The area of the region enclosed by the parabo | vedclass

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(A) The given equations are $(y-2)^2 = x-1$ and $x = 2y-4$.To find the intersection point,substitute $x$ from the line equation into the parabola equa

Vedclass · Accepted Answer

$5$

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(A) The given equations are $(y-2)^2 = x-1$ and $x = 2y-4$.To find the intersection point,substitute $x$ from the line equation into the parabola equation:$(y-2)^2 = (2y-4)-1$$(y-2)^2 = 2(y-2)-1$Let $u = y-2$,then $u^2 = 2u-1$,which gives $u^2-2u+1 = 0$,so $(u-1)^2 = 0$,implying $u=1$.Thus,$y-2 = 1$,so $y=3$. Then $x = 2(3)-4 = 2$.The intersection point is $(2, 3)$.The region is bounded by the $y$-axis $(x=0)$,the $x$-axis $(y=0)$,the line $x = 2y-4$,and the parabola $x = (y-2)^2+1$.Looking at the region,the area is the integral of the parabola from $y=0$ to $y=3$,minus the area of the triangle formed by the line and the axes.Area $= \int_0^3 ((y-2)^2+1) dy - 	ext{Area of triangle with vertices } (0,0), (0,2), (2,3) 	ext{ is not correct, let's re-evaluate.}$Actually,the region is bounded by $x=0$,$y=0$,the parabola,and the line. The area is $\int_0^3 x_{parabola} dy - 	ext{Area of triangle}$.Area $= \int_0^3 ((y-2)^2+1) dy - 	ext{Area of triangle bounded by } x=2y-4, x=0, y=0, y=3$.The line $x=2y-4$ intersects $y$-axis at $(0,2)$ and $x$-axis at $(4,0)$.The area is $\int_0^3 ((y-2)^2+1) dy - 	ext{Area of triangle formed by } (0,2), (2,3), (0,3) = \int_0^3 (y^2-4y+5) dy - \frac{1}{2} 	imes 2 	imes 1 = [\frac{y^3}{3}-2y^2+5y]_0^3 - 1 = (9-18+15) - 1 = 6-1 = 5$.

The area of the region enclosed by the parabola $(y-2)^2=x-1$,the line $x-2y+4=0$,and the positive coordinate axes is

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