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Chapter 10: Problem 7

Graph each hyperbola. Give the domain, range, center, vertices, foci, and equations of the asymptotes for each figure. $$\frac{y^{2}}{25}-\frac{x^{2}}{49}=1$$

Short Answer

Expert verified
Center: (0, 0); Vertices: (0, \pm 5); Foci: (0, \pm \sqrt{74}); Asymptotes: y=\pm \frac{5}{7}x; Domain: (-\infty, \infty); Range: (-\infty, -5] \cup [5, \infty).

Step by step solution

01

Identify the Form of the Hyperbola

The given equation \(\frac{y^2}{25} - \frac{x^2}{49} = 1\) is in the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), which represents a vertical hyperbola.
02

Determine the Center

For the equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the center is at the origin \((0,0)\).
03

Find the Vertices

For a vertical hyperbola, the vertices are located at \((0, \pm a)\). Here, \(a = \sqrt{25} = 5\), so the vertices are \((0, 5)\) and \((0, -5)\).
04

Find the Foci

The foci of the hyperbola are found using \(c = \sqrt{a^2 + b^2}\). Here, \(c = \sqrt{25 + 49} = \sqrt{74}\). Thus, the foci are located at \((0, \pm \sqrt{74})\).
05

Calculate the Equations of the Asymptotes

The equations of the asymptotes for a vertical hyperbola are given by \(y = \pm \frac{a}{b}x\). In this case, \(a = 5\) and \(b = 7\), so the asymptotes are \(y = \pm \frac{5}{7}x\).
06

Determine the Domain and Range

Since the hyperbola opens vertically, the domain includes all real numbers \((-\infty, \infty)\). The range is \((-\infty, -5] \cup [5, \infty)\).
07

Plotting the Hyperbola

Using the center \((0,0)\), vertices \((0, 5)\) and \((0, -5)\), and asymptotes \(y = \pm \frac{5}{7}x\), sketch the hyperbola. Remember, the hyperbola approaches but never touches or crosses the asymptotes.

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Key Concepts

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

domain and range
Understanding the domain and range of a hyperbola is crucial to mastering its graph. The domain refers to all possible x-values, while the range refers to all possible y-values of the hyperbola. For vertical hyperbolas like the one we have here, \(\frac{y^2}{25} - \frac{x^2}{49} = 1\), the hyperbola opens upward and downward. \ \ - **Domain**: Since the hyperbola continues infinitely left and right, the domain includes all real numbers: \(-\backslashinfty, \backslashinfty\). \ \ - **Range**: The y-values are split into two intervals because the hyperbola stretches upward from 5 and downward from -5. Thus, the range is \(-\backslashinfty, -5\] \cup \[5, \backslashinfty\). Knowing these helps in correctly plotting and understanding the hyperbola's behavior on a graph.
hyperbola center
The center of a hyperbola is a crucial reference point when graphing. For the given equation \(\frac{y^2}{25} - \frac{x^2}{49} = 1\), the center is at \(0,0\). \ \ - The center is essentially the origin point from which the hyperbola branches out. \ \ - It serves as the midpoint for measuring the distance to the vertices and foci. Having a clear reference to the center helps when plotting the overall structure of the hyperbola on a coordinate grid and ensures that the graph is accurately represented.
hyperbola vertices
Vertices are essential points on a hyperbola. For vertical hyperbolas, they are aligned along the y-axis. In the example \(\frac{y^2}{25} - \frac{x^2}{49} = 1\): \ \ - The vertices are found at \(0, \backslashpm a\), where \(a = \backslash sqrt{25} = 5\). \ \ - So, the vertices for this hyperbola are at \(0,5\) and \(0,-5\). These points signify the closest points of the hyperbola to the center on either side, acting as starting points for the curve. Correctly identifying and plotting them ensures the hyperbola's correct shape.
hyperbola foci
The foci of a hyperbola are another set of key points that define its shape. They lie along the same axis as the vertices. For our vertical hyperbola \(\frac{y^2}{25} - \frac{x^2}{49} = 1\): \ \ - Foci are calculated using \(c = \backslash sqrt{a^2 + b^2}\). For this equation, \(a = 5\) and \(b = 7\). Hence, \(c = \backslash sqrt{74}\). \ \ - The foci are located at \(0, \backslashpm \backslash sqrt{74}\). These points are critical as the overall shape of the hyperbola symmetrically curves around these points.
asymptotes equations
Asymptotes guide the hyperbola's shape and direction. For vertical hyperbolas like \(\frac{y^2}{25} - \frac{x^2}{49} = 1\): \ \ - Asymptotes are found using the formula \( y = \backslash pm \frac{a}{b}x\). Given \(a = 5\) and \(b = 7\), the equations are \(y = \backslash pm \frac{5}{7}x\). \ \ - These lines indicate the paths the hyperbola will approach but never touch or cross. They act as boundaries that the hyperbola consistently nears as it extends towards infinity. Correctly plotting these will help you sketch an accurate hyperbola.

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Most popular questions from this chapter

Identify which type of conic section is described. The conic section that consists of the set of all points in the plane for which the distance from the point \((3,0)\) is one and one-half times the distance from the line \(x=\frac{4}{3}\)

When a satellite is near Earth, its orbital trajectory may trace out a hyperbola, a parabola, or an ellipse. The type of trajectory depends on the satellite's velocity Vin meters per second. It will be hyperbolic if \(V>\frac{k}{\sqrt{D}},\) parabolic if \(V=\frac{k}{\sqrt{D}}\) and elliptical if \(V<\frac{k}{\sqrt{D}},\) where \(k=2.82 \times 10^{7}\) is a constant and \(D\) is the distance in meters from the satellite to the center of Earth. Use this information If a satellite is scheduled to leave Earth's gravitational influence, its velocity must be increased so that its trajectory changes from elliptical to hyperbolic. Determine the minimum increase in velocity necessary for Explorer IV to escape Earth's gravitational influence when \(D=42.5 \times 10^{6} \mathrm{m}\)

Solve each problem. The coordinates in miles for the orbit of the artificial satellite Explorer VII can be modeled by the equation $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$$ where \(a=4465\) and \(b=4462 .\) Earth's center is located at one focus of the elliptical orbit. (Source: Loh, W., Dynamics and Thermodynamics of Planetary Entry, Prentice-Hall; Thomson, W., Introduction to Space Dynamics, John Wiley and Sons.) (a) Graph both the orbit of Explorer VII and the Earth's surface on the same coordinate axes if the average radius of Earth is 3960 mi. Use the window \([-6750,6750]\) by \([-4500,4500].\) (b) Find the maximum and minimum heights of the satellite above Earth's surface.

$$\text {Graph each horizontal parabola, and give the domain and range.}$$ $$x=(y+1)^{2}$$

Graph each hyperbola. Give the domain, range, center, vertices, foci, and equations of the asymptotes for each figure. $$\frac{(x+6)^{2}}{144}-\frac{(y+4)^{2}}{81}=1$$
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