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Chapter 10: Problem 17

Sketch the parabola with the given equation. Show and label its vertex, focus, axis, and directrix. $$4 x^{2}+4 x+4 y+13=0$$

Short Answer

Expert verified
The parabola opens downwards with vertex \((-\frac{1}{2}, -3)\), focus \((-\frac{1}{2}, -\frac{13}{4})\), and directrix \(y = -\frac{11}{4}\).

Step by step solution

01

Rewrite the Equation in Standard Form

Start by rewriting the given quadratic equation in a more recognizable form. The given equation is \( 4x^2 + 4x + 4y + 13 = 0 \). First, divide the entire equation by 4 to simplify: \( x^2 + x + y + \,\frac{13}{4} = 0 \). Rearrange to isolate \( y \): \( y = -x^2 - x - \frac{13}{4} \).
02

Complete the Square

The equation \( y = -x^2 - x - \frac{13}{4} \) needs the \( x \)-terms to be in a completed square form. Taking \( -x^2 - x \), factor out the negative sign: \( -(x^2 + x) \). To complete the square, take half of the \( x \)-coefficient \((\frac{1}{2})\) and square it \((\frac{1}{4})\). The expression becomes \( -(x^2 + x + \frac{1}{4} - \frac{1}{4}) \). Simplify to \(-(x + \frac{1}{2})^2 + \frac{1}{4}\). Thus, the transformed equation is \( y = -(x+\frac{1}{2})^2 + \frac{1}{4} - \frac{13}{4} \).
03

Find the Vertex

Simplify \( \frac{1}{4} - \frac{13}{4} \) to \( -3 \). Thus, the equation becomes \( y = -(x+\frac{1}{2})^2 - 3 \). The vertex form of the parabola is \( y = -(x-h)^2 + k \). Here, the vertex \( (h, k) \) is \( (-\frac{1}{2}, -3) \).
04

Determine the Focus and Directrix

For a parabola \( y = a(x-h)^2 + k \), the focus is at \( (h, k + \frac{1}{4a}) \) and the directrix is \( y = k - \frac{1}{4a} \). Since \( a = -1 \), the focus is at \( (-\frac{1}{2}, -3 + \frac{1}{-4}) \to (-\frac{1}{2}, -\frac{13}{4}) \) and the directrix is \( y = -3 + \frac{1}{4} \to -\frac{11}{4} \).
05

Sketch the Parabola

Plot the vertex \( (-\frac{1}{2}, -3) \) on the coordinate plane. Mark the focus \( (-\frac{1}{2}, -\frac{13}{4}) \) which is slightly below the vertex. Draw the directrix line \( y = -\frac{11}{4} \), which is slightly above the vertex. Since the \( x \)-term is squared and \( a = -1 \), the parabola opens downwards. Draw a symmetrical curve opening downwards through the plotted points.

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Key Concepts

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

Vertex
A parabola's vertex is a key concept, functioning as the point where the curve's direction changes. For the given equation, after completing the square, we determined the equation in the form: \( y = -(x+\frac{1}{2})^2 - 3 \). From this vertex form of a parabola, we can directly identify the vertex as \((h, k)\). Here, the vertex \( (h, k) \) is at \( (-\frac{1}{2}, -3) \). This point is particularly important because:
  • It represents the highest point on the parabola when it opens downward.
  • It provides symmetry, meaning each side of the vertex mirrors the other.
Locating the vertex helps in understanding the overall behavior and shape of the parabola.
Focus
The focus of a parabola is a unique point that provides a way to describe its geometry and reflective properties. We calculated the location of the focus using the formula: \( (h, k + \frac{1}{4a}) \), where \( a \) is the coefficient of the squared term in the vertex form. For our equation, \( a = -1 \), giving us the focus at \( (-\frac{1}{2}, -3 + \frac{1}{-4}) \). This simplifies to \( (-\frac{1}{2}, -\frac{13}{4}) \). The focus is notable because:
  • All points on the parabola are equidistant from the focus and the directrix.
  • It lies inside the parabola, in the direction in which the parabola opens.
Understanding the focus assists in visualizing how the parabola curves and interacts with its axes and other points.
Axis
The axis of a parabola is an important symmetry line that vertically or horizontally passes through the vertex and focus. For our downward-opening parabola, the axis is vertical. The importance of the axis includes:
  • Serves as a line of symmetry, dividing the parabola into two mirror-image halves.
  • Helps in easily finding equal y-values for x-values equidistant from the axis.
In our specific example, because the parabola's vertex is located at \( x = -\frac{1}{2} \), the axis of symmetry can be represented as the line \( x = -\frac{1}{2} \). This line provides a reference point around which the parabola's shape can be understood more fully.
Directrix
A directrix of a parabola is an essential concept accessed through the relationship with the focus. It is a line perpendicular to the axis of symmetry that all points on the parabola maintain a constant distance from. We determine the directrix using the formula: \( y = k - \frac{1}{4a} \). For the given problem, the directrix line is found at \( y = -3 + \frac{1}{4} \), simplifying to \( y = -\frac{11}{4} \). Noteworthy aspects of the directrix include:
  • It is one of the two geometric entities (the other being the focus) from which every point on the parabola is equidistant.
  • It lies outside of the parabola, opposite the direction that the parabola opens.
This line aids in defining the parabola's openness and its unique distance relationship from the directrix and focus.

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