The general form of an ellipse is 3x + 2y-12x +12y + 12 = 0. What is the standard form of the ellipse? o (-2)? (y+3)? 3D1 (x-2)2 9. (y+3)? =D1 %3D 6 (2-2)? 9. (y-3)2 =D1 %3D o (2+2)? (y+3)2 = 1 9.
The general form of an ellipse is 3x + 2y-12x +12y + 12 = 0. What is the standard form of the ellipse? o (-2)? (y+3)? 3D1 (x-2)2 9. (y+3)? =D1 %3D 6 (2-2)? 9. (y-3)2 =D1 %3D o (2+2)? (y+3)2 = 1 9.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Understanding the Standard Form of an Ellipse
The image presents a mathematical problem involving the conversion of the general form of an ellipse equation into its standard form.
**General Form of the Ellipse:**
\[ 3x^2 + 2y^2 - 12x + 12y + 12 = 0 \]
**Task:**
Identify the equivalent standard form of the ellipse.
**Options Provided:**
1. \(\frac{(x-2)^2}{6} + \frac{(y+3)^2}{9} = 1\)
2. \(\frac{(x-2)^2}{9} + \frac{(y+3)^2}{6} = 1\)
3. \(\frac{(x-2)^2}{9} + \frac{(y-3)^2}{6} = 1\)
4. \(\frac{(x+2)^2}{6} + \frac{(y+3)^2}{9} = 1\)
**Explanation:**
An ellipse's standard form is expressed as:
\[
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1
\]
Where \((h, k)\) is the center of the ellipse, and \(a^2\) and \(b^2\) are the squares of the semi-major and semi-minor axes, respectively.
The task is to rearrange and simplify the given general form equation to match one of the provided standard forms. This involves completing the square for \(x\) and \(y\) terms and adjusting constants accordingly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58402a47-f1ff-47ea-9009-84135bb68162%2F650aee2b-5f77-4fa6-b657-1d8aecccbaa0%2Fzj1a5zh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding the Standard Form of an Ellipse
The image presents a mathematical problem involving the conversion of the general form of an ellipse equation into its standard form.
**General Form of the Ellipse:**
\[ 3x^2 + 2y^2 - 12x + 12y + 12 = 0 \]
**Task:**
Identify the equivalent standard form of the ellipse.
**Options Provided:**
1. \(\frac{(x-2)^2}{6} + \frac{(y+3)^2}{9} = 1\)
2. \(\frac{(x-2)^2}{9} + \frac{(y+3)^2}{6} = 1\)
3. \(\frac{(x-2)^2}{9} + \frac{(y-3)^2}{6} = 1\)
4. \(\frac{(x+2)^2}{6} + \frac{(y+3)^2}{9} = 1\)
**Explanation:**
An ellipse's standard form is expressed as:
\[
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1
\]
Where \((h, k)\) is the center of the ellipse, and \(a^2\) and \(b^2\) are the squares of the semi-major and semi-minor axes, respectively.
The task is to rearrange and simplify the given general form equation to match one of the provided standard forms. This involves completing the square for \(x\) and \(y\) terms and adjusting constants accordingly.
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