CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    If the foci of the ellipse\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]are\[(0,\sqrt{7})\]and\[(0,-\sqrt{7}),\]then the foci of the ellipse\[\frac{{{x}^{2}}}{9+{{t}^{2}}}+\frac{{{y}^{2}}}{16+{{t}^{2}}}=1\,t\in R,\]

    A)  \[(0,\sqrt{7}),(0,-\sqrt{7})\]

    B)  \[(0,7),(0,7)\]

    C)  \[(0,\,2\sqrt{7}),\,(0,\,-2\sqrt{7})\]

    D)  \[(\sqrt{7},0),(-\sqrt{7},0)\]

    E)  \[(\sqrt{7},2\sqrt{7}),(\sqrt{7},-2\sqrt{7})\]

    Correct Answer: A

    Solution :

    Since, the foci of the ellipse\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]are \[(0,\sqrt{7})\]and \[(0,-\sqrt{7})\]. Also, the foci of the ellipse \[\frac{{{x}^{2}}}{9+{{t}^{2}}}+\frac{{{y}^{2}}}{16+{{t}^{2}}}=1\] is\[(0,\sqrt{7})\]and\[(0,-\sqrt{7})\].


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