The area (in square units) of the region bounded by the parabola y...

Given curve is ${y}^{2} = 4\left(x - 2\right) and line$y = 2x - 8$</p><p>Finding the point of intersection of$y = 2x - 4$and${y}^{2} = 4x$.</p><p>$⇒ x = \dfrac{y + 8}{2} = \dfrac{{y}^{2} + 8}{4}$</p><p>$⇒ 2y + 16 = {y}^{2} + 8$</p><p>$⇒ {y}^{2} - 2y - 8 = 0$</p><p>$⇒ \left(y - 4\right)\left(y + 2\right) = 0$</p><p>$⇒ y = 4, - 2$</p><p>$⇒ x = 6, 3$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/4bcc8f87-2e3b-4b16-8c89-38867954f43b.png" style="width: 400px; height: 256px;"/></p><p>Area of the bounded region is given by,$A = {∫}_{- 2}^{4}\left(\dfrac{y + 8}{2} - \left(\dfrac{{y}^{2}}{4} + 2\right)\right)dy$</p><p>$⇒ A = {\left[\dfrac{{y}^{2}}{4} + 2y - \dfrac{{y}^{3}}{12}\right]}_{- 2}^{4}$</p><p>$⇒ A = 4 + 8 - \dfrac{16}{3} - 1 + 4 - \dfrac{2}{3}$</p><p>$⇒ A = 9sq . units$.