Find the area of the parabola y^{2}=4ax bound | vedclass

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(A) The vertex of the parabola $y^{2}=4ax$ is at the origin $(0,0)$.The equation of the latus rectum $LSL^{\prime}$ is $x=a$.The parabola is symmetric

Vedclass · Accepted Answer

$\frac{8}{3}a^{2}$

Vedclass · Answer

(A) The vertex of the parabola $y^{2}=4ax$ is at the origin $(0,0)$.The equation of the latus rectum $LSL^{\prime}$ is $x=a$.The parabola is symmetrical about the $x$-axis.The required area of the region $OLL^{\prime}O$ is given by:Area $= 2 	imes (	ext{Area of the region } OLSO)$Area $= 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{a} \sqrt{4ax} \, dx$Area $= 2 	imes 2 \sqrt{a} \int_{0}^{a} \sqrt{x} \, dx$Area $= 4 \sqrt{a} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{a}$Area $= 4 \sqrt{a} 	imes \frac{2}{3} \left[ x^{3/2} ight]_{0}^{a}$Area $= \frac{8}{3} \sqrt{a} 	imes a^{3/2} = \frac{8}{3} a^{2}$ square units.

Find the area of the parabola $y^{2}=4ax$ bounded by its latus rectum.

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