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BCECE Engineering BCECE Engineering Solved Paper-2009

  • question_answer
    If\[{{F}_{1}}=(3,0),{{F}_{2}}=(-3,0)\] and P is any point on the curve \[16{{x}^{2}}+25{{y}^{2}}=400,\] then \[P{{F}_{1}}+P{{F}_{2}}\]is equal to

    A)  6                            

    B)         8                            

    C)  10                         

    D)         12

    Correct Answer: C

    Solution :

    Given curve is \[16{{x}^{2}}+25{{y}^{2}}=400\] \[\Rightarrow \]\[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\] Here \[a=5,b=4\] \[{{F}_{1}}\]and \[{{F}_{2}}\]are forcus. \[\therefore \]\[P{{F}_{1}}+P{{F}_{2}}=2a=10\]


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