cher
contestada

Determine the vertices, asymptotes, and foci of the hyperbola [tex]576x^2 - 16y^2=144[/tex]

Please show all your work - thanks! :)

Respuesta :

Space

Answer:

Hyperbola Foci: [tex]\displaystyle (\pm \frac{\sqrt{37}}{2}, 0)[/tex]

Hyperbola Vertices: [tex]\displaystyle (\pm \frac{1}{2}, 0)[/tex]

Hyperbola Asymptotes: [tex]\displaystyle y = \pm 6x[/tex]

General Formulas and Concepts:

Math

  • Simplifying

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

Algebra I

  • Coordinates (x, y)

Pre-Calculus

Hyperbola Standard Form: [tex]\displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1[/tex]

  • Center (h, k)
  • Hyperbola Vertices: (±a, 0)
  • Hyperbola Foci: c² = a² + b²

Hyperbola Asymptotes: [tex]\displaystyle (y - k) = \pm \frac{b}{a}(x - h)[/tex]

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle 576x^2 - 16y^2 = 144[/tex]

Step 2: Rewrite

Rewrite equation into hyperbola standard form

  1. [Hyperbola] [Division Property of Equality] Divide 144 on both sides:         [tex]\displaystyle \frac{576x^2 - 16y^2}{144} = 1[/tex]
  2. [Hyperbola] Split fraction:                                                                              [tex]\displaystyle \frac{576x^2}{144} - \frac{16y^2}{144} = 1[/tex]
  3. [Hyperbola] Simplify:                                                                                      [tex]\displaystyle 4x^2 - \frac{y^2}{9} = 1[/tex]
  4. [Hyperbola] Rewrite:                                                                                      [tex]\displaystyle \frac{x^2}{\frac{1}{4}} - \frac{y^2}{9} = 1[/tex]
  5. [Hyperbola] Rewrite a and b:                                                                         [tex]\displaystyle \frac{x^2}{(\frac{1}{2})^2} - \frac{y^2}{3^2} = 1[/tex]

Step 3: Identify Parts

Compare the general standard form and our equation standard form to identify parts of the hyperbola.

[tex]\displaystyle \frac{x^2}{(\frac{1}{2})^2} - \frac{y^2}{3^2} = 1[/tex] ↔ [tex]\displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1[/tex]

Hyperbola Center: (0, 0)

Hyperbola Vertices: [tex]\displaystyle (\pm \frac{1}{2}, 0)[/tex]

Step 4: Identify Other Parts

We go slightly more complex in determining the asymptotes and foci of the hyperbola.

Foci

  1. Substitute in variables [Hyperbola Foci]:                                                      [tex]\displaystyle c^2 = (\frac{1}{2})^2 + 3^2[/tex]
  2. [Hyperbola Foci] Evaluate exponents:                                                          [tex]\displaystyle c^2 = \frac{1}{4} + 9[/tex]
  3. [Hyperbola Foci] Add:                                                                                    [tex]\displaystyle c^2 = \frac{37}{4}[/tex]
  4. [Hyperbola Foci] [Equality Property] Square root both sides:                     [tex]\displaystyle c = \pm \sqrt{\frac{37}{4}}[/tex]
  5. [Hyperbola Foci] Simplify:                                                                              [tex]\displaystyle c = \pm \frac{\sqrt{37}}{2}[/tex]

Our hyperbola foci are [tex]\displaystyle (\frac{\sqrt{37}}{2}, 0)[/tex] and [tex]\displaystyle (-\frac{\sqrt{37}}{2}, 0)[/tex].

Asymptotes

  1. Substitute in variables [Hyperbola Asymptotes]:                                         [tex]\displaystyle (y - 0) = \pm \frac{3}{\frac{1}{2}}(x - 0)[/tex]
  2. [Hyperbola Asymptotes] (Parenthesis) Subtract:                                         [tex]\displaystyle y = \pm \frac{3}{\frac{1}{2}}x[/tex]
  3. [Hyperbola Asymptotes] Divide:                                                                    [tex]\displaystyle y = \pm 6x[/tex]

Our hyperbola asymptotes are y = 6x and y = -6x.

Otras preguntas