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JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
     The eccentricity of the hyperbola \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{25}=1\] is [Karnataka CET 2005]

    A)            3/4 

    B)            3/5

    C)            \[\sqrt{41}/4\]                          

    D)            \[\sqrt{41/5}\]

    Correct Answer: C

    Solution :

                      Equation of hyperbola is \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{25}=1\]                    Eccentricity is \[{{e}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}+1\] i.e., \[{{e}^{2}}=\frac{25}{16}+1\]                    Þ \[{{e}^{2}}=\frac{41}{16}\] Þ \[e=\frac{\sqrt{41}}{4}\].


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