C₁ : y² = ax, C₂ : x² = ay (a > 0)

P : intersection point of y² = ax and x² = ay

$$ \Rightarrow $$ x⁴ = a² y²

$$ \Rightarrow $$ x⁴ = a²ax

$$ \Rightarrow $$ x = a, y = a

$$ \therefore $$ Point P : (a, a)

Line OP : y = x

$$ \Rightarrow $$ Point Q = (b, b)

Area $$\Delta $$OQR = $${1 \over 2}$$

$$ \Rightarrow $$ $${1 \over 2} \times b \times b$$ = $${1 \over 2}$$

$$ \Rightarrow $$ b = 1

As line x = b bisect the area between curve

$$ \therefore $$ $${1 \over 2}\int\limits_0^a {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx$$ = $$\int\limits_0^1 {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx$$

$$ \Rightarrow $$ $$\left[ {{1 \over 2}\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {1 \over 2}{{{x^3}} \over {3a}}} \right]_0^a$$ = $$\left[ {\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {{{x^3}} \over {3a}}} \right]_0^1$$

$$ \Rightarrow $$ $${{{a^2}} \over 3} - {1 \over 2}{{{a^2}} \over 3}$$ - 0 = $${{2\sqrt a } \over 3} - {1 \over {3a}}$$ - 0

$$ \Rightarrow $$ $${{{a^2}} \over 2} + {1 \over a} = 2\sqrt a $$

$$ \Rightarrow $$ $${a^3} + 2 = 4a\sqrt a $$

Squareing both sides, we get

$$ \Rightarrow $$ $${a^6} + 4{a^3} + 4 = 16{a^3}$$

$$ \Rightarrow $$ $${a^6} - 12{a^3} + 4 = $$ 0

Hence $$a$$ satisfy x⁶ – 12x³ + 4 = 0.