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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Ellipse

BITSAT 2013 · Mathematics · Ellipse

S and T are the foci of an ellipse and B is an end of the minor...

S\mathrm{S} and T\mathrm{T} are the foci of an ellipse and B\mathrm{B} is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse isS and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, then the

Options

  1. A14\frac{1}{4}
  2. B13\frac{1}{3}
  3. C12\frac{1}{2}
  4. D23\frac{2}{3}

Correct answer

C. 12\frac{1}{2}

Step-by-step solution

Let eq. of ellipse be x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, S\mathrm{S} is (−ae,0),T(-\mathrm{ae}, 0), \mathrm{T} is (ae,0)(\mathrm{ae}, 0) and B\mathrm{B} is (0,b)(0, \mathrm{ b}) S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, then the
⇒SB=(0+ae)2+\mathrmb2\Rightarrow \mathrm{SB}=\sqrt{(0+\mathrm{ae})^{2}+\mathrmb^{2}} Also SB2=ST2⇒4\mathrma2\mathrme2=\mathrma2\mathrme2+\mathrmb2⇒3a2e2=a2(1−e2)=a2−a2e2⇒4a2e2=a2⇒e2=14⇒e=12\mathrm{SB}^{2}=\mathrm{ST}^{2} \Rightarrow 4 \mathrma^{2} \mathrme^{2}=\mathrma^{2} \mathrme^{2}+\mathrmb^{2}\Rightarrow 3 a^{2} e^{2}=a^{2}\left(1-e^{2}\right)=a^{2}-a^{2} e^{2}\Rightarrow 4 a^{2} e^{2}=a^{2} \Rightarrow e^{2}=\frac{1}{4} \Rightarrow e=\frac{1}{2}
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