Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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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
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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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