Question

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

• A. $x^6 - 12x^3 + 4 = 0$
• B. $x^6 - 12x^3 - 4 = 0$
• C. $x^6 + 6x^3 - 4 = 0$
• D. $x^6 - 6x^3 + 4 = 0$

Answer 1

Answer: (A)

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

Answer 2

$\int\limits^{b}_{{0}}$ $\left(\sqrt{ax}-\frac{x^{2}}{a}\right)dx=\frac{1}{2}\times \frac{16\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)}{3}$
$\Rightarrow \left[\frac{2\sqrt{a}}{3}x^{3/2}-\frac{x^{3}}{3a}\right]^{b}_{_{_0}}=\frac{a^{2}}{6}$
$\Rightarrow \frac{2\sqrt{a}}{3}b^{3/2}-\frac{b^{3}}{3a}=\frac{a^{2}}{6}\,...\left(i\right)$
Also, $\frac{1}{2}\times b^{2}=\frac{1}{2} \Rightarrow b=1$
so, $\frac{2\sqrt{a}}{3}-\frac{1}{3a}=\frac{a^{2}}{6} \Rightarrow a^{3}-4a^{3/2}+2=0$
$\Rightarrow a^{6}+4a^{3}+4=16a^{3} \Rightarrow a^{6}-12a^{3}+4=0$