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 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$∆OQR=\frac{1}{2}$, then$\text{'}a\text{'}$ 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. $x^6-6x+4=0$

Answer 1

The correct option is A

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

Explanation for the correct option:

Step 1: we will find the point of intersections of the given curves and line.

We have given$C_1:y^2=ax$ and$C_2:x^2=ay$ .

Points of intersection after solving these equations are$O(0,0)\&P(a,a)$.

by joining these two point we get the chord, j

therefore, equation of chord is

$$\begin{gathered} \frac{y-0}{x-0}=\frac{a-0}{a-0} \\ \Rightarrow y=x \end{gathered}$$

now, the point of intersection of the line$x=b\&x=y$ is$(b,b)$.

Step 2: draw a diagram of the above information.

Step 3: Find area and solving the given things.

We have given,

area of$∆OQR=\frac{1}{2}$

$\begin{array}{rcl}\frac{1}{2}\times b\times b& =& \frac{1}{2}\\ b^2& =& 1\\ b& =& 1\end{array}$

Now area bounded by the two given curve$C_1:y^2=ax$ and$C_2:x^2=ay$ is

$$\begin{gathered} ={\int }_0^a(\sqrt{ax}-\frac{x^2}{a})dx \\ ={[\frac{2\sqrt{a}{x}^{{\displaystyle \frac{3}{2}}}}{3}-\frac{x^3}{3a}]}_0^a \\ =\frac{1}{3}[2a^2-a^2] \\ =\frac{a^2}{3}......\left(1\right) \end{gathered}$$

the line$x=b$ divides the area bounded by the curves$C_1:y^2=ax$ and$C_2:x^2=ay$ is just half of the original.

$$\begin{gathered} ={\int }_0^b(\sqrt{ax}-\frac{x^2}{a})dx \\ ={[\frac{2\sqrt{a}{x}^{{\displaystyle \frac{3}{2}}}}{3}-\frac{x^3}{3a}]}_0^1=\frac{1}{3}[2a^{\frac{1}{2}}-\frac{1}{a}].......\left(2\right) \end{gathered}$$

From$\left(1\right)\&\left(2\right)$,

$$\begin{gathered} \Rightarrow \frac{1}{3}[2a^{\frac{1}{2}}-\frac{1}{a}]=\frac{1}{2}\times \frac{a^2}{3} \\ \Rightarrow 2a^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1=\frac{a^3}{2} \\ \Rightarrow 4a^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=a^3+2 \\ \Rightarrow 16a^3=a^6+4a^3+4 \\ \Rightarrow a^6-12a^3+4=0 \end{gathered}$$

Hence, option (A) is the correct option.