Question

The line $x=\sqrt{3}y$ intersects with the curve $y^3-x^2-3y^2+8=0$ at the points $A,B$ and $C$. If $O$ be the origin, then $|OA.OB.OC|$ equals

• A. $72$
• B. $64$
• C. $8$
• D. $\frac{8}{\sqrt{3}}$

Answer 1

The correct option is B $64$

The parametric form of the line $x=\sqrt{3}y$ is given by $\frac{x}{\cos {30}^{\circ }}=\frac{y}{\sin {30}^{\circ }}=r$

$\Rightarrow x=\frac{\sqrt{3}r}{2},y=\frac{r}{2}$

Putting the values of $x,y$ in the equation of the curve $y^3-x^2-3y^2+8=0$

$\Rightarrow {\left(\frac{r}{2}\right)}^3-{\left(\frac{\sqrt{3}r}{2}\right)}^2-3{\left(\frac{r}{2}\right)}^2+8=0$
$\Rightarrow r^3-12r^2+64=0$
$\Rightarrow |OA.OB.OC|=\text{product of the roots of the above equation}\Rightarrow |OA.OB.OC|=64$