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BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    The sum of the focal distances of any point on the conic \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\]is

    A)  10                         

    B)  9                            

    C)  41                         

    D)  18

    Correct Answer: A

    Solution :

    Comparing the given equation, \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\] with general equation of ellipse                 \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\]we get                                 \[{{a}^{2}}=25\] \[\Rightarrow \]               \[a=5\] We know that, the sum of focal distances \[=2a\]                                 \[=2\times 5\] \[=10\]


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