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NDA 2020 Maths · Mathematics · Conic Sections · medium

The equation of the hyperbola with foci (±5,0) and transverse axis of length 8 is

Single correct option

The equation of the hyperbola with foci (±5,0)(\pm5,0) and transverse axis of length 88 is
  1. A. x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=1
  2. B. x29−y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1
  3. C. x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}{9}=1
  4. D. x225−y216=1\dfrac{x^2}{25}-\dfrac{y^2}{16}=1
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Answer: (A) x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=1

The foci lie on the x-axis, so the equation is x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1. Transverse axis 2a=82a=8 gives a=4a=4, and c=5c=5 gives b2=c2−a2=25−16=9b^2=c^2-a^2=25-16=9. So the hyperbola is x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=1. Interchanging a2a^2 and b2b^2 gives the first distractor, while the plus sign describes an ellipse, which cannot have a transverse axis.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake