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Magazine Chapter 13: Problem 17
Sketch the graph of each equation. \(x^{2}-4 y^{2}=16\)
Short Answer
Expert verified
The graph is a hyperbola centered at the origin with vertices at \((4, 0)\) and \((-4, 0)\), opening along the x-axis.
Step by step solution
01
Identify the Type of Equation
The given equation is in the form of \(x^2 - 4y^2 = 16\). This equation resembles the standard form of a hyperbola, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). This indicates that we are dealing with a hyperbola.
02
Rewrite Equation in Standard Form
To write the equation in standard form, divide every term by 16: \[\frac{x^2}{16} - \frac{y^2}{4} = 1\]. Now, the equation is \(\frac{x^2}{4^2} - \frac{y^2}{2^2} = 1\), where \(a^2 = 16\) and \(b^2 = 4\).
03
Determine the Orientation
Since the term involving \(x^2\) is positive in the standard form equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the hyperbola opens along the x-axis. The vertices will be on the x-axis.
04
Identify Key Components of the Hyperbola
For the standard form \(\frac{x^2}{4^2} - \frac{y^2}{2^2} = 1\), identify the vertices and the value of \(a\) and \(b\): \(a = 4\), \(b = 2\). The vertices are \((\pm 4, 0)\), and the asymptotes are given by the equations \(y = \pm \frac{b}{a}x = \pm \frac{2}{4}x = \pm \frac{1}{2}x\).
05
Sketch the Graph
Plot the vertices \((4, 0)\) and \((-4, 0)\) on the x-axis. Draw the asymptotes through the origin with slopes \(\pm \frac{1}{2}\). Sketch the graph of the hyperbola opening to the left and right, approaching the asymptotes without touching them.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form
The standard form of a hyperbola is crucial for understanding and sketching its graph accurately. A hyperbola's equation typically appears as either
The terms \( a^2 \) and \( b^2 \) are derived from the denominators of the fractions. In our equation, \( a^2 = 16 \) and \( b^2 = 4 \) which gives us \( a = 4 \) and \( b = 2 \). Correctly rewriting the equation into this form helps us quickly identify other key features and properties of the hyperbola, like vertices and asymptotes.
- \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
- or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \).
The terms \( a^2 \) and \( b^2 \) are derived from the denominators of the fractions. In our equation, \( a^2 = 16 \) and \( b^2 = 4 \) which gives us \( a = 4 \) and \( b = 2 \). Correctly rewriting the equation into this form helps us quickly identify other key features and properties of the hyperbola, like vertices and asymptotes.
Vertices
Vertices are specific points on a hyperbola that help define its shape and are directly influenced by the values of \( a \).
In our equation, the hyperbola is oriented horizontally because of the positive \( x^2 \) term. Therefore, the vertices are located along the x-axis. For hyperbolas of form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at
In our equation, the hyperbola is oriented horizontally because of the positive \( x^2 \) term. Therefore, the vertices are located along the x-axis. For hyperbolas of form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at
- \((a, 0)\) and \((-a, 0)\).
- \((4, 0)\)
- \((-4, 0)\).
Asymptotes
Asymptotes are invisible lines that guide the shape of a hyperbola. The branches of the hyperbola approach these lines but never touch them, shaping the graph.
For hyperbolas in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are given by the equations
For hyperbolas in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are given by the equations
- \( y = \pm \frac{b}{a}x \).
- \( y = \pm \frac{1}{2}x \).
Orientation of Hyperbola
The orientation of a hyperbola determines the direction in which its two main branches open.
This is decided by the sign of the quadratic terms in its standard form equation:
This is decided by the sign of the quadratic terms in its standard form equation:
- If \( x^2 \) is positive, the hyperbola opens horizontally along the x-axis.
- In contrast, if \( y^2 \) is positive, it would open vertically along the y-axis.
