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)

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