CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    The point on the hyperbola \[3{{x}^{2}}-4{{y}^{2}}=72\] which is nearest to the line\[3x+2y+1=0\]is

    A)  \[(-\text{ }6,3)\]        

    B)  (6, 3)

    C)  \[(-\text{ }6,-3)\]         

    D)         \[(\text{ }6,-3)\]

    E)  \[(\sqrt{24},0)\]

    Correct Answer: A

    Solution :

                    First we check which point satisfy the equation of hyperbola. All points in options are satisfied the equation of hyperbola\[3{{x}^{2}}-4{{y}^{2}}=72\]. Now, we find one-by-one the length of perpendicular from point on Ellipse to the line \[3x+2y+1=0\].                 \[{{p}_{(-6,3)}}=\frac{11}{\sqrt{13}}\]                 \[{{p}_{(6,3)}}=\frac{25}{\sqrt{13}}\]                 \[{{p}_{(-6,-3)}}=\frac{23}{\sqrt{13}}\]                 \[{{p}_{(6,-3)}}=\frac{13}{\sqrt{13}}\]                 \[{{p}_{(\sqrt{24},0)}}=\frac{3\sqrt{24}+1}{\sqrt{13}}\] The minimum length is\[{{p}_{(-6,3)}}\]. So, the point \[(-6,3)\] is nearest to the given line.


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