If the area enclosed between the curves y^2 = | vedclass

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(C) Step $1$: Find the points of intersection by substituting $y = kx$ into $y^2 = 4kx$.$(kx)^2 = 4kx \implies k^2x^2 - 4kx = 0 \implies kx(kx - 4) =

Vedclass · Accepted Answer

$4$

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(C) Step $1$: Find the points of intersection by substituting $y = kx$ into $y^2 = 4kx$.$(kx)^2 = 4kx \implies k^2x^2 - 4kx = 0 \implies kx(kx - 4) = 0$.So, $x = 0$ and $x = \frac{4}{k}$.Step $2$: The area $A$ is given by $\int_{0}^{4/k} (\sqrt{4kx} - kx) \, dx = \frac{2}{3}$.Step $3$: Integrate: $2\sqrt{k} \int_{0}^{4/k} x^{1/2} \, dx - k \int_{0}^{4/k} x \, dx = \frac{2}{3}$.$2\sqrt{k} [\frac{2}{3} x^{3/2}]_{0}^{4/k} - k [\frac{x^2}{2}]_{0}^{4/k} = \frac{2}{3}$.$2\sqrt{k} \cdot \frac{2}{3} \cdot (\frac{4}{k})^{3/2} - \frac{k}{2} \cdot (\frac{4}{k})^2 = \frac{2}{3}$.$\frac{4\sqrt{k}}{3} \cdot \frac{8}{k\sqrt{k}} - \frac{k}{2} \cdot \frac{16}{k^2} = \frac{2}{3}$.$\frac{32}{3k} - \frac{8}{k} = \frac{2}{3} \implies \frac{32 - 24}{3k} = \frac{2}{3} \implies \frac{8}{3k} = \frac{2}{3}$.Step $4$: Solving for $k$, $2k = 8 \implies k = 4$.

If the area enclosed between the curves $y^2 = 4kx$ and $y = kx$ for $k > 0$ is $\frac{2}{3}$ sq. units, then $k =$ ?

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