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Magazine Chapter 10: Problem 96
For the following exercises, find the equations of the asymptotes for each hyperbola. $$\frac{(y-3)^{2}}{3^{2}}-\frac{(x+5)^{2}}{6^{2}}=1$$
Short Answer
Expert verified
The asymptotes are \( y = \frac{1}{2}x + \frac{11}{2} \) and \( y = -\frac{1}{2}x + \frac{1}{2} \).
Step by step solution
01
Identify the Standard Form
The given equation of the hyperbola is \( \frac{(y-3)^{2}}{3^{2}}-\frac{(x+5)^{2}}{6^{2}}=1 \). This matches the standard form of a vertical hyperbola \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), where \((h,k)\) is the center, \(a\) represents the semi-major axis (vertical), and \(b\) represents the semi-minor axis (horizontal). Here, \(h = -5\), \(k = 3\), \(a = 3\), and \(b = 6\).
02
Determine the Equation of the Asymptotes
For a vertical hyperbola in the form \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), the equations of the asymptotes are given by the lines \( y = k \pm \frac{a}{b}(x-h) \). Using the identified values, substitute \( h = -5 \), \( k = 3 \), \( a = 3 \), and \( b = 6 \) into the formula: \[ y = 3 \pm \frac{3}{6}(x + 5) \]
03
Simplify the Asymptote Equations
Simplify the expression \( \frac{3}{6} = \frac{1}{2} \). Thus, the equations become:- \( y = 3 + \frac{1}{2}(x + 5) \)- \( y = 3 - \frac{1}{2}(x + 5) \)Simplify further:- \( y = 3 + \frac{1}{2}x + \frac{5}{2} \ = \frac{1}{2}x + \frac{11}{2} \)- \( y = 3 - \frac{1}{2}x - \frac{5}{2} \ = -\frac{1}{2}x + \frac{1}{2} \)
04
Write Final Asymptote Equations
The final equations for the asymptotes of the hyperbola are:- \( y = \frac{1}{2}x + \frac{11}{2} \)- \( y = -\frac{1}{2}x + \frac{1}{2} \)
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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 type of conic section that opens upwards and downwards along the y-axis. This orientation is distinct from a horizontal hyperbola, which opens left and right along the x-axis. Vertical hyperbolas are characterized by their equation having the y-term as positive. For example, in the equation \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), the squared term in y (\((y-k)^2\)) comes first, making it a vertical hyperbola.
- The center of the hyperbola is at point \((h, k)\).
- The value of \(a\) determines the distance from the center to the vertices along the vertical direction.
- \(b\) affects the slope of the asymptotes but doesn't impact the direction in which the hyperbola opens.
Standard Form of Hyperbola
The standard form of a hyperbola is essential for identifying its orientation and key features. For a vertical hyperbola, the standard equation is \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \). If comparing this to other conic sections, a hyperbola's defining property here is the minus sign between the terms, differentiating it from an ellipse which uses a plus sign. Knowing the standard form helps to:
- Determine the direction the hyperbola opens (vertical or horizontal).
- Find the center \((h, k)\).
- Calculate the lengths of the semi-major and semi-minor axes with \(a\) and \(b\) respectively.
Equations of Asymptotes
Asymptotes are straight lines that a hyperbola approaches but never touches. For vertical hyperbolas, the asymptotes' equations provide crucial insight into their behavior at extreme values. The formula for determining the asymptotes of a vertical hyperbola \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \) is \( y = k \pm \frac{a}{b}(x-h) \).To find these equations:
- Substitute the center coordinates \((h, k)\).
- Use \(a\) and \(b\) for calculating slope \(\frac{a}{b}\).
Semi-Major Axis
In hyperbolas, the semi-major axis determines the vertical stretch from the center to the vertices of the curve for a vertical orientation. Specifically, it corresponds to the 'a' value in the equation \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \). For this particular hyperbola, the semi-major axis length is 3 units. Key aspects include:
- It extends vertically from the center \((h, k)\).
- It determines how high and low the hyperbola extends along the y-axis.
Semi-Minor Axis
The semi-minor axis in hyperbolas, though less direct in its implication to the opening, influences the hyperbola's asymptotes. For vertical hyperbolas as \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), the semi-minor axis is the 'b' value. In our equation, \( b = 6 \) indicates a horizontal stretch not of the curve, but crucially in the calculation of slopes for the asymptotes.Roles of semi-minor axis:
- It characterizes the horizontal distance from the center of the hyperbola.
- Plays a vital role in defining the geometry of the asymptotes through slope \(\frac{a}{b}\).
