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Chapter 10: Problem 392

Graph the parabola, labeling the vertex, focus, and directrix. $$(x-1)^{2}=-4(y+3)$$

Short Answer

Expert verified
The vertex is at \((1, -3)\), the focus is \((1, -4)\), and the directrix is \(y = -2\).

Step by step solution

01

Identify the Equation Form

The given equation \((x-1)^2 = -4(y+3)\) is in the form \((x-h)^2 = 4p(y-k)\) which represents a parabola that opens either upward or downward. Here, \((h, k)\) is the vertex and \(p\) determines the direction and width of the parabola.
02

Find the Vertex

In the equation \((x-1)^2 = -4(y+3)\), compare it with \((x-h)^2 = 4p(y-k)\). We find \(h = 1\) and \(k = -3\). Thus, the vertex of the parabola is at the point \((1, -3)\).
03

Determine the Value of \(p\)

From the equation \((x-1)^2 = -4(y+3)\), we identify that \(4p = -4\). Therefore, \(p = -1\), which indicates the parabola opens downward because \(p\) is negative.
04

Find the Focus

The focus of the parabola is \( |p| \) units from the vertex in the direction the parabola opens. Since \(p = -1\), the focus is 1 unit below the vertex. Thus, the focus is at \((1, -4)\).
05

Determine the Directrix

The directrix is a line that is \(|p|\) units in the opposite direction of the focus. Since the parabola opens downward, the directrix is a horizontal line 1 unit above the vertex. Thus, the directrix is \(y = -2\).
06

Plot the Parabola

Using the vertex \((1, -3)\), focus \((1, -4)\), and directrix \(y = -2\), sketch the parabola. The parabola will open downward from the vertex towards the focus, and the directrix serves as a guide that the parabola is equidistant from any point on it to the focus.

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Key Concepts

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

Vertex of a parabola
Every parabola has a point called the vertex, which is essentially the "tip" or the "turning point" of the parabola. It's an important concept because it gives you a very precise location on the parabola. In the equation
  • \((x-h)^2 = 4p(y-k)\),
  • \( (h, k) \) represents the vertex.
      • This means that the vertex is located at this specific coordinate on the graph.
      For example, if we take the equation given in the problem
      • \((x-1)^2 = -4(y+3)\)
      we compare it with the standard form \((x-h)^2 = 4p(y-k)\) and find that
      • h = 1
      • k = -3.
          • So, the vertex is at the point
          • \( (1, -3) \).
              • This point is crucial because the entire parabola is perfectly symmetrical around this point, and it completely determines the shape of the parabola.
Focus of a parabola
The focus of a parabola is like its inner guide. Imagine that a point exists within the parabolic arc, towards which every point on the parabola has some relation. Sounds deep, right? Here's how it works:
In the form \((x-h)^2 = 4p(y-k)\), the focus is found
  • \(|p|\) units from the vertex
  • in the direction the parabola opens.
If the parabola opens downward (as it does in our given equation \((x-1)^2 = -4(y+3)\)), the focus will be beneath the vertex.
For our example,

  • indicating 1 unit below the vertex,
so the focus is found at
  • \((1, -4)\).
The focus controls the "width" or "narrowness" of the parabola, affecting how "wide" the arms of the parabola seem as they stretch away from the vertex.
Directrix of a parabola
The directrix of a parabola might sound complex, but it's basically a line outside the parabola that helps to define it. The directrix is a line that lies opposite to the parabola's opening direction.
To find it, you measure
  • \(|p|\) units from the vertex
  • in the opposite direction of the focus.
      • In our example, where
      • \(p = -1\),
      • the parabola opens downward. Hence, the directrix lies 1 unit above the vertex.
      So for the vertex
      • \((1, -3)\),
      the directrix is the line
      • \(y = -2\).
          • The directrix helps in providing a reference to ensure that the parabola stretches equidistant from any point to both the focus and this line, maintaining its symmetrical nature across the vertex.

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Most popular questions from this chapter

For the following exercises, express the equation for two functions, with \(y\) as a function of \(x\) . Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. $$\frac{(x-2)^{2}}{16}-\frac{(y+3)^{2}}{25}=1$$

For the following exercises, given information about the graph of the hyperbola, find its equation. Vertices at \((3,0)\) and \((-3,0)\) and one focus at \((5,0)\)

An object is projected so as to follow a parabolic path given by \(y=-x^{2}+96 x,\) where \(x\) is the horizontal distance traveled in feet and \(y\) is the height. Determine the maximum height the object reaches.

For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information. The object It enters along a path approximated by the line \(y=3 x-9\) and passes within 1 au of the sun at is closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line \(y=-3 x+9 .\)

For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=\frac{1}{2} x\) and \(y=-\frac{1}{2} x,\) and its closest distance to the centain is 10 yards.
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