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Chapter 11: Problem 21

Graph the parabolas. In each case, specify the focus, the directrix, and the focal width. Also specify the vertex. $$2 x^{2}-16 x-y+33=0$$

Short Answer

Expert verified
Vertex: (4,1). Focus: (4,1.125). Directrix: y=7/8. Focal width: 1/2.

Step by step solution

01

Rearrange the equation

Start by rearranging the given quadratic equation, which is \(2x^2 - 16x - y + 33 = 0\). Move \(y\) to the other side to get \(y = 2x^2 - 16x + 33\).
02

Complete the square for the x-term

The equation can be written as \(y = 2(x^2 - 8x) + 33\). To complete the square inside the parenthesis, take half of the \(-8\), square it, and add it inside the parenthesis: \(x^2 - 8x = (x-4)^2 -16\). So, the equation becomes \(y = 2[(x-4)^2 - 16] + 33\).
03

Simplify the equation

Distribute the 2 inside the bracket: \(y = 2(x-4)^2 - 32 + 33\). Simplify it to get \(y = 2(x-4)^2 + 1\). This equation is in the form \((x-h)^2 = 4p(y-k)\).
04

Identify the vertex

From the equation \(y = 2(x-4)^2 + 1\), the vertex \(V\) can be identified as \((h,k) = (4,1)\).
05

Determine 4p from the equation

Because the equation can be simplified to \((x-4)^2 = \frac{1}{2} (y-1)\), it follows the standard form \((x-h)^2 = 4p(y-k)\). This gives \(4p = \frac{1}{2}\), so \(p = \frac{1}{8}\).
06

Find the focus

Calculate the focus \(F\) using \(p\). Since the parabola opens upwards and \(p = \frac{1}{8}\), the focus is \((4, 1.125)\) which is \((h, k + p)\).
07

Determine the directrix

The directrix is below the vertex for an upward opening parabola. It is given by \(y = k - p\). Substitute \(k = 1\) and \(p = \frac{1}{8}\) to find \(y = 1 - \frac{1}{8} = \frac{7}{8}\).
08

Calculate the focal width

The focal width of the parabola is \(|4p|\). Here, \(4p = \frac{1}{2}\). Hence the focal width is \(\frac{1}{2}\).

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Key Concepts

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

Vertex
The vertex is a crucial point of a parabola, as it represents the "tip" of the parabola.This is the point where the parabola changes direction, and in vertical parabolas, it can represent the minimum or maximum point.In the given equation, once rearranged and simplified, we find that the vertex is at the coordinates
  • \( (h, k) = (4, 1) \)
The vertex is determined by the terms
  • \((x-h)^2\) changes to zero means at \(x=4\)
  • when this occurs, \(y\) becomes \(1\), which tells us \(k=1\)
Thus, the vertex
  • is \((4,1)\)
This point is pivotal when you sketch the parabola.
Focus
The focus of a parabola is a unique point located inside the curve that defines the parabola's shape.Rays emanating parallel to the parabola's axis will reflect off the curve and converge at the focus.For the given parabola, the focus can be calculated using the vertex and the value of \(p\).In this scenario:
  • Since the vertex is at \((4, 1)\)
  • The parabola opens upward, and \(p = \frac{1}{8}\)
The focus is at:
  • \((h, k + p)\)
  • \((4, 1 + \frac{1}{8}) = (4, 1.125)\)
This point lies inside the parabola and acts as a gravitational center for reflecting light or sound waves.
Directrix
The directrix is a line that is positioned entirely outside the parabola.It serves as a reference point, relative to which the parabola's symmetry is balanced.In the formula for a parabola \((x-h)^2 = 4p(y-k)\), the directrix is defined as:
  • \(y = k - p\)
  • For our given parabola, \(k = 1\) and \(p = \frac{1}{8}\)
Thus, we find the equation of the directrix to be:
  • \(y = 1 - \frac{1}{8} = \frac{7}{8}\)
This line stays parallel to the x-axis and outside of the parabola opposite the focus.
Focal Width
The focal width, or latus rectum, is the total width of the parabola at its focus.It lies along a line parallel to the directrix and passes through the focus.This concept directly relates to the value
  • \(4p\)
In our example,
  • \(4p = \frac{1}{2}\)
This means that the total width of the parabola at the level of the focus is:
  • \(\frac{1}{2}\)
A smaller focal width indicates a more "tight" parabola, while a larger focal width signals a "wider" curvature.

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Most popular questions from this chapter

Make the substitutions \(x=x^{\prime} \cos \theta-y^{\prime} \sin \theta\) and \(y=x^{\prime} \sin \theta+y^{\prime} \cos \theta\) in the equation\(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) and show that the result is \(A^{\prime}\left(x^{\prime}\right)^{2}+B^{\prime} x^{\prime} y^{\prime}+C^{\prime}\left(y^{\prime}\right)^{2}+D^{\prime} x^{\prime}+E^{\prime} y^{\prime}+F^{\prime}=0\) where $$\begin{array}{l}A^{\prime}=A \cos ^{2} \theta+B \sin \theta \cos \theta+C \sin ^{2} \theta \\\B^{\prime}=2(C-A) \sin \theta \cos \theta+B\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \\\C^{\prime}=A \sin ^{2} \theta-B \sin \theta \cos \theta+C \cos ^{2} \theta \\\D^{\prime}=D \cos \theta+E \sin \theta \\\E^{\prime}=E \cos \theta-D \sin \theta \\\F^{\prime}=F\end{array}$$

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