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BCECE Engineering BCECE Engineering Solved Paper-2015

  • 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 \[(0,3),\] is

    A)                  4                            

    B) 3      

    C) \[\sqrt{12}\]                      

    D) 7/2

    Correct Answer: A

    Solution :

    Let e be the eccentricity of the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\] Then,    \[e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\] So, coordinates of the foci of the ellipse are \[(\pm \sqrt{7,0})\]. The radius of the circle having centre at (0,3) and passing through foci \[=\sqrt{{{(0-\sqrt{7})}^{2}}+{{(3-0)}^{2}}}=4\]


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