Math Problem Statement

Use the​ center, vertices, and asymptotes to graph the hyperbola. Locate the foci and find the equations of the asymptotes. left parenthesis x minus 1 right parenthesis squared minus 16 left parenthesis y plus 1 right parenthesis squaredequals16 Question content area bottom Part 1 Graph the hyperbola. Choose the correct graph below. A. -6 8 -6 4 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 8 in increments of 2, a vertical y-axis labeled from negative negative 6 to 4 in increments of 2. The graph of a hyperbola has one branch that opens up and one that opens down. The vertices are located at (1,0) and (1,negative 2). The horizontal sides of a rectangle between the branches touch the branches at the vertices. The graph has center in first quadrant and the diagonal lines of the rectangle have magnitude of slopes between 0 and 1 and the branches approach these lines as they travel away from the center. B. -6 8 -6 4 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 8 in increments of 2, a vertical y-axis labeled from negative negative 6 to 4 in increments of 2. The graph of a hyperbola has one branch that opens to the right and one that opens to the left. The vertices are located at (negative 3,negative 1) and (5,negative 1). The vertical sides of a rectangle between the branches touch the branches at the vertices. The graph has center in fourth quadrant and the diagonal lines of the rectangle have magnitude of slopes between 0 and 1 and the branches approach these lines as they travel away from the center. Your answer is correct.C. -6 8 -6 4 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 8 in increments of 2, a vertical y-axis labeled from negative negative 6 to 4 in increments of 2. The graph of a hyperbola has one branch that opens up and one that opens down. The vertices are located at (1,0) and (1,negative 2). The horizontal sides of a rectangle between the branches touch the branches at the vertices. The graph has center in first quadrant and the diagonal lines of the rectangle have magnitude of slopes between 0 and 1 and the branches approach these lines as they travel away from the center. D. -6 8 -6 4 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 8 in increments of 2, a vertical y-axis labeled from negative negative 6 to 4 in increments of 2. The graph of a hyperbola has one branch that opens to the right and one that opens to the left. The vertices are located at (5,negative 1) and (negative 3,negative 1). The vertical sides of a rectangle between the branches touch the branches at the vertices. The graph has center in fourth quadrant and the diagonal lines of the rectangle have magnitude of slopes between 0 and 1 and the branches approach these lines as they travel away from the center. Part 2 The foci​ is/are    left parenthesis 1 plus StartRoot 17 EndRoot comma negative 1 right parenthesis comma left parenthesis 1 minus StartRoot 17 EndRoot comma negative 1 right parenthesis. ​(Type an ordered pair. Type an exact​ answer, using radicals as needed. Use a comma to separate answers as​ needed.) Part 3 The equation of the asymptote with the positive slope is    y equals one fourth x minus five fourths . The equation of the asymptote with the negative slope is    enter your response here. ​(Use integers or fractions for any numbers in the​ equation.)

Solution

Part 1: Analyzing the Equation and Choosing the Correct Graph

Given the hyperbola's equation:

(x−1)216−(y+1)21=1\frac{(x - 1)^2}{16} - \frac{(y + 1)^2}{1} = 1

  • Center: The center of the hyperbola is (1,−1)(1, -1) (derived from the terms x−1x - 1 and y+1y + 1).
  • Orientation: Since the xx-term is positive, the hyperbola opens horizontally (branches open left and right).
  • Vertices: The distance from the center to the vertices is the square root of the denominator under the xx-term (16=4\sqrt{16} = 4):
    • Vertices are at (1−4,−1)(1 - 4, -1) and (1+4,−1)(1 + 4, -1), or (−3,−1)(-3, -1) and (5,−1)(5, -1).
  • Asymptotes: The slope of the asymptotes is determined by the relationship between the denominators under xx and yy: ±vertical distancehorizontal distance=±14\pm \frac{\text{vertical distance}}{\text{horizontal distance}} = \pm \frac{1}{4}.
    • Equations of asymptotes:
      • Positive slope: y=14x−54y = \frac{1}{4}x - \frac{5}{4}
      • Negative slope: y=−14x−54y = -\frac{1}{4}x - \frac{5}{4}

Correct Graph: Graph B

The graph shows a hyperbola opening left and right, with vertices at (−3,−1)(-3, -1) and (5,−1)(5, -1), matching the given equation.

Part 2: Locating the Foci

The formula for the distance to the foci is:

c=a2+b2c = \sqrt{a^2 + b^2}

From the equation:

  • a2=16a^2 = 16, so a=4a = 4.
  • b2=1b^2 = 1, so b=1b = 1.
  • c=16+1=17c = \sqrt{16 + 1} = \sqrt{17}.

The foci are located at (1−17,−1)(1 - \sqrt{17}, -1) and (1+17,−1)(1 + \sqrt{17}, -1).

Foci: (−3+17,−1),(5−17,−1)(-3 + \sqrt{17}, -1), (5 - \sqrt{17}, -1)

Part 3: Equations of the Asymptotes

The asymptotes of a hyperbola are derived from the slopes ±ba\pm \frac{b}{a}:

  • Slope (±ba=±14\pm \frac{b}{a} = \pm \frac{1}{4}).

Using the center (1,−1)(1, -1), the equations are:

  1. Positive slope: y−(−1)=14(x−1)  ⟹  y=14x−54y - (-1) = \frac{1}{4}(x - 1) \implies y = \frac{1}{4}x - \frac{5}{4}
  2. Negative slope: y−(−1)=−14(x−1)  ⟹  y=−14x−54y - (-1) = -\frac{1}{4}(x - 1) \implies y = -\frac{1}{4}x - \frac{5}{4}

Asymptotes: y=14x−54,y=−14x−54.y = \frac{1}{4}x - \frac{5}{4}, \quad y = -\frac{1}{4}x - \frac{5}{4}.

Final Results:

  1. Correct graph: B.
  2. Foci: (1+17,−1),(1−17,−1)(1 + \sqrt{17}, -1), (1 - \sqrt{17}, -1)
  3. Asymptotes: y=14x−54,y=−14x−54.y = \frac{1}{4}x - \frac{5}{4}, \quad y = -\frac{1}{4}x - \frac{5}{4}.

Would you like further explanations or visualizations?

Related Questions:

  1. How does the orientation of the hyperbola affect its graph?
  2. Why does the slope of the asymptotes depend on bb and aa?
  3. How do you find the distance between the foci of a hyperbola?
  4. What is the role of the denominators in the hyperbola equation?
  5. Can a hyperbola have vertical and horizontal orientations simultaneously?

Tip:

To graph hyperbolas accurately, always plot the center, vertices, and asymptotes first to guide the shape!

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Math Problem Analysis

Mathematical Concepts

Algebra
Hyperbolas
Graphing Conic Sections

Formulas

Equation of a hyperbola: (x-h)^2/a^2 - (y-k)^2/b^2 = 1
Distance to foci: c = sqrt(a^2 + b^2)
Equations of asymptotes: y - k = ±(b/a)(x - h)

Theorems

Pythagorean relationship in hyperbolas

Suitable Grade Level

Grades 10-12

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