Textbook Question
Graph each ellipse and give the location of its foci. 9(x − 1)²+4(y+3)² = 36
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 47
Convert each equation to standard form by completing the square on x or y. Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola. x2 + 6x - 4y + 1 = 0
Verified step by step guidance1
Start by rewriting the given equation \(x^2 + 6x - 4y + 1 = 0\) to isolate the \(y\) terms on one side. Move all terms except those involving \(y\) to the other side: \(-4y = -x^2 - 6x - 1\).
Divide both sides of the equation by \(-4\) to solve for \(y\): \(y = \frac{1}{4}x^2 + \frac{3}{2}x + \frac{1}{4}\).
To complete the square for the \(x\) terms, focus on the quadratic expression \(\frac{1}{4}x^2 + \frac{3}{2}x\). Factor out \(\frac{1}{4}\) from the \(x\) terms: \(y = \frac{1}{4}(x^2 + 6x) + \frac{1}{4}\).
Complete the square inside the parentheses: take half of the coefficient of \(x\) (which is 6), square it, and add and subtract that value inside the parentheses. Half of 6 is 3, and \(3^2 = 9\). So, write \(x^2 + 6x + 9 - 9\) inside the parentheses: \(y = \frac{1}{4}((x + 3)^2 - 9) + \frac{1}{4}\).
Simplify the equation by distributing \(\frac{1}{4}\) and combining constants: \(y = \frac{1}{4}(x + 3)^2 - \frac{9}{4} + \frac{1}{4}\). This gives the standard form of the parabola, from which you can identify the vertex, focus, and directrix.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Completing the Square
Completing the square is a method used to rewrite quadratic expressions in the form (x - h)^2 = k, which helps convert equations into standard form. This involves adding and subtracting a constant to create a perfect square trinomial, making it easier to identify key features of conic sections like parabolas.
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Standard Form of a Parabola
The standard form of a parabola's equation reveals its vertex and orientation. For a parabola opening vertically, it is written as (x - h)^2 = 4p(y - k), where (h, k) is the vertex and p relates to the distance between the vertex and the focus or directrix.
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5:33Parabolas as Conic Sections
Vertex, Focus, and Directrix of a Parabola
The vertex is the parabola's turning point, the focus is a fixed point inside the parabola, and the directrix is a line outside it. The parabola is the set of points equidistant from the focus and directrix, and these elements are essential for graphing and understanding its shape.
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Related Practice
Textbook Question
In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes.
Textbook Question
Graph each ellipse and give the location of its foci. (x − 1)²/2 + (y +3)² /5= 1
Textbook Question
In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes.
Textbook Question
Identify each equation without completing the square. y2 - 4x + 2y + 21 = 0
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Textbook Question
In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes.