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Find the equation of hyperbola whose foci are (± 5, 0) and the transverse axis is of length 8?
1. \(\frac{{{x^2}}}{{{6}}} - \frac{{{y^2}}}{{{9}}} = 1\)
2. \(\frac{{{x^2}}}{{{16}}} - \frac{{{y^2}}}{{{9}}} = 1\)
3. \(\frac{{{x^2}}}{{{16}}} - \frac{{{y^2}}}{{{7}}} = 1\)
4. \(\frac{{{x^2}}}{{{6}}} - \frac{{{y^2}}}{{{5}}} = 1\)
5. None of these

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Correct Answer - Option 2 : \(\frac{{{x^2}}}{{{16}}} - \frac{{{y^2}}}{{{9}}} = 1\)

Concept:

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

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


Calculation:

Given: The foci of hyperbola are: (± 5, 0) and the length of transverse axis is 8.

∵ foci lies on the x -axis so, it is horizontal hyperbola.

As we know that for the hyperbola of the form \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) .

The foci are given by: (± c, 0) and length of transverse axis is given by: 2a.

⇒ c = 5 and 2a = 8

⇒ a = 4, a2 = 16 and c2 = 25

As we know that b2 = c2 - a2

By substituting the value of a2 and c2 in the above equation we get,

⇒  b2 = 25 - 16 = 9

Hence, the equation of required hyperbola is: \(\frac{{{x^2}}}{{{16}}} - \frac{{{y^2}}}{{{9}}} = 1\)

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