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JEE Main 2005
Exam
JEE Main
Year
2005
Subject
Mathematics
Chapter
Ellipse

JEE Main 2005 · Mathematics · Ellipse

An ellipse has OB as semi minor axis, F and F^ its focii and the...

An ellipse has OB\mathrm{OB} as semi minor axis, F\mathrm{F} and F′\mathrm{F}^{\prime} its focii and the angle FBF' is a right angle. Then the eccentricity of the ellipse isAn ellipse has OB as semi minor axis, F and F^ its focii and the angle FBF' is a right angle. Then the eccentricity ofAn ellipse has OB as semi minor axis, F and F^ its focii and the angle FBF' is a right angle. Then the eccentricity of

Options

  1. A12\frac{1}{\sqrt{2}}
  2. B12\frac{1}{2}
  3. C14\frac{1}{4}
  4. D13\frac{1}{\sqrt{3}}

Correct answer

A. 12\frac{1}{\sqrt{2}}

Step-by-step solution

An ellipse has OB as semi minor axis, F and F^ its focii and the angle FBF' is a right angle. Then the eccentricity of
∵∠ FBF ′=90∘∴(a2e2+b2)2+(a2e2+b2)2=(2ae)2⇒2(a2e2+b2)=4a2e2⇒e2=b2/a2 Also e2=1−b2/a2=1−e2⇒2e2=1,e=12.\begin{aligned}& \because \angle \text{ FBF }^{\prime}=90^{{}^{\circ}} \\ & \therefore \left(\sqrt{a^2 e^2+b^2}\right)^2+\left(\sqrt{a^2 e^2+b^2}\right)^2=(2 a \mathrm{e})^2 \\ & \Rightarrow 2 \left(a^2 e²+b²\right)=4 a² e² \\ & \Rightarrow \mathrm{e}^2=\mathrm{b}^2 / \mathrm{a}^2 \\ & \text{ Also } \mathrm{e}^2=1-\mathrm{b}^2 / \mathrm{a}^2=1-\mathrm{e}^2 \\ & \Rightarrow 2 \mathrm{e}^2=1, \mathrm{e}=\frac{1}{\sqrt{2}} . \end{aligned}
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