Curves $y^2 = ax$ and $x^2 = ay$ intersect at A and B. The area bounded by these two curves is bisected by the line $x = b$ (given $a>b>0$). If area of the triangle formed by the line AB, $x =b$ and x-axis is 1/2 sq. units, then

$a^6 -12a^3 +4=0$

Two parabolas $y^2 =ax$ and $x^2 =ay$ intersect at A (0,0) and B (a, a). The area bounded by these two curves is

$Δ=\int\limits_0^a\left(\sqrt{ax}-\frac{x^2}{a}\right)dx=\left[\frac{2}{3}\sqrt{a}x^{3/2}=\frac{x^3}{3a}\right]_0^a=\frac{a^2}{3}$

Line x = b bisects the area bounded by the two parabolas.

$∴\int\limits_0^b\left(\sqrt{ax}-\frac{x^2}{a}\right)dx=\frac{a^2}{6}$

$⇒\left[\frac{2}{3}\sqrt{a}x^{3/2}-\frac{x^3}{3a}\right]_0^b=\frac{a^2}{6}$

$⇒\frac{2}{3}a^{1/2}b^{3/2}-\frac{b^3}{3a}=\frac{a^2}{6}$   ...(i)

It is given that area of ΔAPQ is $\frac{1}{2}$

$∴\frac{1}{2}b^2=\frac{1}{2}⇒b=1$

Substituting $b = 1$ in (i), we obtain 

$\frac{2}{3}a^{1/2}-\frac{1}{3a}=\frac{a^2}{6}$

$⇒4a^{3/2}-2=a^3⇒(a^3+2)=16a^3⇒a^6-12a^3+4=0$