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Chapter 11: Problem 17

Graph the parabolas. In each case, specify the focus, the directrix, and the focal width. Also specify the vertex. $$y^{2}+2 y-x+1=0$$

Short Answer

Expert verified
Vertex: \((1, -1)\), Focus: \(\left(\frac{5}{4}, -1\right)\), Directrix: \(x = \frac{3}{4}\), Focal Width: 1.

Step by step solution

01

Identify Standard Form

The equation is given as \(y^{2}+2y-x+1=0\). This can be rearranged into the standard form of a parabola by isolating \(x\). The equation can be rearranged to: \(x = y^2 + 2y + 1\).
02

Complete the Square

To identify the vertex form more clearly, complete the square for the \(y\)-terms. Add and subtract \((2/2)^2 = 1\) from the \(y\)-terms: \(x = (y^2 + 2y + 1) - 1 + 1 = (y+1)^2\).
03

Rewrite in Vertex Form

Now we write the equation in the form \(x = (y + 1)^2\). This indicates a parabolic opening to the right, with vertex at \((1, -1)\).
04

Calculate Focus and Directrix

For the parabola \(x = (y-k)^2 + h\), the focus is at \((h + \frac{1}{4}, k)\) and the directrix is the vertical line \(x = h - \frac{1}{4}\). Thus with \(h = 1\) and \(k = -1\), the focus is at \(\left(1 + \frac{1}{4}, -1\right) = \left(\frac{5}{4}, -1\right)\) and the directrix is \(x = 1 - \frac{1}{4} = \frac{3}{4}\).
05

Determine Focal Width

The focal width of a parabola is given by the absolute value of \(\frac{1}{a}\), where \(a\) is the coefficient of \((y - k)^2\). Here, \(a = 1\), so the focal width is \(|\frac{1}{1}| = 1\).

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Key Concepts

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

Focus and Directrix in Parabolas
The focus and directrix are crucial elements in understanding the geometric properties of a parabola. In general, a parabola is defined as the set of all points that are equidistant from a fixed point, called the focus, and a fixed line, known as the directrix. This unique property helps explain why a parabola looks the way it does.
  • Focus: The focus is a point inside the parabola where all the reflected lines that are parallel to the axis of symmetry will meet. For the parabola described by the equation in vertex form, \[ x = (y - k)^2 + h, \] the focus is located at \( (h + \frac{1}{4}a, k) \).
  • Directrix: This is a vertical line that is perpendicular to the axis of symmetry of the parabola. The equation of the directrix in vertex form is \( x = h - \frac{1}{4}a \).
In our worked problem, based on the equation \( x = (y + 1)^2 + 0 \), the focus is at \( \left( \frac{5}{4}, -1 \right) \), while the directrix is the line \( x = \frac{3}{4} \). These parts are essential for sketching the parabola and for understanding how the structure "hugs" around these mathematical parts.
Vertex of a Parabola
The vertex of a parabola is a point that represents the most extreme point on the curve. In mathematical terms, it is called the "turning point". For a parabola that opens to the right or left, like in our example, the vertex is the point where the curve most sharply changes direction.
  • The vertex form of a parabola's equation is \( x = a(y - k)^2 + h \). In this format,

    Locating the Vertex:

    The vertex is simply \((h, k)\).
  • In our problem, after completing the square, we determined the vertex form to be \( x = (y + 1)^2 \), which means the vertex is at the point \( (1, -1) \).
The vertex is a significant concept because it dictates both the shape and direction of the parabola. For example, the distance between the focus and the vertex helps determine the "stretch" of the parabola, giving it its distinctive shape.
Understanding Focal Width
The focal width, sometimes referred to as the "latus rectum," is an important measure that tells us how wide the parabola is at the focus point. To find the focal width, we use the parameter \( a \) as follows:
  • The focal width is defined as \( \frac{1}{|a|} \). This represents the measure of the line segment that is parallel to the directrix and passes through the focus.
  • In our exercise, \( a = 1 \), so the focal width becomes \( \frac{1}{1} = 1 \).
This measurement has implications for the "spread" of the parabola. A larger focal width suggests a wider parabola, while a smaller one indicates a sharper curve. Understanding the focal width allows us to visualize how "open" a parabola is and provides insight into how it can be applied in real-world contexts, such as satellite dishes or reflective properties in optics.

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Most popular questions from this chapter

In this exercise we graph a hyperbola in which the axes of the curve are not parallel to the coordinate axes. The equation is \(x^{2}+4 x y-2 y^{2}=6\) (a) Use the quadratic formula to solve the equation for \(y\) in terms of \(x .\) Show that the result can be written \(y=x \pm \frac{1}{2} \sqrt{6 x^{2}-12}\) (b) Graph the two equations obtained in part (a). Use the standard viewing rectangle. (c) It can be shown that the equations of the asymptotes are \(y=(1 \pm 0.5 \sqrt{6}) x .\) Add the graphs of these asymptotes to the picture that you obtained in part (b). (d) Change the viewing rectangle so that both \(x\) and \(y\) extend from -50 to \(50 .\) What do you observe?

Length of the transverse axis \(6 ;\) length of the conjugate axis \(2 ;\) foci on the \(y\) -axis; center at the originAsymptotes \(y=\pm 2 x ;\) graph passes through \((1, \sqrt{3})\)

Tangent lines are drawn to the ellipse \(x^{2}+3 y^{2}=12\) at the points (3,-1) and (-3,-1) on the ellipse. (a) Find the equation of each tangent line. Write your answers in the form \(y=m x+b\) (b) Find the point where the tangent lines intersect.

In designing an arch, architects and engineers sometimes use a parabolic arch rather than a semicircular arch. (One reason for this is that, in general, the parabolic arch can support more weight at the top than can the semicircular arch.) In the following figure, the blue arch is a semicircle of radius \(1,\) centered at the origin. The red arch is a portion of a parabola. As is indicated in the figure, the two arches have the same base and the same height. Assume that the unit of distance for each axis is the meter. GRAPH CANT COPY (a) Find the equation of the parabola in the figure. (b) Using calculus, it can be shown that the area under this parabolic arch is \(\frac{4}{3} \mathrm{m}^{2} .\) Assuming this fact, show that the area beneath the parabolic arch is approximately \(85 \%\) of the area beneath the semicircular arch. (c) Using calculus, it can be shown that the length of this parabolic arch is \(\sqrt{5}+\frac{1}{2} \ln (2+\sqrt{5})\) meters. Assuming this fact, show that the length of the parabolic arch is approximately \(94 \%\) of the length of the semicircular arch.

Graph the equations. $$3 x y-4 y^{2}+18=0$$
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