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Chapter 11: Problem 54

Find the center, foci, and vertices of each ellipse. Graph each equation. $$9 x^{2}+y^{2}-18 x=0$$

Short Answer

Expert verified
Center: (1, 0), Vertices: (1, 3), (1, -3), Foci: (1, 2鈭�2), (1, -2鈭�2).

Step by step solution

01

Rewrite the equation in standard form

First, rewrite the given equation by completing the square for the x-terms. Start with the equation: \[ 9x^2 + y^2 - 18x = 0 \]Move the constant term to the right side: \[ 9x^2 - 18x + y^2 = 0 \]Factor out the 9 from the x-terms: \[ 9(x^2 - 2x) + y^2 = 0 \]
02

Complete the square

To complete the square inside the parenthesis, add and subtract the square of half the coefficient of x, which is \( \frac{-2}{2} = -1 \ (-1)^2 = 1 \): \[ 9(x^2 - 2x + 1 - 1) + y^2 = 0 \] Simplify the expression: \[ 9((x - 1)^2 - 1) + y^2 = 0 \]Distribute the 9: \[ 9(x - 1)^2 - 9 + y^2 = 0 \] Move the constants to the right side: \[ 9(x - 1)^2 + y^2 = 9 \]
03

Write the equation in standard ellipse form

Divide through by 9 to normalize the equation: \[ \frac{9(x - 1)^2}{9} + \frac{y^2}{9} = \frac{9}{9} \] This simplifies to: \[ (x - 1)^2 + \frac{y^2}{9} = 1 \] Which is now in the standard form of an ellipse: \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
04

Identify parameters

From the standard form \((x - 1)^2 + \frac{y^2}{9} = 1\), identify \(h = 1, k = 0\), \(a = 1\) and \(b = 3\) since the semi-major axis is along the y-axis: Center: \((h, k) = (1, 0)\)Vertices: The vertices are a distance 3 from the center along the y-axis:\[ (1, 0 \pm 3) = (1, 3) \text{ and } (1, -3) \]
05

Locate the foci

Find the foci using the formula \(c^2 = b^2 - a^2\)\(c^2 = 9 - 1\)\(c^2 = 8\)\(c = \sqrt{8} = 2\sqrt{2}\)The foci are centered at (1, 0), so they are placed at:\[ (1, 0 \pm 2\sqrt{2}) \]
06

Graph the ellipse

Position the center at (1, 0). Plot the vertices at (1, 3) and (1, -3). Also, mark the foci at (1, 2\sqrt{2}) \( \approx (1, 2.828)\) and (1, -2\sqrt{2}) \( \approx (1, -2.828)\). Sketch the ellipse with the given dimensions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
Completing the square is a method used to transform a quadratic equation into a perfect square plus a constant. For an ellipse, this technique helps to rewrite the equation in standard form. Here's how it's done:
  • Start with the quadratic equation: 9x虏 + y虏 - 18x = 0.
  • Reorganize the equation to group x-terms together: 9x虏 - 18x + y虏 = 0.
  • Factor out the coefficient of x虏 terms: 9(x虏 - 2x) + y虏 = 0.
  • Complete the square by adding and subtracting the square of half the coefficient of x: \(\frac{-2}{2} = -1, (-1)^2 = 1\)
  • Add and subtract 1 inside the parenthesis: 9((x - 1)虏 - 1) + y虏 = 0.
  • Distribute and simplify: 9(x - 1)虏 - 9 + y虏 = 0.
  • Move the constant term to the other side, resulting in: 9(x - 1)虏 + y虏 = 9.

This method prepares the equation for conversion into the standard form of the ellipse.
Standard Form of an Ellipse
The standard form of an ellipse is a specific way of writing its equation to easily identify its key properties such as center, major axis, and minor axis. The general form is: \(\frac{(x - h)虏}{a虏} + \frac{(y - k)虏}{b虏} = 1\)
Here's the step-by-step to rewrite our equation in standard form:
  • After completing the square, we have 9(x - 1)虏 + y虏 = 9.
  • Divide every term by 9 to normalize the equation: \(\frac{9(x - 1)虏}{9} + \frac{y虏}{9} = \frac{9}{9}\)
  • Simplify to get: (x - 1)虏 + \(\frac{y虏}{9}\) = 1.

The equation \((x - 1)虏 + \frac{y虏}{9} = 1\) now matches the standard form, where:
  • Center: (h, k) = (1,0)
  • a虏 = 1, so a = 1
  • b虏 = 9, so b = 3
This form makes it easy to understand the ellipse's geometry.
Vertices and Foci of an Ellipse
The vertices and foci are crucial points that help define an ellipse's shape and location. To find them from the standard form equation \((x - 1)虏 + \frac{y虏}{9} = 1\):
  • Identify the center (h, k) = (1, 0).
  • Vertices: These are a distance of 'b' units from the center along the major (longer) axis. Since b = 3, the vertices are at: (1, 3) and (1, -3).
The foci of an ellipse are within the major axis and are found using the formula: \(c虏 = b虏 - a虏 = 9 - 1 = 8\). Thus, \(c = \sqrt{8} = 2\sqrt{2}\):
  • The foci are located at: (1, 2\sqrt{2}) which is approximately (1, 2.828) and (1, -2\sqrt{2}) approximating to (1, -2.828).

These points provide a complete understanding of the ellipse's structure and location.

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