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 3 right parenthesis squared minus 36 left parenthesis y plus 3 right parenthesis squaredequals36 Question content area bottom Part 1 Graph the hyperbola. Choose the correct graph below. A. -6 12 -8 2 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 12 in increments of 2, a vertical y-axis labeled from negative negative 8 to 2 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 (9,negative 3) and (negative 3,negative 3). 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. B. -6 12 -8 2 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 12 in increments of 2, a vertical y-axis labeled from negative negative 8 to 2 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 (3,negative 2) and (3,negative 4). The horizontal 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. C. -6 12 -8 2 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 12 in increments of 2, a vertical y-axis labeled from negative negative 8 to 2 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 (3,negative 2) and (3,negative 4). The horizontal 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. D. -6 12 -8 2 x y

A coordinate system with a horizontal x-axis labeled from negative negative 6 to 12 in increments of 2, a vertical y-axis labeled from negative negative 8 to 2 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 3) and (9,negative 3). 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    enter your response here. ​(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    enter your response here. 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

We are tasked with analyzing and graphing the given hyperbola equation:

(x−3)236−(y+3)236=1.\frac{(x-3)^2}{36} - \frac{(y+3)^2}{36} = 1.

Step 1: Identify the hyperbola's characteristics

  1. Standard form: The hyperbola is of the form: (x−h)2a2−(y−k)2b2=1.\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1. Here:

    • Center (h,k)=(3,−3)(h, k) = (3, -3),
    • a2=36a^2 = 36 ⇒a=6\Rightarrow a = 6,
    • b2=36b^2 = 36 ⇒b=6\Rightarrow b = 6.

  2. Orientation: Since the xx-term comes first, this hyperbola opens horizontally.

  3. Vertices: The vertices are located aa units to the left and right of the center along the xx-axis: (h±a,k)⇒(3±6,−3).(h \pm a, k) \Rightarrow (3 \pm 6, -3). So, the vertices are: (−3,−3) and (9,−3).(-3, -3) \text{ and } (9, -3).

  4. Foci: The distance from the center to each focus is cc, where: c=a2+b2=36+36=72=62.c = \sqrt{a^2 + b^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}. The foci are located cc units to the left and right of the center along the xx-axis: (h±c,k)⇒(3±62,−3).(h \pm c, k) \Rightarrow (3 \pm 6\sqrt{2}, -3). So, the foci are: (3−62,−3) and (3+62,−3).(3 - 6\sqrt{2}, -3) \text{ and } (3 + 6\sqrt{2}, -3).

Step 2: Asymptotes

For a hyperbola, the asymptotes pass through the center and have slopes determined by ±ba\pm \frac{b}{a}: y−k=±ba(x−h).y - k = \pm \frac{b}{a}(x - h). Here: ba=66=1.\frac{b}{a} = \frac{6}{6} = 1. Substitute h=3h = 3 and k=−3k = -3: y+3=±1(x−3).y + 3 = \pm 1(x - 3). Simplify:

  1. Positive slope: y+3=x−3⇒y=x−6.y + 3 = x - 3 \Rightarrow y = x - 6.
  2. Negative slope: y+3=−x+3⇒y=−x+6.y + 3 = -x + 3 \Rightarrow y = -x + 6.

Step 3: Graph

From the vertices, asymptotes, and orientation, the correct graph is A or D. Based on the vertices and the xx-axis opening, the correct answer is D.

Final Answers

  1. Graph: Option D.
  2. Foci: (3−62,−3) and (3+62,−3).(3 - 6\sqrt{2}, -3) \text{ and } (3 + 6\sqrt{2}, -3).
  3. Asymptotes:
    • Positive slope: y=x−6y = x - 6,
    • Negative slope: y=−x+6y = -x + 6.

Would you like me to elaborate on any part of this?

Related Questions

  1. How do we determine the orientation of a hyperbola?
  2. What role does the center (h,k)(h, k) play in hyperbola equations?
  3. How do you calculate the distance to the foci in a hyperbola?
  4. What is the geometric significance of the asymptotes in a hyperbola?
  5. Can the transverse axis ever be vertical for this type of hyperbola?

Tip: Always verify whether the xx-term or yy-term comes first to identify the hyperbola's orientation.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hyperbolas
Conic Sections
Graphing

Formulas

Standard form of 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 Theorem in hyperbolas

Suitable Grade Level

Grades 10-12

Related Recommendation

Graph the Hyperbola: Center, Foci, and Asymptotes

Graphing Hyperbolas: Center, Vertices, Foci, and Asymptotes

Graphing a Hyperbola: Center, Vertices, Foci, and Asymptotes

Graphing Hyperbolas: Equation x^2/9 - (y+2)^2/16 = 1

Graphing the Hyperbola: y²/25 - (x-2)²/1 = 1