The area enclosed by y^{2}=8x and y=sqrt{2}x | vedclass

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(C) First,find the intersection points of $y^{2}=8x$ and $y=\sqrt{2}x$:$(\sqrt{2}x)^{2}=8x \implies 2x^{2}=8x \implies 2x(x-4)=0$.So,$x=0$ and $x=4$.

Vedclass · Accepted Answer

$\frac{13\sqrt{2}}{6}$

Vedclass · Answer

(C) First,find the intersection points of $y^{2}=8x$ and $y=\sqrt{2}x$:$(\sqrt{2}x)^{2}=8x \implies 2x^{2}=8x \implies 2x(x-4)=0$.So,$x=0$ and $x=4$. The intersection points are $(0,0)$ and $(4,4\sqrt{2})$.The total area enclosed by the parabola $y^{2}=8x$ and the line $y=\sqrt{2}x$ is:$A_{total} = \int_{0}^{4} (\sqrt{8x} - \sqrt{2}x) dx = \int_{0}^{4} (2\sqrt{2}\sqrt{x} - \sqrt{2}x) dx$$= 2\sqrt{2} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{4} - \sqrt{2} \left[ \frac{x^{2}}{2} ight]_{0}^{4} = 2\sqrt{2} \cdot \frac{2}{3} \cdot 8 - \sqrt{2} \cdot \frac{16}{2} = \frac{32\sqrt{2}}{3} - 8\sqrt{2} = \frac{8\sqrt{2}}{3}$.The triangle is formed by $y=\sqrt{2}x$,$x=1$,and $y=2\sqrt{2}$.At $x=1$,the line $y=\sqrt{2}x$ gives $y=\sqrt{2}$. The point is $(1, \sqrt{2})$.The horizontal line $y=2\sqrt{2}$ intersects $y=\sqrt{2}x$ at $2\sqrt{2}=\sqrt{2}x \implies x=2$.The vertical line $x=1$ intersects $y=2\sqrt{2}$ at $(1, 2\sqrt{2})$.The vertices of the triangle are $(1, \sqrt{2})$,$(2, 2\sqrt{2})$,and $(1, 2\sqrt{2})$.Area of the triangle $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2-1) 	imes (2\sqrt{2}-\sqrt{2}) = \frac{1}{2} 	imes 1 	imes \sqrt{2} = \frac{\sqrt{2}}{2}$.The required area is the total area minus the area of the triangle:$Area = \frac{8\sqrt{2}}{3} - \frac{\sqrt{2}}{2} = \frac{16\sqrt{2} - 3\sqrt{2}}{6} = \frac{13\sqrt{2}}{6}$.

The area enclosed by $y^{2}=8x$ and $y=\sqrt{2}x$ that lies outside the triangle formed by $y=\sqrt{2}x$,$x=1$,and $y=2\sqrt{2}$ is equal to

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