Trigonometry (11th Edition)
11th Edition
ISBN: 9780134217437
Author: Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher: PEARSON
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![**2. Sketch the ellipse and list its domain and range.**
\[
\frac{(x - 4)^2}{9} + \frac{(y + 2)^2}{4} = 1
\]
**3. Write an equation for the hyperbola with center at (5, −8), focus at (8, −8), and vertex at (7, −8).**
**Explanation:**
- The given equation in Question 2 is in the standard form of an ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) is the center of the ellipse.
- For the given ellipse:
- Center: \((4, -2)\)
- Semi-major axis along the x-direction: \(a = 3\)
- Semi-minor axis along the y-direction: \(b = 2\)
- The domain (x-values): \([4-a, 4+a] = [1, 7]\)
- The range (y-values): \([-2-b, -2+b] = [-4, 0]\)
Graph the ellipse with its center at \((4, -2)\), stretched 3 units horizontally and 2 units vertically.
**Question 3 Explanation:**
- The hyperbola centered at (5, -8) with a vertex at (7, -8) indicates a horizontal orientation.
- The distance between the center and vertex is the semi-major axis length: \(a = 2\)
- The distance between the center and focus (focus at (8, -8)) gives the value for \(c\): \(c = 3\)
- The equation of a hyperbola: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
Find \(b^2\) using the formula \(c^2 = a^2 + b^2\):
\[
9 = 4 + b^2 \Rightarrow b^2 = 5
\]
The equation for the hyperbola is:
\[
\frac{(x-5)^2}{4} - \frac{(y+8)^2}{5} = 1
\]](https://content.bartleby.com/qna-images/question/681a1dd5-ea79-4df0-af09-f7e014c18f3f/c3c995de-b42c-490b-996e-9da22cc4c501/mhhyeqm_processed.jpeg)
Transcribed Image Text:**2. Sketch the ellipse and list its domain and range.**
\[
\frac{(x - 4)^2}{9} + \frac{(y + 2)^2}{4} = 1
\]
**3. Write an equation for the hyperbola with center at (5, −8), focus at (8, −8), and vertex at (7, −8).**
**Explanation:**
- The given equation in Question 2 is in the standard form of an ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) is the center of the ellipse.
- For the given ellipse:
- Center: \((4, -2)\)
- Semi-major axis along the x-direction: \(a = 3\)
- Semi-minor axis along the y-direction: \(b = 2\)
- The domain (x-values): \([4-a, 4+a] = [1, 7]\)
- The range (y-values): \([-2-b, -2+b] = [-4, 0]\)
Graph the ellipse with its center at \((4, -2)\), stretched 3 units horizontally and 2 units vertically.
**Question 3 Explanation:**
- The hyperbola centered at (5, -8) with a vertex at (7, -8) indicates a horizontal orientation.
- The distance between the center and vertex is the semi-major axis length: \(a = 2\)
- The distance between the center and focus (focus at (8, -8)) gives the value for \(c\): \(c = 3\)
- The equation of a hyperbola: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
Find \(b^2\) using the formula \(c^2 = a^2 + b^2\):
\[
9 = 4 + b^2 \Rightarrow b^2 = 5
\]
The equation for the hyperbola is:
\[
\frac{(x-5)^2}{4} - \frac{(y+8)^2}{5} = 1
\]
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