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JEE Main 201511 AprilOnline
Exam
JEE Main
Year
2015
Date
11 April
Mode
Online
Subject
Mathematics
Chapter
Ellipse

JEE Main 2015 · 11 April · Online · Mathematics · Ellipse

If the distance between the foci of an ellipse is half the length...

If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is:

Options

  1. A12\dfrac{1}{2}
  2. B2−1\sqrt{2} - 1
  3. C2−12\dfrac{\sqrt{2} - 1}{2}
  4. D22−12\dfrac{2\sqrt{2} - 1}{2}

Correct answer

B. 2−1\sqrt{2} - 1

Step-by-step solution

2ae=12×2b2a2ae = \dfrac{1}{2} \times \dfrac{2{b}^{2}}{a}

2e=b2a22e = \dfrac{{b}^{2}}{{a}^{2}}

We know that

e=1−b2a2e = \sqrt{1 - \dfrac{{b}^{2}}{{a}^{2}}}

e=1−2ee = \sqrt{1 - 2e}

e2=1−2e{e}^{2} = 1 - 2e

e2+2e−1=0{e}^{2} + 2e - 1 = 0

−2±4+42=−2±222\dfrac{- 2 \pm \sqrt{4 + 4}}{2} = \dfrac{- 2 \pm 2\sqrt{2}}{2}

−1+2 and −1−2- 1 + \sqrt{2} \text{ and } - 1 - \sqrt{2}

∴e=2−1∴ e = \sqrt{2} - 1
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