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Chapter 8: Problem 22

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. \((y+4)^{2}=16(x+4)\)

Short Answer

Expert verified
Vertex \((-4, -4)\), Focus \((0, -4)\), Directrix \(x = -8\).

Step by step solution

01

Rewrite the Equation in Standard Form

The given equation \((y+4)^2 = 16(x+4)\) is in the form of a parabola \((y-k)^2 = 4p(x-h)\), which opens sideways. Compare it to the standard form of a sideways-opening parabola. It is already represented in the standard form where \(h = -4\), \(k = -4\), and \(4p = 16\). Solving \(4p = 16\), we find that \(p = 4\).
02

Find the Vertex

The standard form of a parabola \((y-k)^2 = 4p(x-h)\) gives the vertex as \((h, k)\). From the standard form equation, the vertex \(V\) is \((-4, -4)\).
03

Determine the Focus

For a parabola of the form \((y - k)^2 = 4p(x - h)\), the focus is located at \((h + p, k)\). Here, \(h = -4\), \(k = -4\), and \(p = 4\). Thus, the focus \(F\) is \((-4 + 4, -4) = (0, -4)\).
04

Determine the Directrix

The directrix of a sideways-opening parabola \((y - k)^2 = 4p(x - h)\) is described by the equation \(x = h - p\). Substituting \(h = -4\) and \(p = 4\), we find the directrix \(d\) is \(x = -4 - 4 = -8\).

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Key Concepts

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

standard form
The standard form of a parabola equation provides a clear structure to understand the orientation and key attributes of the parabola. When we talk about a parabola opening sideways, the equation is typically expressed as \[(y - k)^2 = 4p(x - h)\]Here:
  • \(h\) and \(k\) indicate the coordinates of the vertex of the parabola.
  • \(p\) is the focal length, dictating the distance of the vertex from the focus and the directrix.
In our exercise, \[(y+4)^2 = 16(x+4)\]resembles this form, with a sideways opening direction. By comparing to the standard form, we identify
  • \(h = -4\)
  • \(k = -4\)
  • \(4p = 16\) which yields \(p = 4\)
Knowing these parameters allows us to easily compute other characteristics of the parabola such as the vertex, focus, and directrix.
vertex of a parabola
The vertex of a parabola is the point where the parabola turns direction, marking the peak or the trough, depending on its opening direction.Position of the Vertex:- For a parabola expressed as \[(y - k)^2 = 4p(x - h)\],the vertex is positioned at the coordinates \((h, k)\).In the given exercise equation, \[(y+4)^2 = 16(x+4)\],we plug in
  • \(h = -4\)
  • \(k = -4\)
Thus, the vertex \(V\) is located at \((-4, -4)\).This particular vertex tells us that the parabola shifts downward and to the left from the origin. Understanding this helps visualize where the parabola sits in the coordinate plane.
focus of a parabola
The focus of a parabola is a significant point which, along with the directrix, defines the shape of the parabola. Every point on the parabola is equidistant from the focus and the directrix.Location of the Focus:
  • In a parabola with a form like \((y - k)^2 = 4p(x - h)\), the focus lies at \((h + p, k)\).
Given our specific parabola equation, \[(y+4)^2 = 16(x+4)\],we determine:
  • \(h = -4\)
  • \(k = -4\)
  • \(p = 4\)
Substituting these values shows the focus \(F\) is at \((0, -4)\).Understanding where the focus is located is crucial for drawing an accurate parabola, and it shows the precise path a point would take when reflecting off the surface of the parabola.
directrix of a parabola
The directrix of a parabola complements the focus; it is a line that is equidistant from any point on the parabola to the focus.Equation of the Directrix:
  • In a sideways opening parabola like \((y - k)^2 = 4p(x - h)\),the equation for the directrix is \(x = h - p\).
For the given exercise, with equation \[(y+4)^2 = 16(x+4)\],we have:
  • \(h = -4\)
  • \(p = 4\)
By plugging these values in, the directrix \(d\) is defined by \(x = -8\).The line \(x = -8\) runs parallel to the \(y\)-axis and serves as a guiding edge opposite the focus. It aids in ensuring the reflective properties of the parabola remain constant and is integral to graphing and understanding the curve's complete shape.

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Most popular questions from this chapter

For the object from the previous exercise, assume the path followed is given by \(y=-0.5 x^{2}+80 x .\) Determine how far along the horizontal the object traveled to reach maximum height.

For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=-5 ; \quad e=\frac{3}{4}\)

For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=-\frac{1}{4} ; \quad e=\frac{7}{2}\)

For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=-1 ; \quad e=1\)

For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. \(r=\frac{8}{4-3 \quad \cos \theta}\)
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