The area enclosed by the parabola {y^2} = 4ax | vedclass

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(C) The points of intersection of the parabola ${y^2} = 4ax$ and the line $y = 2ax$ are obtained by solving the equations simultaneously.Substituting

Vedclass · Accepted Answer

$\frac{1}{{3a}} 	ext{ sq. unit}$

Vedclass · Answer

(C) The points of intersection of the parabola ${y^2} = 4ax$ and the line $y = 2ax$ are obtained by solving the equations simultaneously.Substituting $y = 2ax$ into ${y^2} = 4ax$,we get:${(2ax)^2} = 4ax$$4{a^2}{x^2} = 4ax$$4ax(ax - 1) = 0$This gives $x = 0$ or $x = \frac{1}{a}$.For $x = 0$,$y = 0$. For $x = \frac{1}{a}$,$y = 2a(\frac{1}{a}) = 2$.The points of intersection are $(0, 0)$ and $(\frac{1}{a}, 2)$.The area enclosed is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = \frac{1}{a}$:$	ext{Area} = \int_0^{1/a} (\sqrt{4ax} - 2ax) dx$$= \int_0^{1/a} (2\sqrt{a}\sqrt{x} - 2ax) dx$$= 2\sqrt{a} [\frac{x^{3/2}}{3/2}]_0^{1/a} - 2a [\frac{x^2}{2}]_0^{1/a}$$= 2\sqrt{a} \cdot \frac{2}{3} (\frac{1}{a})^{3/2} - a (\frac{1}{a})^2$$= \frac{4\sqrt{a}}{3} \cdot \frac{1}{a\sqrt{a}} - \frac{1}{a}$$= \frac{4}{3a} - \frac{1}{a} = \frac{4-3}{3a} = \frac{1}{3a} 	ext{ sq. unit}$.Thus,the correct option is $(C)$.

The area enclosed by the parabola ${y^2} = 4ax$ and the straight line $y = 2ax$ is

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