For  $a>0,$  let the curves  $C_1:y^2=ax$  and  $C_2:x^2=ay$  intersect at origin  $O$  and a point  $P$. Let the line  $x=b(0<b<a)$  intersect the chord  $OP$  and the $x-$ axis at points Q and R, respectively. If the line  $x=b$  bisects the area bounded by the curves, $C_1$  and  $C_2$, and the area of  $\Delta OQR=1/2$, then  $\text{'}a\text{'}$ satisfies the polynomial equation whose degree is 

Given eqns. Are,   $x^2=ay$  and  $y^2=ax$
    
After solving, we get   $x=a,y=a$
Now, coordinates of  $B$  is  $(a,a)$  and  $A$  is  $(0,0)$
Now, coordinates of  $Q$  is  $(b,b)$
 $\therefore \frac{1}{2}{b}^2=\frac{1}{2}\Rightarrow b=1$
Area bounded by curves and  $x=1$ is 
 $\underset{0}{\overset{1}{\int }}(\sqrt{a}{x}^{1/2}-\frac{x^2}{a})dx=\frac{1}{2}\underset{0}{\overset{a}{\int }}(\sqrt{a}{x}^{1/2}-\frac{x^2}{a})dx$
 $\Rightarrow \frac{2}{3}\sqrt{a}-\frac{1}{3a}=\frac{a^2}{6}$
 $$\begin{gathered} \Rightarrow 4a\sqrt{a}-2=a^3\Rightarrow a^6+4a^3+4=16a^3 \\ \Rightarrow a^6-12a^3+4=0\Rightarrow x^6-12x^3+4=0 \end{gathered}$$