JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    The foci of \[16{{x}^{2}}+25{{y}^{2}}=400\] are          [BIT Ranchi 1996]

    A)            \[(\pm 3,\ 0)\]                          

    B)            \[(0,\ \pm 3)\]

    C)            \[(3,\ -3)\]                                

    D)            \[(-3,\ 3)\]

    Correct Answer: A

    Solution :

               The equation of the ellipse is \[16{{x}^{2}}+25{{y}^{2}}=400\] or\[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\].            Here\[{{a}^{2}}=25,{{b}^{2}}=16\Rightarrow e=\frac{3}{5}\].            Hence the foci are \[(\pm 3,0).\]


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