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JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    The radius of the circle passing through the foci of the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\] and having its centre at\[(0,\,\,3)\] is:

    A) \[4\]                                     

    B) \[3\]

    C) \[\sqrt{12}\]                                      

    D) \[\frac{7}{2}\]

    Correct Answer: A

    Solution :

    We have,             \[{{a}^{2}}=16\] and                          \[9={{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] \[\Rightarrow \]               \[9=16(1-{{e}^{2}})\] \[\therefore \]  \[e=\frac{\sqrt{7}}{4}\] Thus, the foci are\[(\pm \sqrt{7},\,\,0)\]. The radius of required circle                 \[=\sqrt{{{(\sqrt{7}-0)}^{2}}+{{3}^{2}}}\]                 \[=\sqrt{7+9}\]                 \[=4\]


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