/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 In this exercise we graph the hy... [FREE SOLUTION] | 91Ó°ÊÓ

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In this exercise we graph the hyperbola $$\frac{(y-3)^{2}}{5^{2}}-\frac{(x-4)^{2}}{3^{2}}=1$$ (a) Solve the equation for \(y\) to obtain $$y=3 \pm 5 \sqrt{1+(x-4)^{2 / 9}}$$ (b) In the standard viewing rectangle, graph the two equations that you obtained in part (a). Then, for a better view, adjust the viewing rectangle so that both \(x\) and \(y\) extend from -20 to 20

Short Answer

Expert verified
Graph both branches using \(y = 3 \pm 5 \sqrt{1+(x-4)^{2/9}}\) with ranges -20 to 20 for both axes.

Step by step solution

01

Understand the Hyperbola Equation

The given hyperbola is in the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\),where the center is \((h, k) = (4, 3)\), \(a = 5\), and \(b = 3\). This is a vertical hyperbola because the \(y\)-term is positive.
02

Solve for y

Rearrange the hyperbola equation to solve for \(y\):\[ \frac{(y-3)^{2}}{5^{2}} = 1 + \frac{(x-4)^{2}}{3^{2}} \]Multiply both sides by 25:\[ (y-3)^2 = 25 \left( 1 + \frac{(x-4)^2}{9} \right) \]Take the square root of both sides:\[ y - 3 = \pm 5 \sqrt{1 + \frac{(x-4)^2}{9}} \]Solve for \(y\):\[ y = 3 \pm 5 \sqrt{1 + \frac{(x-4)^2}{9}} \]
03

Graph the Functions

The two equations \(y = 3 + 5 \sqrt{1 + \frac{(x-4)^2}{9}}\) and \(y = 3 - 5 \sqrt{1 + \frac{(x-4)^2}{9}}\) represent the two branches of the hyperbola.Graph both functions in a plot range for \(x\) and \(y\) from -20 to 20 to visualize the full hyperbola.

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Key Concepts

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

Graphing Functions
Graphing functions is an essential skill in mathematics that allows us to visualize equations and understand their behavior. The act of graphing involves representing algebraic equations as a visual plot on a coordinate plane. It helps in identifying key features like intercepts, increasing or decreasing behavior, and symmetry. To graph a function accurately, consider the following steps:
  • Identify the domain and range of the function.
  • Determine any symmetries; for hyperbolas, check vertical or horizontal symmetry.
  • Plot key points such as the center and vertices.
  • Draw asymptotes, which are lines that the graph approaches but never touches.
By following these steps, you can ensure the graph represents the function correctly and conveys meaningful information visually. Always adjust the viewing window for a clear and complete picture.
Vertical Hyperbola
A vertical hyperbola is one where the equation's structure indicates a vertical orientation. Specifically, if the positive term in the standard form is associated with the variable \(y\), the hyperbola opens upwards and downwards. The standard form is:\[\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\]In this form, the transverse axis is vertical. This sets the direction in which the two branches will open, distinguishing it from a horizontal hyperbola. Key features of vertical hyperbolas include:
  • Center: Located at \((h, k)\).
  • Vertices: Found \(a\) units away from the center along the \(y\)-axis.
  • Asymptotes: Diagonal lines intersecting at the center, guiding the hyperbola's curve.
Understanding these characteristics is crucial for correctly graphing a vertical hyperbola.
Conic Sections
Conic sections are a group of curves derived from the intersection of a plane with a double-napped cone. They include circles, ellipses, parabolas, and hyperbolas. Each type of conic section has distinct geometric properties and applications in fields such as astronomy and engineering. Hyperbolas result from a plane intersecting both nappes of the cone and are characterized by two symmetric open curves known as branches. The location and orientation of a hyperbola are determined by its equation:
  • Ellipses have no negative sign between terms.
  • Hyperbolas involve one term subtracted from another.
When studying conic sections, it's essential to recognize their equations and deduce properties such as foci, directrices, and eccentricity. This understanding aids in solving real-world problems and facilitates communication across mathematical disciplines.
Equation of a Hyperbola
The equation of a hyperbola is foundational for identifying its features and graphing it. The standard forms of hyperbola equations are crucial for distinguishing the type and orientation. For the vertical hyperbola focused on here, the standard form is:\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]This equation reveals:
  • The center \((h, k)\), around which the hyperbola is symmetric.
  • Vertices that define the endpoints of the transverse axis, given by \((h, k\pm a)\).
  • Asymptotes that provide direction, calculated by \( y = k \pm \frac{a}{b}(x - h) \).
Recognizing how to manipulate and transform this equation is vital to effectively analyzing and graphing hyperbolas. It's crucial to solve for \(y\) or \(x\) when further analysis or graphing is required, as seen in rewriting the hyperbola for graphing tasks.

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Most popular questions from this chapter

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.

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