CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    The equation of a hyperbola whose asymptotes are \[3x\pm 5y=0\]and vertices are \[(\pm 5,0)\]is

    A) \[3{{x}^{2}}-5{{y}^{2}}=25\]       

    B) \[5{{x}^{2}}-3{{y}^{2}}=225\]

    C) \[25{{x}^{2}}-9{{y}^{2}}=225\]   

    D) \[9{{x}^{2}}-25{{y}^{2}}=225\]

    Correct Answer: D

    Solution :

    The given equation of asymptotes of hyperbola is \[3x\pm 5y=0\]           ...(i) and vertices   \[(\pm 5,0)\] Now, equation of hyperbola is \[(3x+5y)\,(3x-5y)=\lambda \]                 \[9{{x}^{2}}-25{{y}^{2}}=\lambda \]         ...(ii) Since, \[(\pm \,5,0)\] is the vertices of the hyperbola, then                 \[9{{(5)}^{2}}-0-\lambda \Rightarrow \lambda =225\] From Eq. (ii), \[9{{x}^{2}}-25{{y}^{2}}=225\]


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