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Chapter 11: Problem 50

Find the center, transverse axis, vertices, foci, and asymptotes, Graph each equation. \(\frac{(y+3)^{2}}{4}-\frac{(x-2)^{2}}{9}=1\)

Short Answer

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

Step by step solution

01

Identify the Equation Type and Center

The given equation is \(\frac{(y+3)^{2}}{4}-\frac{(x-2)^{2}}{9}=1\). This is a hyperbola in standard form \(\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1\). Compare it to standard form to identify the center \((h, k)\). Here, \(h = 2\) and \(k = -3\), so the center is \((2, -3)\).
02

Determine the Transverse Axis and Vertices

The transverse axis is vertical because the \(y\)-component comes first in the standard form of a hyperbola. The length of the transverse axis is \2a\. Here, \(a^{2} = 4\), so \(a = 2\). The vertices are located \a\ units away from the center, so they are \((2, -3+2) = (2, -1)\) and \((2, -3-2) = (2, -5)\).
03

Find the Foci

The distance from the center to each focus is \c\, where \(c^{2} = a^{2} + b^{2}\). Here, \(b^{2} = 9\) and \(a^{2} = 4\), so \(c^{2} = 4 + 9 = 13\), thus \(c = \sqrt{13}\). The foci are at \((2, -3 + \sqrt{13})\) and \((2, -3 - \sqrt{13})\).
04

Write the Asymptotes

The equations of the asymptotes for a hyperbola of this form are \(y - k = \pm \frac{a}{b}(x - h)\). Substituting the given values, the equations become \(y + 3 = \pm \frac{2}{3}(x - 2)\).
05

Graph the Hyperbola

To graph the hyperbola, plot the center \((2, -3)\), the vertices \((2, -1)\) and \((2, -5)\), and the foci \((2, -3 + \sqrt{13})\) and \((2, -3 - \sqrt{13})\). Draw the transverse and conjugate axes. Then, sketch the asymptotes using their equations, and finally draw the hyperbola branches approaching the asymptotes.

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Key Concepts

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

Center of Hyperbola
The center of a hyperbola is the midpoint between its vertices. For the equation \(\frac{(y+3)^{2}}{4}-\frac{(x-2)^{2}}{9}=1\), the center can be found by comparing it to the standard form of the hyperbola equation: \(\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1\). This comparison shows that \(h = 2\) and \(k = -3\), thus the center is \((2, -3)\). The center is a crucial reference point for plotting other key features of the hyperbola.
Vertices of Hyperbola
The vertices of the hyperbola are points where the hyperbola intersects its transverse axis. Since the transverse axis is vertical here, the vertices are \((h, k \pm a)\). From the standard form, \(a^{2} = 4\), so \(a = 2\). Therefore, the vertices are \((2, -3+2) = (2, -1)\) and \((2, -3-2) = (2, -5)\). These points help in drawing the shape of the hyperbola.
Foci of Hyperbola
The foci (or focuses) are points located along the transverse axis, outside of the vertices. The distance from the center to a focus is given by \(c\), where \(c^{2} = a^{2} + b^{2}\). Here, \(a^{2} = 4\) and \(b^{2} = 9\), so \(c^{2} = 4 + 9 = 13\), thus \(c = \sqrt{13}\). The foci are located at \((2, -3 + \sqrt{13})\) and \((2, -3 - \sqrt{13})\). These points are essential for understanding how 'open' or 'spread out' the hyperbola is.
Asymptotes of Hyperbola
Asymptotes are lines that the hyperbola approaches as it extends to infinity. For a hyperbola in the form \(\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1\), the equations for the asymptotes are \(y - k = \pm \frac{a}{b}(x - h)\). Substituting the given values, the asymptotes are: \(y + 3 = \pm \frac{2}{3}(x - 2)\). These lines help guide the drawing of the hyperbola branches, showing the general direction they will follow.
Graphing Hyperbolas
To graph a hyperbola, start by plotting its center, vertices, and foci. For \(\frac{(y+3)^{2}}{4}-\frac{(x-2)^{2}}{9}=1\), plot the center at \((2, -3)\), the vertices at \((2, -1)\) and \((2, -5)\), and the foci at \((2, -3 + \sqrt{13})\) and \((2, -3 - \sqrt{13})\). Next, draw the transverse and conjugate axes. Use the asymptote equations (\

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