If the area included between the two parabolas y² = 4a(x + a) and y² = 4b(b – x) is 16/3. Then the product of A.M. and the G.M. of a and b is

Equating the values of y² from the two equations of parabolas we get  4a(x + a) = 4b(b – x) or x=b-a the abscissa of the point of intersection is 

b-a and both the curves are symmetrical about x- axis.

$\mathrm{Required} \mathrm{area}==2[\text{ Area APM }+\text{ Area PMB }]\text{, by symmetry }$

$=2\left[{\int }_{-a}^{b-a} ydx,\text{ for the parabola }{y}^2=4a(x+a)\right.\left.+{\int }_{b-a}^b ydx,\text{ for the parabola }{y}^2=4b(b-x)\right]$

$$\begin{gathered} 2{\int }_{-\mathrm{a}}^{\mathrm{b}-\mathrm{a}} \sqrt{4\mathrm{a}(\mathrm{x}+\mathrm{a})}\mathrm{dx}+2{\int }_{\mathrm{b}-\mathrm{a}}^{\mathrm{b}} \sqrt{4\mathrm{b}(\mathrm{b}-\mathrm{a})}\mathrm{dx}=\frac{16}{3} \\ \begin{array}{l}4\sqrt{a}{\int }_{-a}^{b-a} (x+a{)}^{1/2}dx+4\sqrt{b}{\int }_{b-a}^b (b-x{)}^{1/2}dx=\frac{16}{3}\\ 4\sqrt{\mathrm{a}}{\left[\frac{2}{3}(x-a{)}^{3/2}\right]}_{-a}^{b-a}-4\sqrt{b}{\left[\frac{2}{3}(b-x{)}^{3/2}\right]}_{\mathrm{b}-\mathrm{a}}^{\mathrm{b}}=\frac{16}{3}\\ \frac{1}{3}8\sqrt{a}\cdot {\mathrm{b}}^{3/2}+\frac{1}{3}8{\sqrt{\mathrm{ba}}}^{3/2}=\frac{16}{3}\\ \frac{8}{3}(a+b)\sqrt{\left(ab\right)}.=\frac{16}{3}\\ \frac{(a+b)}{2}\sqrt{\left(ab\right)}=1\end{array} \end{gathered}$$