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Chapter 13: Problem 69

Sketch the graph \(\frac{(x+5)^{2}}{16}-\frac{(y+2)^{2}}{25}=1\)

Short Answer

Expert verified
The graph is a hyperbola centered at (-5, -2) with vertices at (-1, -2) and (-9, -2).

Step by step solution

01

Identify the type of conic section

The given equation \(\frac{(x+5)^{2}}{16}-\frac{(y+2)^{2}}{25}=1\) is a hyperbola. This is because it is in the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), indicating a horizontally oriented hyperbola.
02

Determine the center

The center of the hyperbola is given by the coordinates \((h, k)\). From the equation \((x+5)^2/16 - (y+2)^2/25 = 1\), we identify \(h = -5\) and \(k = -2\). Therefore, the center is \((-5, -2)\).
03

Calculate the vertices

For a horizontally oriented hyperbola in the form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), the vertices are located \(a\) units from the center along the x-axis. Here, \(a^2 = 16\) so \(a = 4\). Thus, the vertices are \((-5+4, -2) = (-1, -2)\) and \((-5-4, -2) = (-9, -2)\).
04

Find the asymptotes

The asymptotes for a hyperbola in this form are given by the equations \(y = k \pm \frac{b}{a}(x-h)\). Here, \(b^2 = 25\) so \(b = 5\), and \(a = 4\). Thus the asymptotes are \(y = -2 + \frac{5}{4}(x+5)\) and \(y = -2 - \frac{5}{4}(x+5)\).
05

Sketch the graph

Plot the center at \((-5, -2)\). Draw the vertices at \((-1, -2)\) and \((-9, -2)\). Use the slopes of the asymptotes, \(\frac{5}{4}\), to draw asymptotes passing through the center. Sketch the branches of the hyperbola opening left and right, approaching but never touching the asymptotes.

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Key Concepts

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

Conic Sections
Conic sections are curves that can be formed by the intersection of a plane and a cone. The four basic types of conic sections are circles, ellipses, parabolas, and hyperbolas. Each carries unique characteristics, making them foundational concepts in geometry and algebra.
When examining the equation \[ \frac{(x+5)^{2}}{16} - \frac{(y+2)^{2}}{25} = 1 \], we can identify it as a hyperbola. Clearly, this relates to one of the four conic sections.
  • A hyperbola consists of two parts called branches, which are mirror images.
  • Conic sections can describe various natural phenomena, such as planetary orbits (ellipses) or satellite dishes (parabolas).
  • Critical to understanding is recognizing the general format of each conic section, making it easier to identify and graph them.
Center of Hyperbola
The center of a hyperbola is a vital reference point for graphing. It serves as the midpoint between the hyperbola's vertices and as the point from which other features are measured. For hyperbolas of the form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the center is \((h, k)\).
In our equation, \( (x+5)^2/16 - (y+2)^2/25 = 1 \), we find:
  • \(h = -5\)
  • \(k = -2\)
  • Therefore, the center is \((-5, -2)\).
Establishing this center allows for placing other elements of the hyperbola accurately. This arrangement is crucial for drawing and interpreting its graph correctly.
Vertices of Hyperbola
The vertices of a hyperbola are points where the hyperbola intersects its principal axis. These points are essential as they denote the closest approach of the branches of the hyperbola. For a horizontally oriented hyperbola, the equation is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]. Here, the vertices can be found \(a\) units from the center, along the x-axis.
In our exercise:
  • \(a^2 = 16\), so \(a = 4\).
  • The center is \((-5, -2)\).
  • Hence, the vertices are \((-5+4, -2) = (-1, -2)\) and \((-5-4, -2) = (-9, -2)\).
Plotting these vertices helps give shape to the hyperbola, making it easier to visualize and graph accurately.
Asymptotes of Hyperbola
The asymptotes of a hyperbola are crucial lines that indicate the hyperbola's trajectory. They act as invisible guides that the branches of the hyperbola approach but never cross. For a hyperbola in the form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the asymptotes can be represented by the equations:
  • \(y = k \pm \frac{b}{a}(x-h)\)
For our specific equation:
  • Given \(b^2 = 25\), \(b= 5\).
  • \(a = 4\)
  • The center is \((-5, -2)\)
  • The asymptote equations are: \(y = -2 + \frac{5}{4}(x + 5)\) and \(y = -2 - \frac{5}{4}(x + 5)\).
These asymptotes are essential when sketching the graph since they define the direction in which the hyperbola opens. They help in drawing a more accurate representation of the hyperbola by guiding the curves from each vertex. By understanding asymptotes, one can appreciate the infinite nature of hyperbolas.

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Most popular questions from this chapter

If you are given a list of equations of circles and parabolas and none are in standard form, explain how you would determine which is an equation of a circle and which is an equation of a parabola. Explain also how you would distinguish the upward or downward parabolas from the left-opening or right- opening parabolas.

Cindy Brown, an architect, is drawing plans on grid paper for a circular pool with a fountain in the middle. The paper is marked off in centimeters, and each centimeter represents 1 foot. On the paper, the diameter of the "pool" is 20 centimeters, and "fountain" is the point \((0,0)\) a. Sketch the architect's drawing. Be sure to label the axes. b. Write an equation that describes the circular pool. c. Cindy plans to place a circle of lights around the fountain such that each light is 5 feet from the fountain. Write an equation for the circle of lights and sketch the circle on your drawing.

Sketch the graph of each equation. If the graph is a parabola, find irs vertex. If the graph is a circle, find its center and radius. $$\frac{x^{2}}{8}+\frac{y^{2}}{8}=2$$

Sketch the graph of each equation. If the graph is a parabola, find irs vertex. If the graph is a circle, find its center and radius. $$x=y^{2}+2$$

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=16 \\ y &=-\frac{1}{4} x^{2}+4 \end{aligned}\right. $$
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