Chapter 10: Problem 28
Graph each hyperbola. See Example 3. $$ \frac{(y-2)^{2}}{4}-\frac{(x+1)^{2}}{1}=1 $$
Short Answer
Expert verified
Center at (-1, 2), vertices at (-1, 4) and (-1, 0), asymptotes: \( y=2x+4 \) and \( y=-2x \).
Step by step solution
01
Identify the form of the hyperbola
The given equation is \( \frac{(y-2)^2}{4} - \frac{(x+1)^2}{1} = 1 \). This is in the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), indicating a vertical hyperbola.
02
Determine the center of the hyperbola
For the equation \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the center is at \((h, k)\). Here, \( h = -1 \) and \( k = 2 \), so the center is at \( (-1, 2) \).
03
Identify a and b values
From the equation \( \frac{(y-2)^2}{4} - \frac{(x+1)^2}{1} = 1 \), determine \( a^2 = 4 \) so \( a = 2 \), and \( b^2 = 1 \) so \( b = 1 \).
04
Find the vertices
The vertices are found \( a \) units above and below the center. Thus, for this hyperbola, the vertices are located at \( (-1, 2+2) = (-1, 4) \) and \( (-1, 2-2) = (-1, 0) \).
05
Calculate the slopes of the asymptotes
The slopes of the asymptotes for a vertical hyperbola are given by \( \pm \frac{a}{b} = \pm \frac{2}{1} = \pm 2 \).
06
Write the equations of the asymptotes
Using the center \( (-1, 2) \) and the slopes \( m = \pm 2 \), the equations of the asymptotes are \( y - 2 = 2(x + 1) \) which simplifies to \( y = 2x + 4 \), and \( y - 2 = -2(x + 1) \) which simplifies to \( y = -2x \).
07
Sketch the graph
Plot the center of the hyperbola at \( (-1, 2) \), then draw the asymptotes using their equations. Mark the vertices at \( (-1, 4) \) and \( (-1, 0) \). This will help outline the shape of the hyperbola, opening upwards and downwards through the vertices, approaching the asymptotes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertical Hyperbola
A vertical hyperbola is a specific type of hyperbola that opens up and down. This is determined by its standard equation form which is \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]In this equation:
- The center of the hyperbola is \((h, k)\)
- The variable \(a\) represents the distance from the center to the vertices along the y-axis, indicating an up-down orientation.
- If the equation was reversed to \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) it would signify a horizontal hyperbola.
Conic Sections
Conic sections are figures created by the intersection of a plane with a cone. They encompass different shapes:
Hyperbolas have different properties, such as two foci and two directrices, compared to the single focus and directrix of a parabola. Understanding conic sections helps in grasping concepts of geometry, especially when relating equations to graph shapes.
When analyzing a hyperbola, recognizing it as a conic section provides a mathematical framework for understanding its features and transformations.
- Circle
- Ellipse
- Parabola
- Hyperbola
Hyperbolas have different properties, such as two foci and two directrices, compared to the single focus and directrix of a parabola. Understanding conic sections helps in grasping concepts of geometry, especially when relating equations to graph shapes.
When analyzing a hyperbola, recognizing it as a conic section provides a mathematical framework for understanding its features and transformations.
Asymptotes
Asymptotes are lines that a curve approaches as it heads towards infinity. For hyperbolas, asymptotes are lines that help guide the shape and direction.
For a vertical hyperbola, the slopes of the asymptotes are determined by \(\pm \frac{a}{b} \) in the equation \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]Here,
For a vertical hyperbola, the slopes of the asymptotes are determined by \(\pm \frac{a}{b} \) in the equation \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]Here,
- The slopes of the asymptotes become \(\pm\) the ratio of \(a\) to \(b\).
- In the provided problem, the calculated slopes are \(2\) and \(-2\).
Vertices of Hyperbola
Vertices are pivotal points of a hyperbola that indicate where the curve turns and changes direction. For a vertical hyperbola, the vertices are determined by the value \(a\) in the standard equation:\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]
- The vertices lie along the \(y\)-axis, precisely \(a\) units above and below the center \((h, k)\).
- With the center at \((-1, 2)\) and \(a = 2\), the vertices are positioned at \((-1, 4)\) and \((-1, 0)\).