The area of the region bounded by the parabol | vedclass

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(A) Given parabola: $(y-2)^{2} = x-1 \Rightarrow x = (y-2)^{2} + 1$.At ordinate $y=3$,$x = (3-2)^{2} + 1 = 2$. So,the point is $(2, 3)$.Differentiatin

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$9$

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(A) Given parabola: $(y-2)^{2} = x-1 \Rightarrow x = (y-2)^{2} + 1$.At ordinate $y=3$,$x = (3-2)^{2} + 1 = 2$. So,the point is $(2, 3)$.Differentiating $(y-2)^{2} = x-1$ with respect to $y$,we get $2(y-2) = \frac{dx}{dy}$.At $y=3$,$\frac{dx}{dy} = 2(3-2) = 2$.The equation of the tangent at $(2, 3)$ is $x - 2 = 2(y - 3) \Rightarrow x = 2y - 4$.The region is bounded by the parabola $x = (y-2)^{2} + 1$,the tangent $x = 2y - 4$,and the $x$-axis $(y=0)$.Integrating with respect to $y$ from $y=0$ to $y=3$:Area $= \int_{0}^{3} [((y-2)^{2} + 1) - (2y - 4)] dy$$= \int_{0}^{3} (y^{2} - 4y + 4 + 1 - 2y + 4) dy = \int_{0}^{3} (y^{2} - 6y + 9) dy$$= \int_{0}^{3} (y-3)^{2} dy = \left[ \frac{(y-3)^{3}}{3} ight]_{0}^{3} = 0 - (\frac{-27}{3}) = 9 	ext{ sq. units.}$

The area of the region bounded by the parabola $(y-2)^{2}=(x-1)$,the tangent to it at the point whose ordinate is $3$,and the $x$-axis is:

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