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 $x=b(0<b<a)$ intersects 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

The correct option is A $x^6-12x^3+4=0$

Given, ar$\left(\Delta OQR\right)=\frac{1}{2}$
$\Rightarrow \frac{1}{2}\times b\times b=\frac{1}{2}$
$\Rightarrow b=1(\because b>0)$


As per the question,
$\underset{0}{\overset{1}{\int }}(\sqrt{ax}-\frac{x^2}{a})dx=\frac{1}{2}\underset{0}{\overset{a}{\int }}(\sqrt{ax}-\frac{x^2}{a})dx$
$\Rightarrow \frac{2}{3}\sqrt{a}-\frac{1}{3a}=\frac{1}{2}[\frac{2}{3}{a}^2-\frac{a^3}{3a}]$
$\Rightarrow \frac{2}{3}\sqrt{a}-\frac{1}{3a}=\frac{a^2}{6}$
$\Rightarrow 4a\sqrt{a}=2+a^3$
$\Rightarrow 16a^3=4+a^6+4a^3$
$\Rightarrow a^6-12a^3+4=0.$