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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Ellipse

BITSAT 2013 · Mathematics · Ellipse

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

An ellipse has OBO B as semi minor axis, FF and F′F^{\prime} its focii and the angle FBF′F B F^{\prime} is a right angle. Then the eccentricity of the ellipse isAn ellipse has O B as semi minor axis, F and F^ its focii and the angle F B F^ is a right angle. Then the eccentricity

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

∵∠FBF′=90∘⇒FB2+F′B2=FF′2∴(a2e2+b2)2+(a2e2+b2)2=(2ae)2⇒2(a2e2+b2)=4a2e2⇒e2=b2a2…(1)\begin{aligned} & \because \angle F B F^{\prime}=90^{{}^{\circ}} \Rightarrow F B^{2}+F^{\prime} B^{2}=F F^{\prime 2} \\ \therefore &\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}+\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}=(2 a e)^{2} \\ & \Rightarrow 2\left(a^{2} e^{2}+b^{2}\right)=4 a^{2} e^{2} \Rightarrow e^{2}=\frac{b^{2}}{a^{2}} \ldots(1) \end{aligned} An ellipse has O B as semi minor axis, F and F^ its focii and the angle F B F^ is a right angle. Then the eccentricity
Also, e2=1−b2/a2=1−e2e^{2}=1-b^{2} / a^{2}=1-e^{2} (By using equation (i)) ⇒2e2=1⇒e=12\Rightarrow 2 e^{2}=1 \Rightarrow e=\frac{1}{\sqrt{2}}
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