Q32 / 40 · NTA Abhyas (JEE Main) 2020 · Mathematics · Hyperbola

A hyperbola having the transverse axis of length 2 units has the same focii as that of ellipse 3x2+4y2=12, then its equation is

  • Correct answer
    A
    2x2-2y2=1
  • B
    2x2-2y2=3
  • C
    x2-y2=-2
  • D
    x2-y2=2

Solution

For the hyperbola x2a2-y2b2=1, length of semi transverse axis a=12,

For the ellipse x24+y23=1, the foci is ±1,0

i.e. ±1=±a2+b2

⇒b2=1-a2=1-12=12 

Hence, the equation of the required hyperbola is x212-y212=1

Hence, the correct answer is option: 'A'.

More Hyperbola Questions