If the area (in sq. units) bounded by the par | vedclass

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(D) Given the equations of the parabola $y^2 = 4\lambda x$ and the line $y = \lambda x$.To find the points of intersection,substitute $y = \lambda x$

Vedclass · Accepted Answer

$24$

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(D) Given the equations of the parabola $y^2 = 4\lambda x$ and the line $y = \lambda x$.To find the points of intersection,substitute $y = \lambda x$ into the parabola equation:$(\lambda x)^2 = 4\lambda x$$\lambda^2 x^2 - 4\lambda x = 0$$\lambda x(\lambda x - 4) = 0$Since $\lambda > 0$,we have $x = 0$ and $x = \frac{4}{\lambda}$.The area bounded by the curves is given by:$A = \int_{0}^{4/\lambda} (\sqrt{4\lambda x} - \lambda x) dx = \frac{1}{9}$$A = 2\sqrt{\lambda} \int_{0}^{4/\lambda} \sqrt{x} dx - \lambda \int_{0}^{4/\lambda} x dx$$A = 2\sqrt{\lambda} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4/\lambda} - \lambda \left[ \frac{x^2}{2} \right]_{0}^{4/\lambda}$$A = \frac{4}{3} \sqrt{\lambda} \left( \frac{4}{\lambda} \right)^{3/2} - \frac{\lambda}{2} \left( \frac{4}{\lambda} \right)^2$$A = \frac{4}{3} \sqrt{\lambda} \cdot \frac{8}{\lambda \sqrt{\lambda}} - \frac{\lambda}{2} \cdot \frac{16}{\lambda^2}$$A = \frac{32}{3\lambda} - \frac{8}{\lambda} = \frac{32 - 24}{3\lambda} = \frac{8}{3\lambda}$Given $A = \frac{1}{9}$,we have:$\frac{8}{3\lambda} = \frac{1}{9}$$3\lambda = 72$$\lambda = 24$

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

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