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  • Can someone help me with this, - Integral Calculus - JEE Main-14

If the area ( in sq. units) bounded by the parabola y^{2}=4\lambda x and 

the line y=\lambda x , \lambda>0 , is \frac{1}{9} , then \lambda is equal to :  

  • Option 1)

    2\sqrt6

  • Option 2)

    48

  • Option 3)

    24

  • Option 4)

    4\sqrt3

 

Answers (1)

best_answer

the parabola y^{2}=4\lambda x and  the line y=\lambda x

If \lambda>0, then

Area = \int_{0}^{4/\lambda}(2\sqrt\lambda\sqrt x-\lambda x)dx=\frac{1}{9}

        => [\frac{2\sqrt \lambda x^{\frac{3}{2}}}{3/2}-\frac{\lambda x^{2}}{2}]^{4/\lambda}_{0}=\frac{1}{9}

       => \frac{4}{3}\sqrt\lambda \frac{8}{\lambda ^{\frac{3}{2}}}-\lambda \frac{8}{\lambda ^{2}}=\frac{1}{9}

      => \frac{32}{3\lambda }-\frac{8}{\lambda }=\frac{1}{9}

       => \frac{8}{3\lambda }=\frac{1}{9} 

       => \lambda=24

    


Option 1)

2\sqrt6

Option 2)

48

Option 3)

24

Option 4)

4\sqrt3

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