Question

Find the area of the circle ${4x}^2+{4y}^2=9$ which is interior to the parabola $y^2=4x$.

Answer 1

Equations are $y^2=4x$ …………$\left(1\right)$

and $4x^2+4y^2=9$

i.e., $x^2+y^2={\left(\frac{3}{2}\right)}^2$ ……….$\left(2\right)$

From $\left(1\right)$ or $\left(2\right)$

$x^2+4x=\frac{9}{4}$

$\Rightarrow x=\frac{1}{2},\frac{-9}{2}$

$\therefore y=\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}$

From P, draw PM $\perp$ to x-axis

$OA=3/2$

Required Area $=$ Area of shaded Portion

$=2$(area of OAPO)

Area of $OAPO=\left[{\int }_0^{1/2}2\sqrt{x}+{\int }_{1/2}^{3/2}\sqrt{\frac{9}{4}-x^2}\right]=2{\int }_0^{1/2}\sqrt{x}dx+{\int }_{1/2}^{3/2}\sqrt{\frac{9}{4}-x^2}dx$

$=\frac{2[x^{3/2}{]}^{1/2}}{3/2}$ $+\frac{1}{2}$ ${[x\sqrt{\frac{9}{4}-x^2}+(3/2{)}^2{\sin }^{-1}\left(\frac{x}{3/2}\right)]}_{1/2}^{3/2}$

Required area $=\frac{\sqrt{2}}{6}+\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\left(\frac{1}{3}\right)$.