The area bounded by the parabola y^{2}=16x an | vedclass

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(B) The given parabola is $y^{2}=16x$. Comparing this with $y^{2}=4ax$,we get $4a=16$,which implies $a=4$.The latus rectum is the line $x=a$,so $x=4$.

Vedclass · Accepted Answer

$\frac{64}{3}$ sq. units

Vedclass · Answer

(B) The given parabola is $y^{2}=16x$. Comparing this with $y^{2}=4ax$,we get $4a=16$,which implies $a=4$.The latus rectum is the line $x=a$,so $x=4$.The points of intersection of the parabola and the latus rectum are $(4, 8)$ and $(4, -8)$.In the first quadrant,the area is bounded by the curve $y=\sqrt{16x}=4\sqrt{x}$,the $x$-axis,and the line $x=4$.The required area is given by the integral:$	ext{Area} = \int_{0}^{4} y \, dx = \int_{0}^{4} 4\sqrt{x} \, dx$$= 4 \int_{0}^{4} x^{\frac{1}{2}} \, dx$$= 4 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} ight]_{0}^{4}$$= 4 	imes \frac{2}{3} \left[ x^{\frac{3}{2}} ight]_{0}^{4}$$= \frac{8}{3} \left[ 4^{\frac{3}{2}} - 0^{\frac{3}{2}} ight]$$= \frac{8}{3} 	imes 8 = \frac{64}{3} 	ext{ sq. units}$.

The area bounded by the parabola $y^{2}=16x$ and its latus rectum in the first quadrant is

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