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Find the vertices of the hyperbola 3y2 - x2 = 108 ?
1. (0, ± 5)
2. (± 5, 0)
3. (0, ± 6)
4. (± 6, 0)
5. None of these

1 Answer

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Best answer
Correct Answer - Option 3 : (0, ± 6)

CONCEPT:

The properties of a vertical hyperbola \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) are:

  • Its centre is given by: (0, 0)
  • Its foci are given by: (0, - ae) and (0, ae)
  • Its vertices are given by: (0, - a)  and (0, a)
  • Its eccentricity is given by: \(e = \frac{{\sqrt {{a^2}\ +\ {b^2}} }}{a}\)
  • Length of transverse axis = 2a and its equation is x = 0.
  • Length of conjugate axis = 2b and its equation is y = 0.
  • Length of its latus rectum is given by: \(\frac{2b^2}{a}\)


CALCULATION:

Given: Equation of hyperbola is 3y2 - x2 = 108

The given equation of hyperbola can be re-written as: \(\frac{{{y^2}}}{{{36}}} - \frac{{{x^2}}}{{{108}}} = 1\)

As we can see that, the given hyperbola is a vertical hyperbola.

So, by comparing the given equation of hyperbola with \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) we get,

⇒ a2 = 36 and b2 = 108

As we know that, vertices of a vertical hyperbola is given by: (0, ± a)

So, the vertices of the given hyperbola is: (0, ± 6)

Hence, option C is the correct answer.

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