Question

For $a>0$, let the curves $C_1:y^2=ax$ and $C_2:x^2=ay$ intersect at origin and a point .

Let the line $x=b(0<b<a)$ intersect the chord and the x-axis at points and , 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

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

Explanation for the correct option:

Draw the graph of the given equations:

Given, $\Delta OQR=\frac{1}{2}$

$\Rightarrow$$\frac{1}{2}\times b\times b=\frac{1}{2}$

$\Rightarrow$ $b=1$

Now, According to the question

${\int }_0^1\left(\sqrt{ax}-\frac{x^2}{a}\right)dx=\frac{1}{2}{\int }_0^a\left(\sqrt{ax}-\frac{x^2}{a}\right)dx$

$\Rightarrow$ $\frac{2}{3}\sqrt{a}-\frac{1}{3}a=\frac{a^2}{6}$

$\Rightarrow$ $2\sqrt{a}-\frac{1}{a}=\frac{a^2}{2}$

$\Rightarrow$ $4\sqrt{a}-\frac{2}{a}=a^2$

$\Rightarrow$ $4a\sqrt{a}-2=a^3$

$\Rightarrow$ $4a\sqrt{a}=2+a^3$

$\Rightarrow$ ${\left(4a\sqrt{a}\right)}^2={\left(2+a^3\right)}^2$

$\Rightarrow$ $16a^3=4+a^6+4a^3$

$\Rightarrow$ ${\mathit{a}}^{\mathbf{6}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{12}{\mathit{a}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence, Option ‘A’ is Correct.