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Magazine Chapter 11: Problem 30
Find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation. \(\frac{y^{2}}{16}-\frac{x^{2}}{4}=1\)
Short Answer
Expert verified
Center: (0,0), Vertices: (0, 卤4), Foci: (0, 卤2鈭�5), Asymptotes: y=卤2x.
Step by step solution
01
Identify the Form of the Equation
The given equation is \(\frac{y^{2}}{16}-\frac{x^{2}}{4}=1\). This is a hyperbola in the form \(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\). Identify \(a^{2}\) and \(b^{2}\).
02
Determine the Values of a and b
From the equation: \(a^{2} = 16\) and \(b^{2} = 4\). So, \(a = 4\) and \(b = 2\).
03
Locate the Center
For hyperbolas in this standard form, the center is at the origin, \((0, 0)\).
04
Identify the Transverse Axis
Since the \(y^2\) term is first, the transverse axis is vertical.
05
Find the Vertices
Vertices occur at \((0, \pm 4)\).
06
Determine the Foci
Calculate the foci using the formula \(c = \sqrt{a^{2} + b^{2}}\), where \(a = 4\) and \(b = 2\). So, \(c = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\). The foci are located at \((0, \pm 2\sqrt{5})\).
07
Determine the Equations of the Asymptotes
Asymptotes of the hyperbola take the form \( y = \pm \frac{a}{b} x\). Here, this is \( y = \pm \frac{4}{2} x\) or \( y = \pm 2x\).
08
Graph the Hyperbola
Plot the center at \(0, 0\) and the vertices at \((0, \pm 4)\). Draw the asymptotes \( y = 2x\) and \( y = -2x\). Then sketch the hyperbola opening upward and downward along the transverse axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
hyperbola equation
A hyperbola is a type of conic section that resembles two mirrored curves. Its general equation is written as: \[\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\] or \[\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\].
The given equation in our exercise is \[\frac{y^{2}}{16} - \frac{x^{2}}{4} = 1\]. This equation matches the first form where the equation is rearranged such that \(\frac{y^{2}}{a^{2}}\) comes before \(\frac{x^{2}}{b^{2}}\).
Identifying the forms of \(a^{2}\) and \(b^{2}\) lets us proceed to the other steps. Here, \(a^{2} = 16\) and \(b^{2} = 4\). By taking the square root, we find that \(a = 4\) and \(b = 2\).
The given equation in our exercise is \[\frac{y^{2}}{16} - \frac{x^{2}}{4} = 1\]. This equation matches the first form where the equation is rearranged such that \(\frac{y^{2}}{a^{2}}\) comes before \(\frac{x^{2}}{b^{2}}\).
Identifying the forms of \(a^{2}\) and \(b^{2}\) lets us proceed to the other steps. Here, \(a^{2} = 16\) and \(b^{2} = 4\). By taking the square root, we find that \(a = 4\) and \(b = 2\).
transverse axis
The transverse axis of a hyperbola is the line segment that passes through the center and the vertices. It is a measure of the length between the vertices.
For a hyperbola of the form \(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\), the transverse axis is vertical.
Conversely, if the hyperbola were given by \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\), the transverse axis would be horizontal.
In our case, since \(y^{2}\) comes first in the equation, the transverse axis is vertical, passing through \((0, 0)\) and moving up and down.
For a hyperbola of the form \(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\), the transverse axis is vertical.
Conversely, if the hyperbola were given by \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\), the transverse axis would be horizontal.
In our case, since \(y^{2}\) comes first in the equation, the transverse axis is vertical, passing through \((0, 0)\) and moving up and down.
vertices
Vertices are the points where the hyperbola crosses its transverse axis. These points lie at a distance of \(a\) from the center.
Given our standard form \(\frac{y^{2}}{16} - \frac{x^{2}}{4} = 1\), the vertices are calculated using \(a = 4\).
Since the center is at \((0, 0)\) and the axis is vertical, the vertices are located at \((0, \pm 4)\).
Given our standard form \(\frac{y^{2}}{16} - \frac{x^{2}}{4} = 1\), the vertices are calculated using \(a = 4\).
Since the center is at \((0, 0)\) and the axis is vertical, the vertices are located at \((0, \pm 4)\).
foci
The foci (singular: focus) of a hyperbola are points located along the transverse axis, further from the center than the vertices. The distance from the center to each focus is denoted as \(c\). To calculate \(c\), we use the formula \(c = \sqrt{a^{2} + b^{2}}\). \
For our hyperbola: \[c = \sqrt{4^{2} + 2^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\]. \
Hence, the foci are located at \((0, \pm 2\sqrt{5})\).
For our hyperbola: \[c = \sqrt{4^{2} + 2^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\]. \
Hence, the foci are located at \((0, \pm 2\sqrt{5})\).
asymptotes
Asymptotes are straight lines that the hyperbola approaches but never touches as the distance from the center increases. In the case of a hyperbola, the equations for the asymptotes can be derived using the slopes \( \pm \frac{a}{b}\).
For our particular hyperbola, the slopes are \( \pm \frac{4}{2} = \pm 2\).
Therefore, the equations are \(y = 2x\) and \(y = -2x\). \
These lines aid in sketching the hyperbola accurately by acting as guides for the two curves.
For our particular hyperbola, the slopes are \( \pm \frac{4}{2} = \pm 2\).
Therefore, the equations are \(y = 2x\) and \(y = -2x\). \
These lines aid in sketching the hyperbola accurately by acting as guides for the two curves.
