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Ch. 7 - Conic Sections
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 7 - Conic SectionsProblem 51
Chapter 8, Problem 51

Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 9x2 +25y² - 36x + 50y – 164 = 0

Verified step by step guidance
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Start with the given equation: \(9x^{2} + 25y^{2} - 36x + 50y - 164 = 0\).
Group the \(x\) terms and \(y\) terms together and move the constant to the other side: \(9x^{2} - 36x + 25y^{2} + 50y = 164\).
Factor out the coefficients of the squared terms from each group: \(9(x^{2} - 4x) + 25(y^{2} + 2y) = 164\).
Complete the square for each group inside the parentheses: - For \(x^{2} - 4x\), take half of \(-4\) which is \(-2\), square it to get \(4\), and add inside the parentheses. - For \(y^{2} + 2y\), take half of \(2\) which is \(1\), square it to get \(1\), and add inside the parentheses. Remember to balance the equation by adding the equivalent values multiplied by the factored coefficients to the right side.
Rewrite the equation as perfect square trinomials: \(9(x - 2)^{2} + 25(y + 1)^{2} = \text{new constant}\), where the new constant is calculated by adding \(9 \times 4\) and \(25 \times 1\) to 164.

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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 transform the given equation into a standard form, making it easier to identify the center and dimensions of conic sections like ellipses.
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Solving Quadratic Equations by Completing the Square

Standard Form of an Ellipse

The standard form of an ellipse equation is (x - h)²/a² + (y - k)²/b² = 1, where (h, k) is the center, and a and b represent the lengths of the semi-major and semi-minor axes. Converting to this form allows for straightforward graphing and analysis of the ellipse's properties.
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Graph Ellipses at Origin

Foci of an Ellipse

The foci are two fixed points inside an ellipse such that the sum of the distances from any point on the ellipse to the foci is constant. Their locations depend on the values of a, b, and c, where c² = |a² - b²|, and are essential for understanding the ellipse's shape and orientation.
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Foci and Vertices of an Ellipse
Related Practice
Textbook Question

Identify each equation without completing the square.

9x2+25y2−54x−200y+256=09x^2+25y^2-54x-200y+256=0

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Textbook Question

Identify each equation without completing the square.

y2−4x−4y=0y^2-4x-4y=0

Textbook Question

Graph each ellipse and give the location of its foci. 9(x − 1)²+4(y+3)² = 36

Textbook Question

Identify each equation without completing the square. 4x2 - 9y2 - 8x - 36y - 68 = 0

Textbook Question

In Exercises 51–56, graph each relation. Use the relation's graph to determine its domain and range. x29+y216=1\(\frac{x^2}{9}\) + \(\frac{y^2}{16}\) = 1

Textbook Question

In Exercises 51–56, graph each relation. Use the relation's graph to determine its domain and range. x29−y216=1\(\frac{x^2}{9}\) - \(\frac{y^2}{16}\) = 1