Textbook Question
Graph each ellipse and give the location of its foci. (x +3)²/9 + (y -2)² = 1
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 45
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. y2 - 2y + 12x - 35 = 0
Verified step by step guidance1
Start by rewriting the given equation: \(y^2 - 2y + 12x - 35 = 0\). Group the \(y\) terms together and move the \(x\) term and constant to the other side: \(y^2 - 2y = -12x + 35\).
Complete the square for the \(y\) terms. Take half of the coefficient of \(y\), which is \(-2\), divide by 2 to get \(-1\), then square it to get \(1\). Add \(1\) to both sides to maintain equality: \(y^2 - 2y + 1 = -12x + 35 + 1\).
Rewrite the left side as a perfect square: \((y - 1)^2 = -12x + 36\). Then isolate \(x\) by moving terms: \((y - 1)^2 = -12x + 36\) becomes \((y - 1)^2 = -12(x - 3)\).
Identify the vertex from the equation in standard form \((y - k)^2 = 4p(x - h)\), where the vertex is at \((h, k)\). Here, the vertex is at \((3, 1)\).
Determine the value of \(p\) by comparing \(-12\) to \$4p\(, so \(4p = -12\) which gives \)p = -3\(. Use \)p\( to find the focus at \)(h + p, k)\( and the directrix as the vertical line \)x = h - p$.
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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 or (y - k)^2 by adding and subtracting terms. This technique helps convert equations into standard form, making it easier to identify key features like the vertex of a parabola.
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Standard Form of a Parabola
The standard form of a parabola's equation reveals its geometric properties clearly. For parabolas opening horizontally or vertically, the equation is written as (y - k)^2 = 4p(x - h) or (x - h)^2 = 4p(y - k), where (h, k) is the vertex and p relates to the distance between the vertex and the focus.
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Vertex, Focus, and Directrix of a Parabola
The vertex is the parabola's turning point, the focus is a fixed point inside the parabola used to define it, and the directrix is a line perpendicular to the axis of symmetry. Knowing these helps in graphing the parabola and understanding its shape and orientation.
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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
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
Graph each ellipse and give the location of its foci. x²/25 + (y -2)² /36= 1