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Chapter 10: Problem 11

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$\frac{(y+2)^{2}}{9}-\frac{(x+2)^{2}}{4}=1$$

Short Answer

Expert verified
Vertices: \((-2, 1)\), \((-2, -5)\); Foci: \((-2, -2 \pm \sqrt{13})\); Asymptotes: \(y + 2 = \frac{3}{2}(x + 2)\), \(y + 2 = -\frac{3}{2}(x + 2)\).

Step by step solution

01

Identify the Hyperbola Form and Values

The given equation is \(\frac{(y+2)^2}{9} - \frac{(x+2)^2}{4} = 1\). This is in the standard form for a vertical hyperbola, \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), where the center is \((h, k)\), \(a^2 = 9\), and \(b^2 = 4\). Thus, \(a = 3\), \(b = 2\), and the center \((h, k)\) is \((-2, -2)\).
02

Determine the Vertices

For a vertical hyperbola centered at \((-2, -2)\), the vertices are located \(a\) units up and down from the center. Therefore, the vertices are \((-2, -2+3) = (-2, 1)\) and \((-2, -2-3) = (-2, -5)\).
03

Calculate the Foci

The distance from the center to each focus is given by \(c\) where \(c^2 = a^2 + b^2\). Thus, \(c^2 = 9 + 4 = 13\), so \(c = \sqrt{13}\). The foci are \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\).
04

Find the Equations of the Asymptotes

The asymptotes for a vertical hyperbola \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) are given by the equations \((y-k) = \pm\frac{a}{b}(x-h)\). Thus, substituting the values, the equations are \[y + 2 = \pm\frac{3}{2}(x + 2)\].
05

Sketch the Hyperbola and Key Features

Plot the center \((-2, -2)\), the vertices \((-2, 1)\) and \((-2, -5)\), and the foci \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\). Draw the asymptotes following the equations \(y + 2 = \frac{3}{2}(x + 2)\) and \(y + 2 = -\frac{3}{2}(x + 2)\) as dashed lines. Sketch the hyperbola opening vertically with respect to the center, 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.

Vertices of a Hyperbola
The vertices of a hyperbola are crucial points that help in understanding its shape and orientation. For the hyperbola given by \(\frac{(y+2)^2}{9} - \frac{(x+2)^2}{4} = 1\), the vertices can be easily determined once you know the center and the orientation.
Given that the center of the hyperbola is \((-2, -2)\) and it is oriented vertically,
  • The distance from the center to each vertex is the value \(a\), which is 3 in this case.
  • This means the vertices are located 3 units above and below the center along the y-axis.
    The two vertices are thus at \((-2, 1)\) and \((-2, -5)\).
Foci of a Hyperbola
The foci of a hyperbola are fixed points inside each branch of the hyperbola. They play a critical role in defining the shape and the curves of the hyperbola.
To find the foci for our hyperbola
  • We compute the distance \(c\) using the relationship \(c^2 = a^2 + b^2\).
  • Given that \(a = 3\) and \(b = 2\), we calculate \(c^2 = 9 + 4 = 13\) thus, \(c = \sqrt{13}\).
  • The foci, located \(c\) units up and down from the center along the y-axis, are at \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\).
Asymptotes of a Hyperbola
The asymptotes are lines that the hyperbola approaches but never touches. They define the direction in which the branches of the hyperbola extend.
For a vertical hyperbola like ours, \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), the equations of the asymptotes are:
  • \((y-k) = \pm \frac{a}{b}(x-h)\), where \((h, k)\) is the center.
  • Substituting our values, we get:
    \(y + 2 = \pm \frac{3}{2}(x + 2)\).
    These are the equations of the asymptotes for this hyperbola.
Equation of a Hyperbola
The equation of a hyperbola helps to specify its geometry and orientation. It consists of a standard form that identifies key parameters.
Our hyperbola is given by the equation \(\frac{(y+2)^2}{9} - \frac{(x+2)^2}{4} = 1\), which is characteristic of a vertically oriented hyperbola:
  • The center \((h, k)\) is at \((-2, -2)\).
  • Parameters \(a^2 = 9\) and \(b^2 = 4\) give us \(a = 3\) and \(b = 2\) respectively, directing the distances to vertices and foci.Understanding how these parameters fit into the formula allows for easy determination of other features like vertices and asymptotes.
Graphing Hyperbolas
Graphing a hyperbola involves identifying and plotting several key components. Follow these steps to sketch your hyperbola:
  • Begin by plotting the center \((-2, -2)\).
  • Next, draw the vertices \((-2, 1)\) and \((-2, -5)\). These determine the general "spine" of the hyperbola.
  • Locate the foci \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\) within each curve branch.
  • Draw the asymptotes given by \(y + 2 = \frac{3}{2}(x + 2)\) and \(y + 2 = -\frac{3}{2}(x + 2)\) as dashed lines. These lines give approximate guides for the hyperbola's approach but will never intersect the graph.
  • Finally, sketch the two branches of the hyperbola, making sure they curve around but never touch the asymptotes.
This comprehensive visual process reinforces understanding and aids in accurate representation of hyperbolas.

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Most popular questions from this chapter

Lissajous figures are used in the study of electrical circuits to determine the phase difference \(\phi\) between a known voltage \(V_{1}(t)=A \sin (\omega t)\) and an unknown voltage \(V_{2}(\vec{t})=B \sin (\omega t+\phi)\) having the same frequency. The voltages are graphed parametrically as \(x=V_{1}(t)\) and \(y=V_{2}(t)\) If \(\phi\) is acute, then $$\phi=\sin ^{-1} \frac{y_{\mathrm{int}}}{y_{\max }}$$ where \(y_{\text {int }}\) is the nonnegative \(y\) -intercept and \(y_{\max }\) is the maximum \(y\) -value on the curve. (a) Graph the parametric curve \(x=V_{1}(t)\) and \(y=V_{2}(t)\) for the specified range of \(t\) (b) Use the graph to approximate \(\phi\) in degrees. $$\begin{aligned}V_{1}(t)=80 \sin (60 \pi t), & V_{2}(t)=70 \cos (60 \pi t-\pi / 3) & 0 \leq t \leq 0.035\end{aligned}$$

Find an equation for the indicated part of the hyperbola. Lower halves of the branches of \(\frac{x^{2}}{9}-\frac{y^{2}}{49}=1\)

Find an equation in \(x\) and \(y\) whose graph contains the points on the curve \(C\). Sketch the graph of \(C\), and indicate the orientation. $$x=\tan t, \quad y=1 ; \quad-\pi / 2

Find a polar equation of the parabola with focus at the pole and the given vertex. $$V\left(4, \frac{\pi}{2}\right)$$

Exer. \(67-68\) : The planets move around the sun in elliptical orbits. Given the semimajor axis a in millions of kilometers and eccentricity \(e,\) graph the orbit for the planet. Center the major axis on the \(x\) -axis, and plot the location of the sum at one focus. $$\text { Earth's path } a=149.6, e=0.093$$
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