The area (in sq units) between the curve y^2 | vedclass

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(A) The given parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$. The latus rectum of the parabola is the line $x = a

Vedclass · Accepted Answer

$\frac{32}{3}$

Vedclass · Answer

(A) The given parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$. The latus rectum of the parabola is the line $x = a = 2$. The area bounded by the curve and the latus rectum is symmetric about the $x$-axis. The required area is $A = 2 \int_0^2 y \, dx = 2 \int_0^2 \sqrt{8x} \, dx$. $A = 2 	imes 2\sqrt{2} \int_0^2 x^{1/2} \, dx = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} ight]_0^2$. $A = 4\sqrt{2} 	imes \frac{2}{3} 	imes (2)^{3/2} = \frac{8\sqrt{2}}{3} 	imes 2\sqrt{2} = \frac{16 	imes 2}{3} = \frac{32}{3}$ sq units.

The area (in sq units) between the curve $y^2 = 8x$ and its latus rectum is

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