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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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 13
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. y2 - 6x = 0
Verified step by step guidance1
Rewrite the given equation \(y^2 - 6x = 0\) in the standard form of a parabola. Start by isolating the \(x\) term: \(y^2 = 6x\).
Recognize that the equation \(y^2 = 4px\) represents a parabola that opens to the right with vertex at the origin \((0,0)\). Here, \(4p = 6\), so solve for \(p\) by dividing both sides by 4: \(p = \frac{6}{4} = \frac{3}{2}\).
Identify the focus of the parabola. Since the parabola opens to the right, the focus is located at \((p, 0)\), which means the focus is at \((\frac{3}{2}, 0)\).
Find the equation of the directrix. For a parabola opening to the right, the directrix is a vertical line given by \(x = -p\). Substitute \(p = \frac{3}{2}\) to get the directrix: \(x = -\frac{3}{2}\).
To graph the parabola, plot the vertex at the origin \((0,0)\), the focus at \((\frac{3}{2}, 0)\), and draw the directrix line \(x = -\frac{3}{2}\). Sketch the parabola opening to the right, equidistant from the focus and directrix.
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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 to identify its geometric properties easily. For a parabola that opens horizontally, the form is (y - k)^2 = 4p(x - h), where (h, k) is the vertex and p determines the distance to the focus and directrix. Rewriting the given equation into this form helps locate these 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 any point on the curve. For the form (y - k)^2 = 4p(x - h), the focus is at (h + p, k) and the directrix is the vertical line x = h - p. Understanding these definitions is key to finding their coordinates.
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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. Knowing the orientation (horizontal or vertical) and the value of p helps determine the shape and width of the parabola. This visual representation aids in understanding the parabola’s properties.
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Related Practice
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
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. x2 = - 16y
Textbook Question
Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. x2/9−y2/25=1
Textbook Question
Graph each ellipse and locate the foci. 25x²+4y² = 100
Textbook Question
Graph each ellipse and locate the foci. x² = 1 – 4y²
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