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(B) The area $A$ bounded by the parabola $y^2 = 4kx$ and the line $x = k$ is given by $A = 2 \int_{0}^{k} \sqrt{4kx} \, dx$.$A = 2 \cdot 2\sqrt{k} \in

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$\frac{\pi}{6}$

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(B) The area $A$ bounded by the parabola $y^2 = 4kx$ and the line $x = k$ is given by $A = 2 \int_{0}^{k} \sqrt{4kx} \, dx$.$A = 2 \cdot 2\sqrt{k} \int_{0}^{k} x^{1/2} \, dx = 4\sqrt{k} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{k} = 4\sqrt{k} \cdot \frac{2}{3} k^{3/2} = \frac{8}{3} k^2$.Given $A = \frac{128}{3}$, so $\frac{8}{3} k^2 = \frac{128}{3} \implies k^2 = 16 \implies k = 4$ (since $k > 0$).Now, $\sin^{-1} \left( \frac{2}{k} \right) = \sin^{-1} \left( \frac{2}{4} \right) = \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6}$.

If the area of the region bounded by the parabola $y^2 = 4kx$ and the line $x = k$, (where $k > 0$) is $\frac{128}{3}$ sq. units, then the value of $\sin^{-1} \left( \frac{2}{k} \right)$ is equal to...

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