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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Ellipse

NDA 2017 · Mathematics · Ellipse

What is the equation of the ellipse having foci ( ± 2,0) and the...

What is the equation of the ellipse having foci (±2,0)( \pm 2,0) and the eccentricity 14\frac{1}{4} ?

Options

  1. Ax264+y260=1\frac{x^2}{64}+\frac{y^2}{60}=1
  2. Bx260+y264=1\frac{x^2}{60}+\frac{y^2}{64}=1
  3. Cx220+y224=1\frac{x^2}{20}+\frac{y^2}{24}=1
  4. Dx224+y220=1\frac{\mathrm{x}^2}{24}+\frac{\mathrm{y}^2}{20}=1

Correct answer

A. x264+y260=1\frac{x^2}{64}+\frac{y^2}{60}=1

Step-by-step solution

foci: (±2,0),e=14c=2,e=14=ca⇒2a=14⇒a=8( \pm 2,0), \mathrm{e}=\frac{1}{4}\mathrm{c}=2, \mathrm{e}=\frac{1}{4}=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{2}{\mathrm{a}}=\frac{1}{4} \Rightarrow \mathrm{a}=8 We know, a2−b2=c2⇒b2=a2−c2=82−22=64−4=60\mathrm{a}^2-\mathrm{b}^2=\mathrm{c}^2\Rightarrow b^2=a^2-c^2=8^2-2^2=64-4=60 Eq qnq^n of ellipse ⇒x2a2+y2b2=1⇒x264+y260=1\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow \frac{x^2}{64}+\frac{y^2}{60}=1
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