The area enclosed by y ^ 2 = 8x and y = 2 x that lies outside the...

Intersection point of ${y}^{2} = 8x \text{ and } y = \sqrt{2}x$ is \left(0, 0\right) \text{ and } \left(2, 4\sqrt{2}\right)$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/a48a9bc1-bf0f-4448-8c8d-66154c4f9153-image.png" style="width: 297px; height: 229px;" /></p><p>Area of$∆ ABC = \dfrac{1}{2}\left(\sqrt{2}\right) \cdot 1 = \dfrac{1}{\sqrt{2}}$</p><p>So required area=area between the curve${y}^{2} = 8x$and the line$y = \sqrt{2}x$$- ar △ ABC$

$= \left[{∫}_{0}^{4}\left(\sqrt{8x} - \sqrt{2}x\right)dx\right] - \dfrac{1}{\sqrt{2}}$

$= \dfrac{32\sqrt{2}}{3} - 8\sqrt{2} - \dfrac{1}{\sqrt{2}} = \dfrac{13\sqrt{2}}{6}