Chapter11: Conics
Section11.2: Parabolas
Problem 97E: Write the equation of a parabola that opens up or down in standard form and the equation of a...
Related questions
Question
![### 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
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Trigonometry (MindTap Course List)
Trigonometry
ISBN:
9781337278461
Author:
Ron Larson
Publisher:
Cengage Learning
College Algebra
Algebra
ISBN:
9781305115545
Author:
James Stewart, Lothar Redlin, Saleem Watson
Publisher:
Cengage Learning