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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Hyperbola

NDA 2015 · Mathematics · Hyperbola

The eccentricity of the hyperbola 16 x^2-9 y^2=1 is

The eccentricity of the hyperbola 16x2−9y2=116 x^2-9 y^2=1 is

Options

  1. A3/5
  2. B5/3
  3. C4/5
  4. D5/4

Correct answer

B. 5/3

Step-by-step solution

∵16x2−9y2=1 or  x2(14)2−y2(13)2=1\begin{aligned} & \because 16 x^2-9 y^2=1 \\ & \text{\text{ or } } \frac{x^2}{\left(\frac{1}{4}\right)^2}-\frac{y^2}{\left(\frac{1}{3}\right)^2}=1 \end{aligned} Comparing with x2a2−y2b2=1\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{ b}^2}=1∴a2=116b2=19∵e=1+b2a2∴e=1+1/91/16⇒e=53\begin{aligned} & \therefore \mathrm{a}^2=\frac{1}{16} \\ & & \mathrm{ b}^2=\frac{1}{9} \\ & \because \mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}} \\ & \therefore \mathrm{e}=\sqrt{1+\frac{1 / 9}{1 / 16}} \\ & \Rightarrow \quad \mathrm{e}=\frac{5}{3} \end{aligned}$$
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