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 $$\triangle OQR = \frac{1}{2}$$, then 'a' satisfies the equation:

For a>0, the curves are
 $$C1\ :y^{2\ }=\ ax\ \ and\ C2:\ x^2\ =\ ay.$$

At  intersection (other than origin):
From $$y^{2\ }=\ ax\ \ and\ \ x^2\ =\ ay,$$

      $$y\ =\ \sqrt{\ ax},\ y\ =\ \frac{x^2}{a}$$

Equating:

$$\sqrt{\ ax}\ =\ \frac{x^2}{a}\ $$

$$a\sqrt{ax}\ =\ x^2$$

$$a^2x\ =\ x^4$$

x = a

Thus P(a,a), so chord OP is y=x.

Line x=b intersects:

• OP at Q(b,b)
• x-axis at R(b,0)

Area(△OQR) = $$\frac{1}{2}b^2$$

Given:

$$\frac{b^2}{2}\ =\ \frac{1}{2}$$

$$b\ =1$$

Area between curves:

Since x=1 bisects the area:

Multiply by 6:

Multiply by a:

Rearrange:
 

Now squaring on both sides gives:

Bring all the terms to one side :

Hence the final answer is 

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