Textbook Question
Graph each ellipse and locate the foci. x2/(9/4) +y2/(25/4) = 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 9
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. x2 = 12y
Verified step by step guidance1
Identify the form of the given parabola equation. The equation is \(x^2 = 12y\), which matches the standard form of a vertical parabola: \(x^2 = 4py\), where the vertex is at the origin \((0,0)\) and the parabola opens upward if \(p > 0\) or downward if \(p < 0\).
Compare the given equation \(x^2 = 12y\) with the standard form \(x^2 = 4py\) to find the value of \(p\). Set \(4p = 12\), then solve for \(p\).
Use the value of \(p\) to determine the focus of the parabola. For a parabola in the form \(x^2 = 4py\), the focus is located at \((0, p)\).
Find the equation of the directrix. The directrix is a horizontal line given by \(y = -p\) for this form of parabola.
Summarize the results: the vertex is at the origin \((0,0)\), the focus is at \((0, p)\), and the directrix is the line \(y = -p\). Use these to sketch the parabola, noting it opens upward since \(p\) is positive.
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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 x² = 4py or y² = 4px, where p represents the distance from the vertex to the focus and directrix. Recognizing the equation's form helps identify the parabola's orientation and key features.
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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 the vertex. For x² = 4py, the focus is at (0, p) and the directrix is the line y = -p, which guide the parabola's shape.
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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 symmetry and the parabola's opening direction (upward or downward) is essential for accurate graphing.
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Related Practice
Textbook Question
Find the standard form of the equation of each hyperbola satisfying the given conditions. Endpoints of transverse axis: (0, −6), (0, 6); asymptote: y=2x
Textbook Question
Find the standard form of the equation of each hyperbola satisfying the given conditions. Foci: (−4, 0), (4, 0); vertices:(−3, 0), (3, 0)
Textbook Question
Find the standard form of the equation of each hyperbola satisfying the given conditions. Center: (4, −2); Focus: (7, −2); vertex: (6, −2)
Textbook Question
Graph each ellipse and locate the foci. x² = 1 – 4y²
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Textbook Question
Graph each ellipse and locate the foci. x2/49 +y2/81 = 1