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NDA 2019 Maths · Mathematics · Conic Sections · medium

The foci of the conic 9x^2-16y^2=144 are at the points

Single correct option

The foci of the conic 9x2−16y2=1449x^2-16y^2=144 are at the points
  1. A. (±4,0)(\pm4,0)
  2. B. (0,±5)(0,\pm5)
  3. C. (±7,0)(\pm\sqrt7,0)
  4. D. (±5,0)(\pm5,0)
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Answer: (D) (±5,0)(\pm5,0)

Dividing by 144144 gives x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1, a hyperbola with a2=16a^2=16 and b2=9b^2=9. Then c2=a2+b2=25c^2=a^2+b^2=25, so c=5c=5 and the foci are (±5,0)(\pm5,0) on the x-axis. The value 7=a2−b2\sqrt7=\sqrt{a^2-b^2} comes from using the ellipse formula, and (±4,0)(\pm4,0) are the vertices.
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Original question written in the NDA 2019 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