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Magazine Chapter 7: Problem 7
In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ y^{2}-5 x^{2}=20 $$
Short Answer
Expert verified
The center is (0,0), vertices are (0,卤2鈭�5), foci are (0,卤2鈭�6), asymptotes are y=卤鈭�5x, and eccentricity is 鈭�(6/5).
Step by step solution
01
Write the Equation in Standard Form
The given equation is \( y^{2} - 5x^{2} = 20 \). To write this equation in the standard form of a hyperbola, divide every term by 20, the number on the right side. This gives us \( \frac{y^{2}}{20} - \frac{5x^{2}}{20} = 1 \), simplifying further, \( \frac{y^{2}}{20} - \frac{x^{2}}{4} = 1 \).
02
Identify the Hyperbola Type and Locate the Center
From the standard form \( \frac{y^{2}}{20} - \frac{x^{2}}{4} = 1 \), we recognize that the hyperbola's transverse axis is vertical since the \( y^{2} \) term is positive. The center of the hyperbola is at \((h, k)\), which is \((0, 0)\) for this equation.
03
Find Vertices of the Hyperbola
For a vertical hyperbola \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), the vertices are at \( (h, k \pm a) \). Here, \( a^{2} = 20 \) so \( a = \sqrt{20} = 2\sqrt{5} \). Therefore, the vertices are located at \((0, \pm 2\sqrt{5})\).
04
Calculate the Foci
The foci of a vertical hyperbola \((centered~on~h,~,k)\) are given by \( (h, k \pm c) \), where \( c = \sqrt{a^{2} + b^{2}} \). Here, \( b^{2} = 4 \), so \( c = \sqrt{20 + 4} = \sqrt{24} = 2\sqrt{6} \). Therefore, the foci are \((0, \pm 2\sqrt{6})\).
05
Determine Asymptotes
The asymptotes of a vertical hyperbola are given by the lines \( y = \pm \frac{a}{b}x \). Here, \( a = \sqrt{20} = 2\sqrt{5} \) and \( b = 2 \). Therefore, the equations of the asymptotes are \( y = \pm \frac{2\sqrt{5}}{2}x = \pm \sqrt{5}x \).
06
Calculate the Eccentricity
The eccentricity \( e \) of a hyperbola is found using \( e = \frac{c}{a} \). Here, \( c = 2\sqrt{6} \) and \( a = 2\sqrt{5} \), so \( e = \frac{2\sqrt{6}}{2\sqrt{5}} = \frac{\sqrt{6}}{\sqrt{5}} \approx 1.095 \).
07
Graph the Hyperbola
Draw the coordinate axes and plot the center at \( (0, 0) \). Plot the vertices at \( (0, \pm 2\sqrt{5}) \) and the foci at \((0, \pm 2\sqrt{6})\). Draw dashed lines for the asymptotes \( y = \pm \sqrt{5}x \) to guide the shape. Sketch the hyperbola opening upwards and downwards in line with these guidelines.
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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 analyze a hyperbola thoroughly, it's crucial to express its equation in the standard form. This form helps in identifying its center and axis orientation, among other characteristics. The standard form of a hyperbola can look different based on its orientation:
- Vertical transverse axis: \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \)
- Horizontal transverse axis: \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \)
Vertices of Hyperbola
Vertices serve as vital reference points determining the extent of the hyperbola along its transverse axis. For a vertical hyperbola in the form \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), the vertices are located at \( (h, k \pm a) \).
From our equation, with \( a^2 = 20 \), we find \( a = \sqrt{20} = 2\sqrt{5} \). This means the vertices for our hyperbola are
From our equation, with \( a^2 = 20 \), we find \( a = \sqrt{20} = 2\sqrt{5} \). This means the vertices for our hyperbola are
- \((0, +2\sqrt{5})\)
- \((0, -2\sqrt{5})\)
Foci of Hyperbola
The foci are fixed points that determine the shape and the extent of the hyperbola. For a hyperbola, the formula to find the foci is dependent on both \( a \) and \( b \). Notably, for a vertical hyperbola such as ours, the foci are positioned at \( (h, k \pm c) \), where \( c = \sqrt{a^{2} + b^{2}} \).
Given \( a^2 = 20 \) and \( b^2 = 4 \), we find \( c = \sqrt{20 + 4} = \sqrt{24} = 2\sqrt{6} \). Thus, the foci are positioned at
Given \( a^2 = 20 \) and \( b^2 = 4 \), we find \( c = \sqrt{20 + 4} = \sqrt{24} = 2\sqrt{6} \). Thus, the foci are positioned at
- \((0, +2\sqrt{6})\)
- \((0, -2\sqrt{6})\)
Asymptotes of Hyperbola
Asymptotes provide a guiding framework for the hyperbola's general shape, especially at its outermost reaches. For vertical hyperbolas, asymptotes are expressed as \( y = \pm \frac{a}{b}x \).
With our hyperbola, using \( a = 2\sqrt{5} \) and \( b = 2 \), the asymptotes become the lines
With our hyperbola, using \( a = 2\sqrt{5} \) and \( b = 2 \), the asymptotes become the lines
- \( y = +\sqrt{5}x \)
- \( y = -\sqrt{5}x \)
Eccentricity of Hyperbola
The eccentricity, denoted as \( e \), measures how much a conic section deviates from being circular, and helps in assessing the "strength" or "sharpness" of the hyperbola's curves. For hyperbolas, it's given by the formula \( e = \frac{c}{a} \).
With our numbers, \( c = 2\sqrt{6} \) and \( a = 2\sqrt{5} \), thus the eccentricity is calculated as
With our numbers, \( c = 2\sqrt{6} \) and \( a = 2\sqrt{5} \), thus the eccentricity is calculated as
- \( e = \frac{\sqrt{6}}{\sqrt{5}} \approx 1.095 \)
