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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

A hyperbola having its centre at the origin is passing through the...

A hyperbola having its centre at the origin is passing through the point (5,2)(5,2) and has transverse axis of length 8 along the X\mathrm{X}-axis. Then the eccentricity of its conjugate hyperbola is

Options

  1. A133\frac{\sqrt{13}}{3}
  2. B133\sqrt{\frac{13}{3}}
  3. C132\frac{\sqrt{13}}{2}
  4. D132\sqrt{\frac{13}{2}}

Correct answer

C. 132\frac{\sqrt{13}}{2}

Step-by-step solution

Let equation of hyperbola is x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 Given, 2a=8⇒a=42 \mathrm{a}=8 \Rightarrow \mathrm{a}=4 It passes through (5,2)(5,2) Hence, on solving we get, b=83\mathrm{b}=\frac{8}{3} Now, eccentricity of its conjugate hyperbola is =a2+b2 b2=132\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}}=\frac{\sqrt{13}}{2}
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