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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2024 · 30 January · Morning Shift · Mathematics · Ellipse

If the length of the minor axis of ellipse is equal to half of the...

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

Options

  1. A53\dfrac{\sqrt{5}}{3}
  2. B32\dfrac{\sqrt{3}}{2}
  3. C13\dfrac{1}{\sqrt{3}}
  4. D25\dfrac{2}{\sqrt{5}}

Correct answer

D. 25\dfrac{2}{\sqrt{5}}

Step-by-step solution

The length of minor axis is given by (2b) andthedistancebetweenfociisgivenby (2ae)\left(2b\right) and the distance between foci is given by \left(2ae\right)

Given: 2b=12×2ae2b = \dfrac{1}{2} \times 2ae

⇒2b=ae⇒ 2b = ae

⇒4b2=a2e2⇒ 4{b}^{2} = {a}^{2}{e}^{2}

⇒4b2a2=e2⇒ \dfrac{4{b}^{2}}{{a}^{2}} = {e}^{2}

⇒4(1−e2)=e2⇒ 4\left(1 - {e}^{2}\right) = {e}^{2}

⇒4−4e2=e2⇒ 4 - 4{e}^{2} = {e}^{2}

⇒e=25⇒ e = \dfrac{2}{\sqrt{5}}

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