Math Problem Statement

Given equation x^2-4y^2-6x-8y-31=0. (1) write the equation in standard form. (2) find the vertices and foci. (3) find equations of asymptotes. (4) sketch the graph, using the center, vertices and asymptotes.

Solution

Here is the graph of the hyperbola, with the following elements highlighted:

  • Center at (3,1)(3, -1), marked in red.
  • Vertices at (9,1)(9, -1) and (3,1)(-3, -1), marked in green.
  • Foci at (3+35,1)(3 + 3\sqrt{5}, -1) and (335,1)(3 - 3\sqrt{5}, -1), marked in purple.
  • Asymptotes represented by the dashed red lines, corresponding to the equations y+1=12(x3)y + 1 = \frac{1}{2}(x - 3) and y+1=12(x3)y + 1 = -\frac{1}{2}(x - 3).

Summary of Findings:

  1. Standard Form: (x3)236(y+1)29=1\frac{(x - 3)^2}{36} - \frac{(y + 1)^2}{9} = 1
  2. Vertices: (9,1)(9, -1) and (3,1)(-3, -1)
  3. Foci: (3+35,1)(3 + 3\sqrt{5}, -1) and (335,1)(3 - 3\sqrt{5}, -1)
  4. Asymptotes: y+1=12(x3)y + 1 = \frac{1}{2}(x - 3) and y+1=12(x3)y + 1 = -\frac{1}{2}(x - 3)

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

  1. How does the distance between the foci affect the shape of a hyperbola?
  2. How would the equation of the hyperbola change if the transverse axis were vertical instead of horizontal?
  3. How do the asymptotes help in sketching the graph of a hyperbola?
  4. What are the coordinates of the co-vertices for this hyperbola?
  5. How would you convert this hyperbola's equation into parametric form?

Tip: When identifying asymptotes of a hyperbola, the slopes are determined by the ratio ba\frac{b}{a}, giving insight into the orientation and steepness of the asymptotes.

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Math Problem Analysis

Mathematical Concepts

Conic Sections
Hyperbola
Equation of a Hyperbola
Vertices
Foci
Asymptotes

Formulas

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Theorems

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Suitable Grade Level

Grades 11-12