CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    The area cut off by the latus rectum from the parabola\[{{y}^{2}}=4ax\]is:

    A)  \[(8/3)\text{ }a\text{ }sq\text{ }unit\]                  

    B)  \[(8/3)\,\sqrt{a}\,sq\,\] unit

    C)                         \[(3/8)\text{ }{{a}^{2}}\text{ }sq\text{ }unit\]   

    D)  \[(8/3)\text{ }{{a}^{3}}sq\text{ }unit\]

    E)  \[(8/3)\text{ }{{a}^{2}}\text{ }sq\text{ }unit\]

    Correct Answer: E

    Solution :

    Required area\[=2\int_{0}^{a}{2\sqrt{a}}\sqrt{x}dx\] \[=2\times 2\sqrt{a}\left( \frac{{{x}^{3/2}}}{3/2} \right)_{0}^{a}\] \[=\frac{8}{3}{{a}^{2}}\]sq unit


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