Question

The area common to the parabola $y^2=ax$ and the circle $x^2+y^2=4ax$

• A. $(3\sqrt{3}-\frac{4}{3}\pi )a^2$
• B. $(-3\sqrt{3}+\frac{4}{3}\pi )a^2$
• C. $(-3\sqrt{3}-\frac{4}{3}\pi )a^2$
• D. $(3\sqrt{3}+\frac{4}{3}\pi )a^2$

Answer 1

The correct option is D $(3\sqrt{3}+\frac{4}{3}\pi )a^2$

Given parabola is $y^2=ax$ ...................$\left(1\right)$
and the circle is $x^2+y^2=4ax$ .......................$\left(2\right)$
Both these curves are symmetrical about $x-$axis.
Solving $\left(1\right)$ and $\left(2\right)$ for $x$, we have
$x^2+ax=4ax$ or $x(x-3a)=0$
or $x=0,3a$
Thus, the two curves intersect at the points where $x=0$ and $x=3a.$
Also eqn$\left(2\right)$ meets the $x-$axis at $A(4a,0)$
Area common to $\left(1\right)$ and $\left(2\right)$ i.e., the shaded area
$=2$Area$\left[ORP\right]+2$Area$\left[PRA\right]$
$=2\left[{\int }_0^{3a}y.dx\right]$ from $\left(1\right)+\left[{\int }_{3a}^{4a}y.dx\right]$,from $\left(2\right)$
$=2[{\int }_0^{3a}\sqrt{ax}dx+{\int }_{3a}^{4a}\sqrt{4ax-x^2}dx]$
$=2\sqrt{a}{\left|\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right|}_0^{3a}+2{\int }_{3a}^{4a}\sqrt{[4a^2-{(x-2a)}^2]}dx$
$\frac{4\sqrt{a}}{3}{\left(3a\right)}^{\frac{3}{2}}+2[\frac{1}{2}(x-2a)\sqrt{4a^2-{(x-2a)}^2}+\frac{4a^2}{2}{\sin }^{-1}\left(\frac{x-2a}{2a}\right)]$
$=4\sqrt{3}{a}^2+2[0-\frac{1}{2}a\sqrt{3}a]+2a^2[\frac{\pi }{2}-\frac{\pi }{6}]$
$=4\sqrt{3}{a}^2-\sqrt{3}{a}^2+\frac{4}{3}\pi a^2$
$=(3\sqrt{3}+\frac{4}{3}\pi )a^2$