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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    If P is any point on the ellipse \[\frac{{{x}^{2}}}{36}+\frac{{{y}^{2}}}{16}=1,\] and S and S' are the foci, then \[PS+PS'\]is equal to

    A)  \[4\]                                    

    B)  \[8\]

    C)  \[10\]                                  

    D)  \[12\]

    Correct Answer: D

    Solution :

    Given, ellipse is \[\frac{{{x}^{2}}}{36}+\frac{{{y}^{2}}}{16}=1\] Here, \[{{a}^{2}}=36,\] \[{{b}^{2}}=16\] Since, \[a>b,\] so the sum of the focal distance of any point P on the ellipse is \[PS+PS'=2a\] \[\Rightarrow \]               \[PS+PS'=2\times 6\] \[\Rightarrow \]               \[PS+PS'=12\]


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