For a0, let the curves C_1: y^2=a x and C | vedclass

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(A) The curves are $C_{1}: y^{2}=ax$ and $C_{2}: x^{2}=ay.$ Solving them,we get $x^{4}/a^{2} = ax \Rightarrow x(x^{3}-a^{3})=0.$ Thus,the intersection

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

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(A) The curves are $C_{1}: y^{2}=ax$ and $C_{2}: x^{2}=ay.$ Solving them,we get $x^{4}/a^{2} = ax \Rightarrow x(x^{3}-a^{3})=0.$ Thus,the intersection points are $O(0,0)$ and $P(a,a).$
The chord $OP$ has the equation $y=x.$ The line $x=b$ intersects $OP$ at $Q(b,b)$ and the $x$-axis at $R(b,0).$
The area of $\Delta OQR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes b 	imes b = \frac{b^{2}}{2}.$ Given $	ext{Area} = \frac{1}{2},$ we have $b^{2}=1 \Rightarrow b=1.$
The total area bounded by the curves is $\int_{0}^{a} (\sqrt{ax} - x^{2}/a) dx = [\frac{2}{3}\sqrt{a}x^{3/2} - x^{3}/(3a)]_{0}^{a} = \frac{2}{3}a^{2} - \frac{1}{3}a^{2} = \frac{a^{2}}{3}.$
The line $x=b$ bisects this area,so $\int_{0}^{b} (\sqrt{ax} - x^{2}/a) dx = \frac{1}{2} 	imes \frac{a^{2}}{3} = \frac{a^{2}}{6}.$
Substituting $b=1$: $\frac{2}{3}\sqrt{a} - \frac{1}{3a} = \frac{a^{2}}{6}.$
Multiplying by $6a$: $4a^{3/2} - 2 = a^{3} \Rightarrow a^{3} + 2 = 4a^{3/2}.$
Squaring both sides: $(a^{3}+2)^{2} = 16a^{3} \Rightarrow a^{6} + 4a^{3} + 4 = 16a^{3} \Rightarrow a^{6} - 12a^{3} + 4 = 0.$

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