Question

The area bounded by the curve $y^2=4a(a-|x-a|);a>0$, is

• A. $\frac{8a^2}{3}$
• B. $\frac{16a^2}{3}$
• C. $\frac{19a^2}{3}$
• D. $\frac{13a^2}{3}$

Answer 1

The correct option is B $\frac{16a^2}{3}$


$y^2=4a(a-|x-a|)y^2=\{\begin{array}{cc}4ax;& x<a\\ 4a(2a-x);& x\ge a\end{array}$
$\therefore$ Required area
$=4{\int }_0^a\sqrt{4ax}=\frac{16a^2}{3}$