The area bounded by the parabola y^2=4ax and | vedclass

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(A) The parabola is $y^2=4ax$. The latus-rectum is the line $x=a$.To find the area,we integrate with respect to $x$ from $0$ to $a$.Since the parabola

Vedclass · Accepted Answer

$\frac{8}{3} a^2$ sq. units

Vedclass · Answer

(A) The parabola is $y^2=4ax$. The latus-rectum is the line $x=a$.To find the area,we integrate with respect to $x$ from $0$ to $a$.Since the parabola is symmetric about the $x$-axis,the total area $A$ is twice the area in the first quadrant.$A = 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{a} \sqrt{4ax} \, dx$$A = 2 \int_{0}^{a} 2\sqrt{a} \sqrt{x} \, dx = 4\sqrt{a} \int_{0}^{a} x^{1/2} \, dx$$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{a} = 4\sqrt{a} \cdot \frac{2}{3} \cdot a^{3/2}$$A = \frac{8}{3} \sqrt{a} \cdot a \sqrt{a} = \frac{8}{3} a^2 	ext{ sq. units}$

The area bounded by the parabola $y^2=4ax$ and its latus-rectum $x=a$ is

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