Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Finding the Foci of a Hyperbola
To find the foci of the hyperbola represented by the equation:
\[ -4x^2 + y^2 - 8y - 20 = 0 \]
follow these steps:
1. **Rewrite the Hyperbola Equation in Standard Form**:
- First, isolate the terms involving \(y\):
\[ y^2 - 8y - 4x^2 = 20 \]
- Next, complete the square for the \(y\) terms:
\[ y^2 - 8y \rightarrow (y^2 - 8y + 16 - 16) \right] = (y - 4)^2 - 16 \]
So, the equation becomes:
\[ (y - 4)^2 - 16 - 4x^2 = 20 \]
- Simplify it:
\[ (y - 4)^2 - 4x^2 = 36 \]
- Divide the entire equation by 36 to normalize it:
\[ \frac{(y - 4)^2}{36} - \frac{4x^2}{36} = 1 \]
\[ \frac{(y - 4)^2}{36} - \frac{x^2}{9} = 1 \]
Now the equation is in the standard form of a hyperbola:
\[ \frac{(y - 4)^2}{6^2} - \frac{x^2}{3^2} = 1\]
2. **Identify Key Constants**:
- Compare with the standard form of a hyperbola \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\)
In this hyperbola:
- \(a^2 = 36 \rightarrow a = 6\)
- \(b^2 = 9 \rightarrow b = 3\)
- The center \((h, k)\) is at \((0, 4)\)
3. **Calculate the Foci**:
- For a vertical hyperbola, use the formula for the distance from the center to each focus, \(c\), found by:
\[ c = \sqrt{a^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d7590ab-1e8b-4f01-9940-34be16d61a30%2F9d72dc35-2e25-4e12-ab8d-c496e1379b9f%2Fan7ex3f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Foci of a Hyperbola
To find the foci of the hyperbola represented by the equation:
\[ -4x^2 + y^2 - 8y - 20 = 0 \]
follow these steps:
1. **Rewrite the Hyperbola Equation in Standard Form**:
- First, isolate the terms involving \(y\):
\[ y^2 - 8y - 4x^2 = 20 \]
- Next, complete the square for the \(y\) terms:
\[ y^2 - 8y \rightarrow (y^2 - 8y + 16 - 16) \right] = (y - 4)^2 - 16 \]
So, the equation becomes:
\[ (y - 4)^2 - 16 - 4x^2 = 20 \]
- Simplify it:
\[ (y - 4)^2 - 4x^2 = 36 \]
- Divide the entire equation by 36 to normalize it:
\[ \frac{(y - 4)^2}{36} - \frac{4x^2}{36} = 1 \]
\[ \frac{(y - 4)^2}{36} - \frac{x^2}{9} = 1 \]
Now the equation is in the standard form of a hyperbola:
\[ \frac{(y - 4)^2}{6^2} - \frac{x^2}{3^2} = 1\]
2. **Identify Key Constants**:
- Compare with the standard form of a hyperbola \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\)
In this hyperbola:
- \(a^2 = 36 \rightarrow a = 6\)
- \(b^2 = 9 \rightarrow b = 3\)
- The center \((h, k)\) is at \((0, 4)\)
3. **Calculate the Foci**:
- For a vertical hyperbola, use the formula for the distance from the center to each focus, \(c\), found by:
\[ c = \sqrt{a^2
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