Solved papers for JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

done JEE Main Paper (Held On 19 May 2012) Total Questions - 1

• question_answer1) If the foci of the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]coincide with the foci of the hyperbola \[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25},\]then \[{{b}^{2}}\] is equal to     JEE Main  Online Paper (Held On 19  May  2012)
  A)
  8                                             
  done clear
  B)
                         10
  done clear
  C)
                         7     
  done clear
  D)
                         9
  done clear
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