The area of the region bounded by the parabol | vedclass

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Question

(B) The area $A$ of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum $x = a$ is given by the integral:$A = 2 \int_{0}^{a} \sqrt{4ax

Vedclass · Accepted Answer

$\pm 3$

Vedclass · Answer

(B) The area $A$ of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum $x = a$ is given by the integral:$A = 2 \int_{0}^{a} \sqrt{4ax} \, dx$$A = 2 	imes 2\sqrt{a} \int_{0}^{a} x^{1/2} \, dx$$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{a}$$A = 4\sqrt{a} 	imes \frac{2}{3} 	imes a^{3/2} = \frac{8}{3} a^2$Given that the area is $24$ sq. units,we have:$\frac{8}{3} a^2 = 24$$a^2 = 24 	imes \frac{3}{8} = 9$$a = \pm 3$Thus,the correct option is $B$.

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is $24$ sq. units. Then,$a = $ . . . . . . .

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