Verified Solution • 2025–26 Session moderate

QJEEM20Ma2f04: JEE Main 2020 Mathematics - Class 11 JEEM20Ma2f04 Solution: A hyperbola having the transverse axis of length has the same…

A hyperbola having the transverse axis of length 2\sqrt 2 has the same foci as that of the ellipse 3x2 + 4y2 = 12, then this hyperbola does not pass through which of the following points?

Multiple Choice Options

  • A (1,−12)\left( {1, - {1 \over {\sqrt 2 }}} \right)
  • B (32,12)\left( {\sqrt {{3 \over 2}} ,{1 \over {\sqrt 2 }}} \right)
  • C (−32,1)\left( { - \sqrt {{3 \over 2}} ,1} \right)
  • D (12,0)\left( {{1 \over {\sqrt 2 }},0} \right)

Step-by-Step Verified Solution

Ellipse : x24+y23=1{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1

eccentricity = 1−34=12\sqrt {1 - {3 \over 4}} = {1 \over 2}

∴\therefore foci = ( ±\pm 1, 0)

for hyperbola, given 2a=2⇒a=122a = \sqrt 2 \Rightarrow a = {1 \over {\sqrt 2 }}

∴\therefore hyperbola will be

x21/2−y2b2=1{{{x^2}} \over {1/2}} - {{{y^2}} \over {{b^2}}} = 1

eccentricity = 1+2b2\sqrt {1 + 2{b^2}}

∴\therefore foci =(±1+2b22,0)= \left( { \pm \sqrt {{{1 + 2{b^2}} \over 2}} ,0} \right)

∵\because Ellipse and hyperbola have same foci

⇒1+2b22=1\Rightarrow \sqrt {{{1 + 2{b^2}} \over 2}} = 1

⇒b2=12\Rightarrow {b^2} = {1 \over 2}

∴\therefore Equation of hyperbola : x21/2−y21/2=1{{{x^2}} \over {1/2}} - {{{y^2}} \over {1/2}} = 1

⇒\Rightarrow x2−y2=12{x^2} - {y^2} = {1 \over 2}

Clearly (32,12)\left( {\sqrt {{3 \over 2}} ,{1 \over {\sqrt 2 }}} \right) does not lie on it.

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring