The area of the region enclosed by the curves | vedclass

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(A) Given curves are $y = (x-2)^2$ and $y^2 = -8(x-2)$.Let $X = x-2$ and $Y = y$. Then the equations become $Y = X^2$ and $Y^2 = -8X$.These are standa

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$\frac{8}{3}$

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(A) Given curves are $y = (x-2)^2$ and $y^2 = -8(x-2)$.Let $X = x-2$ and $Y = y$. Then the equations become $Y = X^2$ and $Y^2 = -8X$.These are standard parabolas $Y = X^2$ and $Y^2 = 4aX$ where $4a = -8$,so $a = -2$ (magnitude $|a| = 2$).The area enclosed by parabolas $y^2 = 4ax$ and $x^2 = 4by$ is given by $\frac{16}{3} |a| |b|$.Here,$Y = X^2$ (i.e.,$X^2 = 1Y$,so $4b = 1 \implies b = \frac{1}{4}$) and $Y^2 = -8X$ (i.e.,$4a = -8 \implies a = -2$).Area $= \frac{16}{3} 	imes |\frac{1}{4}| 	imes |-2| = \frac{16}{3} 	imes \frac{1}{4} 	imes 2 = \frac{8}{3}$ square units.

The area of the region enclosed by the curves $y=x^2-4x+4$ and $y^2=16-8x$ is:

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