Q207 / 270 · NDA 2017 (Phase 1) · Mathematics

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

  • Correct answer
    A
    $\frac{x^2}{64}+\frac{y^2}{60}=1$
  • B
    $\frac{x^2}{60}+\frac{y^2}{64}=1$
  • C
    $\frac{x^2}{20}+\frac{y^2}{24}=1$
  • D
    $\frac{\mathrm{x}^2}{24}+\frac{\mathrm{y}^2}{20}=1$

Solution

foci: $( \pm 2,0), \mathrm{e}=\frac{1}{4}$ $\mathrm{c}=2, \mathrm{e}=\frac{1}{4}=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{2}{\mathrm{a}}=\frac{1}{4} \Rightarrow \mathrm{a}=8$ We know, $\mathrm{a}^2-\mathrm{b}^2=\mathrm{c}^2$ $\Rightarrow b^2=a^2-c^2=8^2-2^2=64-4=60$ Eq $q^n$ of ellipse $\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $\Rightarrow \frac{x^2}{64}+\frac{y^2}{60}=1$

Hence, the correct answer is option: 'A'.

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