Problem 22
Question
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. \((y+4)^{2}=16(x+4)\)
Step-by-Step Solution
Verified Answer
Vertex \((-4, -4)\), Focus \((0, -4)\), Directrix \(x = -8\).
1Step 1: 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 \(p = 4\).
2Step 2: 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 \((-4, -4)\).
3Step 3: 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 \((-4 + 4, -4) = (0, -4)\).
4Step 4: 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 \(x = -4 - 4 = -8\).
Key Concepts
standard formvertex of a parabolafocus of a paraboladirectrix of a parabola
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.
- \(h = -4\)
- \(k = -4\)
- \(4p = 16\) which yields \(p = 4\)
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
- \(h = -4\)
- \(k = -4\)
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 \((h + p, k)\).
- \(h = -4\)
- \(k = -4\)
- \(p = 4\)
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 \(x = h - p\).
- \(h = -4\)
- \(p = 4\)
Other exercises in this chapter
Problem 22
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For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations
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For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. \(4
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