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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Hyperbola

NTA Abhyas 2020 · Mathematics · Hyperbola

A hyperbola having the transverse axis of length 2 units has the...

A hyperbola having the transverse axis of length 2unitshasthesamefociiasthatofellipse3x2+4y2=12\sqrt{2} units has the same focii as that of ellipse 3x^{2} + 4y^{2} = 12, then its equation is

Options

  1. A2x2−2y2=12x^{2} - 2y^{2} = 1
  2. B2x2−2y2=32x^{2} - 2y^{2} = 3
  3. Cx2−y2=−2x^{2} - y^{2} = - 2
  4. Dx2−y2=2x^{2} - y^{2} = 2

Correct answer

A. 2x2−2y2=12x^{2} - 2y^{2} = 1

Step-by-step solution

For the hyperbola& x2a2−y2b2=1,\dfrac{x2}{a2} - \dfrac{y2}{b2} = 1,~length of semi transverse axis& a=12,a = \dfrac{1}{2},

For the ellipse~x24+y23=1,\dfrac{x2}{4} + \dfrac{y2}{3} = 1,~the foci is& \left(\pm 1,0\right)<br/><br/>i.e. <br /><br />i.e.~\pm 1 = \pm \sqrt{a^{2} + b^{2}}<br/><br/><br /><br />⇒ b^{2} = 1 - a^{2} = 1 - \dfrac{1}{2} = \dfrac{1}{2}</p><p>Hence,theequationoftherequiredhyperbolais </p><p>Hence, the equation of the required hyperbola is~\dfrac{x2}{\dfrac{1}{2}} - \dfrac{y2}{\dfrac{1}{2}} = 1$

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