BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer The area bounded by the parabolas \[{{y}^{2}}=4a(x+a)\]and \[{{y}^{2}}=-4a(x-a)\]is

    A)  \[\frac{16}{3}{{a}^{2}}\text{sq units}\]

    B)         \[\frac{8}{3}\text{sq}\,\text{units}\]    

    C)         \[\frac{\text{4}}{\text{3}}{{\text{a}}^{\text{2}}}\text{sq}\,\text{units}\]

    D)         None of these

    Correct Answer: A

    Solution :

    The focus of the parabolas \[{{y}^{2}}=4a(x+a)\]and\[{{y}^{2}}=-4a(x-a)\]  are  \[(-a,0)\]   and   \[(a,0)\] respectively. The required area is shown in the figure which is given by as following.                 Area of \[ABCD=2.\]area of \[AOCB\]                                                 \[=4.\]area of AOB                                                 \[=4.\int_{-a}^{0}{\sqrt{4a(x+a)}}dx\]                                                 \[=4.2\sqrt{a}\int_{-a}^{0}{\sqrt{x+a}}\,dx\]                                                 \[=8\sqrt{a}\left[ \frac{2}{3}{{(x+a)}^{3/2}} \right]_{-a}^{0}\]                 \[\Rightarrow \]Area of \[ABCD=8\sqrt{a}.\frac{2}{3}({{a}^{3/2}}-0)\]                                                 \[=\frac{16}{3}\sqrt{a}.{{a}^{3/2}}\]                                                        \[=\frac{16}{3}{{a}^{2}}\]                 which is the required area.

adversite


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