CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    The length of transverse axis of the hyperbola \[3{{x}^{2}}-4{{y}^{2}}=32\] is:

    A)  \[\frac{8\sqrt{2}}{\sqrt{3}}\]                     

    B)  \[\frac{16\sqrt{2}}{\sqrt{3}}\]

    C)  \[\frac{64}{3}\]

    D)  \[\frac{3}{32}\]

    Correct Answer: A

    Solution :

    Equation of hyperbola \[3{{x}^{2}}-4{{y}^{2}}=32\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{\frac{32}{3}}-\frac{{{y}^{2}}}{\frac{32}{8}}=1\Rightarrow \frac{{{x}^{2}}}{\left( \frac{4\sqrt{2}}{\sqrt{3}} \right)}-\frac{{{y}^{2}}}{{{(2)}^{2}}}=1\] \[a=\frac{4\sqrt{2}}{\sqrt{3}},\]therefore length of transverse axis of a hyperbola \[=2a=\frac{2\times 4\sqrt{2}}{\sqrt{3}}=\frac{8\sqrt{2}}{\sqrt{3}}\]


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