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Find the equation of the hyperbola whose foci are at (0, ± 6) and the length of whose conjugate axis is 2√11 ?
1. \(\frac{{{y^2}}}{{{21}}} - \frac{{{x^2}}}{{{11}}} = 1\)
2. \(\frac{{{y^2}}}{{{25}}} - \frac{{{x^2}}}{{{11}}} = 1\)
3. \(\frac{{{y^2}}}{{{16}}} - \frac{{{x^2}}}{{{11}}} = 1\)
4. \(\frac{{{y^2}}}{{{5}}} - \frac{{{x^2}}}{{{11}}} = 1\)
5. None of these

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Correct Answer - Option 2 : \(\frac{{{y^2}}}{{{25}}} - \frac{{{x^2}}}{{{11}}} = 1\)

Concept:

The hyperbola of the form \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) has:

  • Centre is given by: (0, 0)
  • Vertices are given by: (0, ± a)
  • Foci are given by: (0, ± c)
  • Length of transverse axis is given by: 2a
  • Length of conjugate axis is given by: 2b
  • Eccentricity is given by: \(e = √ {1 + \frac{{{b^2}}}{{{a^2}}}} \)
  • b2 = c2 - a2


Calculation:

Given: The foci and the length of conjugate axis are: (0, ± 6) and 2√11 respectively.

∵ The foci of the given hyperbola are of the form (0, ± c), it is a vertical hyperbola i.e it is of the form:

\(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\)

⇒ c = 6 and c2 = 36.

As we know that he length of conjugate axis of the hyperbola of the form \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) is given by: 2b

⇒ 2b = 2√11

⇒ b = √11 and b2 = 11.

As we know that, b2 = c2 - a2

⇒ a2 = c2 - b2 = 36 - 11 = 25

Hence, the equation of required hyperbola is: \(\frac{{{y^2}}}{{{25}}} - \frac{{{x^2}}}{{{11}}} = 1\)

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