The area of the region bounded by the parabol | vedclass

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(C) Given the parabola $y^{2}=8x$. Comparing this with $y^{2}=4ax$,we get $4a=8$,so $a=2$.The latus rectum is the line $x=a$,which is $x=2$.The points

Vedclass · Accepted Answer

$\frac{32}{3}$ sq. units

Vedclass · Answer

(C) Given the parabola $y^{2}=8x$. Comparing this with $y^{2}=4ax$,we get $4a=8$,so $a=2$.The latus rectum is the line $x=a$,which is $x=2$.The points of intersection of the parabola and the latus rectum are $(2, 4)$ and $(2, -4)$.The area $A$ bounded by the parabola and the latus rectum is symmetric about the $x$-axis.$A = 2 \int_{0}^{2} y \, dx = 2 \int_{0}^{2} \sqrt{8x} \, dx$$A = 2 	imes 2\sqrt{2} \int_{0}^{2} x^{1/2} \, dx = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{2}$$A = 4\sqrt{2} 	imes \frac{2}{3} 	imes (2)^{3/2} = \frac{8\sqrt{2}}{3} 	imes 2\sqrt{2} = \frac{8 	imes 2 	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

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