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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Hyperbola

NTA Abhyas 2020 · Mathematics · Hyperbola

If the eccentricity of the hyperbola x^ 2 16 - y^ 2 b^ 2 = - 1 is...

If the eccentricity of the hyperbola x216−y2b2=−1\dfrac{x^{2}}{16} - \dfrac{y^{2}}{b^{2}} = - 1 is 54,\dfrac{5}{4}, then b2b^{2} is equal to

Options

  1. A2569\dfrac{256}{9}
  2. B163\dfrac{16}{3}
  3. C99
  4. D33

Correct answer

A. 2569\dfrac{256}{9}

Step-by-step solution

y2b2−x216=1\dfrac{y2}{b2} - \dfrac{x2}{16} = 1

For the hyperbola, e=1+16b2e = \sqrt{1 + \dfrac{16}{b2}}
⇒1+16b2=54⇒b2=2569⇒ \sqrt{1 + \dfrac{16}{b2}} = \dfrac{5}{4} ⇒ b^{2} = \dfrac{256}{9}

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