VIT Engineering VIT Engineering Solved Paper-2008

  • question_answer
    The length of the parabola \[{{y}^{2}}=12x\] cut off by the latusrectum is

    A)  \[6\,[\sqrt{2}+\log (1+\sqrt{2})]\]

    B)  \[3\,[\sqrt{2}+\log (1+\sqrt{2})]\]

    C)  \[6\,[\sqrt{2}-\log (1+\sqrt{2})]\]

    D)  \[3\,[\sqrt{2}-\log (1+\sqrt{2})]\]

    Correct Answer: A

    Solution :

    Given equation of parabola is \[{{y}^{2}}=12x\] ?(i) and equation of latusrectum is \[x=3\] ?(ii) From Eqs. (i) and (ii), we get \[{{y}^{2}}=36\] \[\Rightarrow \] \[y=\pm \,6\] \[\therefore \]Coordinates of end points of a latusrecturm are (3, 6) and (3, - 6). \[\therefore \]Required length\[=2\int_{0}^{3}{\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}dx}\] \[=2\int_{0}^{3}{\sqrt{1+{{\left( \frac{6}{y} \right)}^{2}}}}=2\int_{0}^{3}{\sqrt{\frac{12x+36}{12x}}}dx\] \[=2\int_{\,0}^{\,3}{\frac{x+3}{\sqrt{{{x}^{3}}+3x}}dx}\] \[=2\left[ \sqrt{{{x}^{2}}+3x}+\frac{3}{2}\log \left| \left( x+\frac{3}{2} \right)+\sqrt{{{x}^{2}}+3x} \right| \right]_{0}^{3}\] \[=2\left[ 3\sqrt{2}+\frac{3}{2}\log \left( \frac{9}{2}+3\sqrt{2} \right)-\frac{3}{2}\log \left( \frac{3}{2} \right) \right]\] \[=2\,[3\sqrt{2}+3\log (\sqrt{2}+1)\] \[=6\,[\sqrt{2}+\log \,(1+\sqrt{2})]\]


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