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JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    If the foci of the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] and the hyperbola \[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\] coincide, then the value of \[{{b}^{2}}\] is [MNR 1992; UPSEAT 2001; AIEEE 2003; Karnataka CET 2004; Kerala (Engg.) 2005]

    A)            1     

    B)            5

    C)            7     

    D)            9

    Correct Answer: C

    Solution :

                      Hyperbola is \[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\]            \[a=\sqrt{\frac{144}{25}},\,\,b=\sqrt{\frac{81}{25}},\,\,{{e}_{1}}=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4}\]                   Therefore, foci \[=(a{{e}_{1}},0)=\left( \frac{12}{5}.\frac{5}{4},0 \right)=(3,\,0)\]                   Therefore, focus of ellipse \[=(4e,0)\] i.e. \[(3,\,0)\]                    Þ \[{{x}^{2}}+\] Hence\[{{b}^{2}}=16\left( 1-\frac{9}{16} \right)=7\].


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