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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    If the distance between the foci and the distance between the directories of the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] are in the ratio \[3:2\], then \[a:b\] is :

    A)  \[\sqrt{2}:1\]                   

    B)  \[\sqrt{3}:\sqrt{2}\]

    C)  \[1:2\]                                 

    D)  \[2:1\]

    Correct Answer: A

    Solution :

    Equation of hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Distance between foci \[=2ae\] and distance between directories \[=\frac{2a}{e}\] According to the question                 \[\frac{2ae}{2a/e}=\frac{3}{2}\] \[\Rightarrow \]               \[{{e}^{2}}=\frac{3}{2}\] We know,                 \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)\] \[\Rightarrow \]               \[\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{3}{2}-1=\frac{1}{2}\] \[\Rightarrow \]               \[\frac{b}{a}=\frac{1}{\sqrt{2}}\] \[\therefore \] Required ratio \[a:b\] is  \[\sqrt{2}:1,\]


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