Finding the Equation of a Hyperbola with Given Foci

Analyze the Given Information:

• We have two foci: ( F_1 = (0, 0) ) and ( F_2 = (0, -\frac{4}{3}) ).
• The distance between the foci is the focal distance, ( 2c ).

Determine the Focal Distance:

• Calculate the distance between the foci:
  [
  c = \frac{1}{2} \times \text{distance between } F_1 \text{ and } F_2 = \frac{1}{2} \times \left| -\frac{4}{3} – 0 \right| = \frac{2}{3}
  ]

Determine the Type of Hyperbola:

• Since both foci are on the y-axis, the hyperbola will be vertically oriented.

Standard Form of a Vertically Oriented Hyperbola:

• The standard form of a hyperbola with a vertical transverse axis centered at ( (h, k) ) is:
  [
  \frac{(x – h)^2}{b^2} – \frac{(y – k)^2}{a^2} = -1
  ]

Center of the Hyperbola:

• The center ((h, k)) is the midpoint of the line segment joining the foci:
  [
  h = \frac{0 + 0}{2} = 0, \quad k = \frac{0 – \frac{4}{3}}{2} = -\frac{2}{3}
  ]

Relation Between a, b, and c:

• For a hyperbola, the relationship between (a), (b), and (c) is (c^2 = a^2 + b^2).

Find the Value of (a):

• The distance between the vertices of a hyperbola along the transverse axis is (2a).
• Since we are only given the foci, without additional specifics, we can assume (a) as a variable for now.
• Assume the constant difference mentioned in the hyperbola definition, i.e., (2a).

Equation of the Hyperbola:

• Using the center ((0, -\frac{2}{3})) and the relationship (c^2 = a^2 + b^2), we can write:
  [
  c^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}
  ]
• If (a^2) is a known constant (or assumed), then:
  [
  b^2 = c^2 – a^2
  ]
• Without a specific (a), the equation is:
  [
  \frac{x^2}{b^2} – \frac{(y + \frac{2}{3})^2}{a^2} = -1
  ]

Conclusion:

• Since (a) is not specified, this general form assumes a known (a), allowing us to solve for specific cases.
• The transformation (y + \frac{2}{3}) recenters the hyperbola at the midpoint of the foci.

Understand the Asymptotes of a Hyperbola:

• For a hyperbola centered at ((h, k)) with a vertical transverse axis, the asymptotes are given by:
  [
  y = k \pm \frac{a}{b}(x – h)
  ]

Determine the Angle Between Asymptotes:

• The angle ( \theta ) between the asymptotes of a hyperbola is given by:
  [
  2\theta = \tan^{-1}\left(\frac{a}{b}\right) + \tan^{-1}\left(-\frac{a}{b}\right)
  ]
• Simplifying gives:
  [
  2\theta = 2\tan^{-1}\left(\frac{a}{b}\right) = 60^\circ
  ]
• Therefore, ( \theta = 30^\circ ), and thus:
  [
  \tan(30^\circ) = \frac{a}{b}
  ]

Calculate ( \tan(30^\circ) ):

• (\tan(30^\circ) = \frac{1}{\sqrt{3}}).
• Therefore, (\frac{a}{b} = \frac{1}{\sqrt{3}}).

Use the Relationship (c^2 = a^2 + b^2):

• From the properties of hyperbolas, we know:
  [
  c^2 = a^2 + b^2
  ]
• Given (c = \frac{2}{3}), we have:
  [
  \left(\frac{2}{3}\right)^2 = a^2 + b^2
  ]
  [
  \frac{4}{9} = a^2 + b^2
  ]

Express (a) and (b) in Terms of Each Other:

• From ( \frac{a}{b} = \frac{1}{\sqrt{3}} ), we have:
  [
  a = \frac{b}{\sqrt{3}}
  ]

Solve for (a) and (b):

• Substitute (a = \frac{b}{\sqrt{3}}) into the equation:
  [
  \frac{4}{9} = \left(\frac{b}{\sqrt{3}}\right)^2 + b^2
  ]
  [
  \frac{4}{9} = \frac{b^2}{3} + b^2
  ]
  [
  \frac{4}{9} = \frac{4b^2}{3}
  ]

• Solve for (b^2):
  [
  b^2 = \frac{4}{12} = \frac{1}{3}
  ]

• Solve for (a^2):
  [
  a^2 = \left(\frac{b}{\sqrt{3}}\right)^2 = \frac{b^2}{3} = \frac{1/3}{3} = \frac{1}{9}
  ]

Write the Equation of the Hyperbola:

• With (a^2 = \frac{1}{9}) and (b^2 = \frac{1}{3}), the equation becomes:
  [
  \frac{x^2}{\frac{1}{3}} – \frac{(y + \frac{2}{3})^2}{\frac{1}{9}} = -1
  ]
• Simplify:
  [
  3x^2 – 9(y + \frac{2}{3})^2 = -1
  ]