Question

If the line $y=\sqrt{3}x$ cuts the curve $x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0$ at points $A,B,C$ then $\left(\text{where O is origin}\right)$

• A. $OA+OB+OC=3\sqrt{3}-1$
• B. $OA+OB+OC=-(1+3\sqrt{3})$
• C. $OA.OB.OC=\frac{4}{13}(3\sqrt{3}-1)$
• D. $OA.OB.OC=\frac{4}{13}(3\sqrt{3}+1)$

Answer 1

Answer: (B), (C)

$OA.OB.OC=\frac{4}{13}(3\sqrt{3}-1)$


Any point on line $y=\sqrt{3}x$ at a distance of $r$ units from origin $=(0+\frac{r}{2},0+\frac{r\sqrt{3}}{2})$
$\because$ It is also lying on the given curve,
$\Rightarrow {\left(\frac{r}{2}\right)}^3+{\left(\frac{r\sqrt{3}}{2}\right)}^3+3\left(\frac{r}{2}\right)\left(\frac{r\sqrt{3}}{2}\right)+5{\left(\frac{r}{2}\right)}^2+3{\left(\frac{r\sqrt{3}}{2}\right)}^2+4\left(\frac{r}{2}\right)+5\left(\frac{r\sqrt{3}}{2}\right)-1=0$
$\Rightarrow \frac{r^3}{8}(1+3\sqrt{3})+\frac{r^2}{4}(14+3\sqrt{3})+\frac{r}{2}(4+5\sqrt{3})-1=0$
Cubic equation in $r$, roots $r_1,r_2,r_3$ i.e.,$OA,OB,OC.$
So, $OA+OB+OC=\frac{-2(14+3\sqrt{3})}{1+3\sqrt{3}}=-(1+3\sqrt{3})$
$OA\cdot OB\cdot OC=\frac{8}{1+3\sqrt{3}}=\frac{4}{13}(3\sqrt{3}-1)$