Question

If the line $y=x\sqrt{3}$ cuts the curve $x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0$ at the points $A,B$ and $C$,then $OA.OB.OC$ is equal to (where '$O$' is origin)

• A. $\frac{4}{13}(3\sqrt{3}-1)$
• B. $(3\sqrt{3}-1)$
• C. $\frac{1}{\sqrt{3}}(2+7\sqrt{3})$
• D. $\frac{4}{13}(3\sqrt{3}+1)$

Answer 1

The correct option is A $\frac{4}{13}(3\sqrt{3}-1)$

Coordinates of a point on the line $y=x\sqrt{3}$ at a distance r from origin
is $(r\cos \theta ,r\sin \theta )$
$\therefore \tan \theta =\sqrt{3}$
$\therefore (\frac{r}{2},\frac{r\sqrt{3}}{2})$ lies on the given curve
$\Rightarrow \frac{r^3}{8}+\frac{r^3.3\sqrt{3}}{8}+3.\frac{r}{2}.\frac{r\sqrt{3}}{2}+5.\frac{r^2}{4}+3.\frac{r^2.3}{4}+4.\frac{r}{2}+5.\frac{r\sqrt{3}}{2}-1=0$
$\Rightarrow \left(\frac{1+3\sqrt{3}}{8}\right)r^3+\frac{r^2}{4}(3\sqrt{3}+14)+\frac{r}{2}(5\sqrt{3}+4)-1=0$
$\Rightarrow r_1.r_2.r_3=\frac{8}{3\sqrt{3}+1}$
$=\frac{8}{27-1}\times (3\sqrt{3}-1)$
$=\frac{4}{13}(3\sqrt{3}-1)$