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

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$x^{6} + 6 x^{3} - 4 = 0$

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

$x^{6} - 12 x^{3} - 4 = 0$

$x^{6} - 6 x^{3} + 4 = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} \int_{0}^{b} (\sqrt{a} x - \frac{x^{2}}{a}) d x = \frac{1}{2} \frac{16 (\frac{a}{4}) (\frac{a}{4})}{3} \\ \Rightarrow \frac{2 \sqrt{a} \sqrt{b^{3}}}{3} - \frac{b^{3}}{3 a} = \frac{a^{2}}{6} \\ Also \frac{b^{2}}{2} = \frac{1}{2} \Rightarrow b = 1 \\ ∴ \frac{2 \sqrt{a}}{3} - \frac{1}{3 a} = \frac{a^{2}}{6} \\ 4 \sqrt{a} = a^{2} + \frac{2}{a} \Rightarrow 16 a = a^{4} + \frac{4}{a^{2}} + 4 a \Rightarrow a^{6} - 12 a^{3} + 4 = 0 \end{array}$