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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Ellipse

NDA 2022 · Mathematics · Ellipse

What is the distance between the foci of the ellipse x^2+2 y^2=1 ?

What is the distance between the foci of the ellipse x2+2y2=1x^2+2 y^2=1 ?

Options

  1. A1
  2. B2\sqrt{2}
  3. C2
  4. D222 \sqrt{2}

Correct answer

B. 2\sqrt{2}

Step-by-step solution

Given equation of ellipse x2+2y2=1⇒x21+y212=1⇒a2=1 and b2=12\begin{aligned} & x^2+2 y^2=1 \Rightarrow \frac{x^2}{1}+\frac{y^2}{\frac{1}{2}}=1 \\ & \Rightarrow a^2=1 \text{ and } b^2=\frac{1}{2}\end{aligned}∵b2=a2−c2c2=a2−b2=1−12=12c=12\begin{aligned} & \because \mathrm{b}^2=\mathrm{a}^2-\mathrm{c}^2 \\ & \mathrm{c}^2=\mathrm{a}^2-\mathrm{b}^2=1-\frac{1}{2}=\frac{1}{2} \\ & \mathrm{c}=\frac{1}{\sqrt{2}}\end{aligned}ThedistancebetweenfociofellipseThe distance between foci of ellipse=2 \mathrm{C}=\frac{2}{\sqrt{2}}=\sqrt{2}$
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