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Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    The radius of the circle passing through the foci of the ellipse \[{{T}^{2}}{{\left[ \frac{\delta (G/T)}{\delta T} \right]}_{v}}\] and having its centre at (0,3), is

    A)  4     

    B) 3     

    C) \[-{{T}^{2}}{{\left[ \frac{\delta (G/T)}{\delta T} \right]}_{v}}\]                                    

    D) \[C{{l}_{2}}O,IC{{l}^{-}}_{2}\]

    Correct Answer: A

    Solution :

    We have,\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\] The eccentricity e of the ellipse is g^ea by \[e=\sqrt{1-\frac{9}{16}}+\frac{\sqrt{7}}{4}\] So, the coordinates of the foci are \[(\pm \sqrt{7},0).\] \[\therefore \]Radius of the circle \[=\sqrt{{{(7-0)}^{2}}+{{(0-3)}^{2}}}\] \[=\sqrt{7+9}=\sqrt{16}=4\]                                                


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