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Magazine Chapter 13: Problem 21
Graph each hyperbola. See Example 2. $$ y^{2}-4 x^{2}=16 $$
Short Answer
Expert verified
The hyperbola opens vertically with center at (0,0), vertices at (0,4) and (0,-4), and asymptotes y=2x and y=-2x.
Step by step solution
01
Identify the General Form
The general form of a hyperbola is \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) or \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). The given equation is \(y^2 - 4x^2 = 16\).
02
Rearrange to Standard Form
To match the general form, divide all terms by 16: \(\frac{y^2}{16} - \frac{4x^2}{16} = 1\). Simplify this to \(\frac{y^2}{16} - \frac{x^2}{4} = 1\).
03
Identify the Hyperbola's Parts
Now, \(\frac{y^2}{16} - \frac{x^2}{4} = 1\) represents a vertical hyperbola because \(y^2\) is the positive term. Here, \(a^2 = 16\) so \(a = 4\) and \(b^2 = 4\) so \(b = 2\). The center is at \((0, 0)\).
04
Determine Vertices and Asymptotes
For a vertical hyperbola, vertices are at \((0, a)\) and \((0, -a)\), which are \((0, 4)\) and \((0, -4)\). Asymptotes are lines through the center with slopes \(\pm \frac{a}{b}\), thus \(y = \pm \frac{4}{2}x = \pm 2x\).
05
Sketch the Graph
Plot the center at \((0,0)\), draw the vertices at \((0,4)\) and \((0,-4)\), and draw asymptotes \(y = 2x\) and \(y = -2x\). Sketch the hyperbola's branches opening vertically through these vertices.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form of Hyperbola
To graph a hyperbola, understanding its standard form is crucial. The standard form of a hyperbola can be recognized as either \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) or \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). This form helps determine the hyperbola's orientation and components:
This matches the vertical hyperbola's form because \(y^2\) is the positive term. Identifying this standard form is our key starting point for graphing.
- In the first form, the hyperbola is vertical, meaning it opens up and down.
- In the second form, the hyperbola is horizontal, which implies it opens left and right.
This matches the vertical hyperbola's form because \(y^2\) is the positive term. Identifying this standard form is our key starting point for graphing.
Vertices of Hyperbola
The vertices of a hyperbola play a significant role in determining its shape and placement on the graph. When dealing with the standard form of a vertical hyperbola, like in our exercise, vertices can be found in the following way:
- Calculate \(a\) by taking the square root of \(a^2\), which is beneath the \(y^2\) term in the equation.
- For \(a^2 = 16\), we find \(a = 4\).
- The center of our hyperbola is \((h, k)\), which is \((0, 0)\) here due to the absence of shifting terms.
- The vertices, therefore, lie at \((h, k + a)\) and \((h, k - a)\).
Asymptotes of Hyperbola
Asymptotes are critical lines that guide the shape of a hyperbola. For a vertical hyperbola in its standard form, these asymptotes are calculated using the relationship \(y = \pm \frac{a}{b}x\). Let's break down this concept further:
- First, identify \(b\) by taking the square root of \(b^2\) found under the \(x^2\) term.
- In our case, since \(b^2 = 4\), then \(b = 2\).
- Substitute \(a\) and \(b\) into the asymptote formula to find their equations.
- This results in \(y = \pm 2x\).
Vertical Hyperbola
A vertical hyperbola is characterized by its branches opening in the vertical direction. This is quite distinct from a horizontal hyperbola, which has branches opening horizontally. Our exercise helps illustrate this vertical alignment:
- In the equation \(\frac{y^2}{16} - \frac{x^2}{4} = 1\), the positive \(y^2\) term dictates a vertical opening.
- Its center being at \((0, 0)\) means the symmetry is equally spread along the y-axis.
- Notice the vertices \((0, 4)\) and \((0, -4)\) aligning along the vertical axis, reinforcing this orientation.
