in broader contexts.

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Most popular questions from this chapter

Use a graphing device to graph the conic. $$9 x^{2}+36=y^{2}+36 x+6 y$$

Determine the equation of the given conic in \(X Y\) -coordinates when the coordinate axes are rotated through the indicated angle. $$x^{2}+2 \sqrt{3} x y-y^{2}=4, \quad \phi=30^{\circ}$$

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