The area of the region bounded by the parabol | vedclass

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(A) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,which implies $a = 2$.The focus of the parabola is $(a

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$\frac{32}{3}$

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(A) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,which implies $a = 2$.The focus of the parabola is $(a, 0) = (2, 0)$.The equation of the latus rectum is $x = a = 2$.The parabola is symmetric about the $x$-axis.The area of the region bounded by the parabola and its latus rectum is given by $A = 2 \int_{0}^{2} y \, dx$.Since $y^2 = 8x$,we have $y = \sqrt{8x} = 2\sqrt{2} \sqrt{x}$.Substituting this into the integral,we get $A = 2 \int_{0}^{2} 2\sqrt{2} \sqrt{x} \, dx = 4\sqrt{2} \int_{0}^{2} x^{1/2} \, dx$.Evaluating the integral: $A = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{2} = 4\sqrt{2} 	imes \frac{2}{3} 	imes (2)^{3/2} = \frac{8\sqrt{2}}{3} 	imes 2\sqrt{2} = \frac{16 	imes 2}{3} = \frac{32}{3}$ sq. units.

The area of the region bounded by the parabola $y^2 = 8x$ and its latus rectum is . . . . . . sq. units.

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