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JEE Main 20149 AprilOnline
Exam
JEE Main
Year
2014
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Ellipse

JEE Main 2014 · 9 April · Online · Mathematics · Ellipse

If OB is the semi-minor axis of an ellipse, F _ 1 and F _ 2 are its...

If OBOB is the semi-minor axis of an ellipse, F1{F}_{1} and F2{F}_{2} are its focii and the angle between F1B{F}_{1}B and F2B{F}_{2}B is a right angle, then the square of the eccentricity of the ellipse isIf OB is the semi-minor axis of an ellipse, F _ 1 and F _ 2 are its focii and the angle between F _ 1 B and F _ 2 B is

Options

  1. A14\dfrac{1}{4}
  2. B12\dfrac{1}{\sqrt{2}}
  3. C12\dfrac{1}{2}
  4. D122\dfrac{1}{2 \sqrt{2}}

Correct answer

C. 12\dfrac{1}{2}

Step-by-step solution

In standard ellipse x2a2+y2b2=1\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1

Semi-major axis OA=aOA = a

Semi-minor axis OB=bOB = b

Focus F1≡(−ae,0){F}_{1} ≡ \left(- ae, 0\right)

F2≡(ae,0){F}_{2} ≡ \left(ae, 0\right)

If OB is the semi-minor axis of an ellipse, F _ 1 and F _ 2 are its focii and the angle between F _ 1 B and F _ 2 B is

∠F1BF2=90°∠ {F}_{1}B{F}_{2} = 90 °

OF1=OF2⇒∠F1BO=∠F2BO=45°O{F}_{1} = O{F}_{2} ⇒ ∠ {F}_{1}BO = ∠ {F}_{2}BO = 45 °

⇒aeb=tan⁡45°=1⇒ \dfrac{ae}{b} = \tan 45 ° = 1

⇒b=ae⇒ b = ae

Also, b2=a2(1−e2){b}^{2} = {a}^{2}\left(1 - {e}^{2}\right)

⇒a2e2=a2(1−e2)⇒ {a}^{2}{e}^{2} = {a}^{2}\left(1 - {e}^{2}\right)

⇒2e2=1⇒ 2{e}^{2} = 1

⇒e2=12⇒ {e}^{2} = \dfrac{1}{2}

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