The area of the figure bounded by the parabol | vedclass

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(B) The given equations of the parabolas are:$y^2 = -8(x-2) \quad \dots(1)$$y^2 = 24(x+2) \quad \dots(2)$To find the intersection points, equate the e

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$\frac{32}{3} \sqrt{6} 	ext{ sq. unit}$

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(B) The given equations of the parabolas are:$y^2 = -8(x-2) \quad \dots(1)$$y^2 = 24(x+2) \quad \dots(2)$To find the intersection points, equate the expressions for $y^2$:$-8(x-2) = 24(x+2)$$-8x + 16 = 24x + 48$$-32x = 32 \implies x = -1$At $x = -1$, $y^2 = 24(-1+2) = 24$, so $y = \pm \sqrt{24} = \pm 2\sqrt{6}$.The area is symmetric about the $x$-axis, so the total area is twice the area above the $x$-axis:$	ext{Area} = 2 \left[ \int_{-2}^{-1} \sqrt{24(x+2)} \, dx + \int_{-1}^{2} \sqrt{-8(x-2)} \, dx ight]$$	ext{Area} = 2 \left[ 2\sqrt{6} \int_{-2}^{-1} \sqrt{x+2} \, dx + 2\sqrt{2} \int_{-1}^{2} \sqrt{2-x} \, dx ight]$$	ext{Area} = 4 \left[ \sqrt{6} \left( \frac{2}{3}(x+2)^{3/2} ight)_{-2}^{-1} + \sqrt{2} \left( -\frac{2}{3}(2-x)^{3/2} ight)_{-1}^{2} ight]$$	ext{Area} = 4 \left[ \sqrt{6} \left( \frac{2}{3}(1) - 0 ight) + \sqrt{2} \left( 0 - (-\frac{2}{3}(3)^{3/2}) ight) ight]$$	ext{Area} = 4 \left[ \frac{2\sqrt{6}}{3} + \frac{2\sqrt{2}}{3} \cdot 3\sqrt{3} ight] = 4 \left[ \frac{2\sqrt{6}}{3} + 2\sqrt{6} ight] = 4 \left[ \frac{8\sqrt{6}}{3} ight] = \frac{32}{3} \sqrt{6} 	ext{ sq. unit}$.

The area of the figure bounded by the parabolas $y^2+8x=16$ and $y^2-24x=48$ is

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