Textbook Question
In Exercises 13–26, use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. x2/100−y2/64=1
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Ch. 7 - Conic Sections
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Chapter 8, Problem 15
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. 8x2 + 4y = 0
Verified step by step guidance1
Rewrite the given equation \(8x^2 + 4y = 0\) in a form that isolates \(y\). Start by subtracting \$8x^2$ from both sides to get \(4y = -8x^2\).
Divide both sides of the equation by 4 to solve for \(y\): \(y = -2x^2\). This is now in the form \(y = ax^2\), which represents a vertical parabola.
Recall the standard form of a vertical parabola with vertex at the origin is \(y = \frac{1}{4p} x^2\), where \(p\) is the distance from the vertex to the focus (and also to the directrix).
Compare \(y = -2x^2\) with \(y = \frac{1}{4p} x^2\) to identify \(\frac{1}{4p} = -2\). Solve for \(p\) by setting \(4p = \frac{1}{-2}\), which gives \(p = -\frac{1}{8}\).
Use the value of \(p\) to find the focus and directrix: since \(p\) is negative, the parabola opens downward. The focus is at \((0, p)\), and the directrix is the line \(y = -p\). Write these explicitly using the value of \(p\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Standard Form of a Parabola
A parabola can be expressed in standard form as either (x - h)^2 = 4p(y - k) for vertical parabolas or (y - k)^2 = 4p(x - h) for horizontal parabolas. Converting the given equation into one of these forms helps identify the vertex, focus, and directrix.
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Parabolas as Conic Sections
Focus and Directrix of a Parabola
The focus is a fixed point inside the parabola, and the directrix is a line outside it, both equidistant from any point on the parabola. The parameter p in the standard form determines the distance from the vertex to the focus and directrix.
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Graphing a Parabola
Graphing involves plotting the vertex, focus, and directrix, then sketching the curve that is equidistant from the focus and directrix. Understanding the orientation (vertical or horizontal) and the width of the parabola is essential for an accurate graph.
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Related Practice
Textbook Question
Graph the hyperbola. Locate the foci and find the equations of the asymptotes. (x^2)/16 - y^2 = 1
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Textbook Question
Graph each ellipse and locate the foci.4x²+16y² = 64
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Textbook Question
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. y2 - 6x = 0
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Textbook Question
Graph the hyperbola. Locate the foci and find the equations of the asymptotes. 4y^2 - x^2 = 16
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Textbook Question
Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y2/16−x2/36=1
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