Question

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

• A. $\frac{4}{13}(3\sqrt{3}-1)$
• B. $3\sqrt{3}+1$
• C. $\frac{2}{\sqrt{3}}+7$
• D. None of these

Answer 1

Answer: (A)

Any point on the line $y=\sqrt{3}x$ at a distance r from the origin is $\left(\frac{r}{2},\frac{\sqrt{3}r}{2}\right)$ . This point lies on the given curve if
$\frac{r^3}{8}+\frac{3\sqrt{3}{r}^3}{8}+\frac{3\sqrt{3}{r}^2}{4}+\frac{5r^2}{4}+\frac{9r^2}{4}+2r+\frac{5\sqrt{3}r}{2}-1=0$
$\Rightarrow \left(3\sqrt{3}+1\right)r^3+2\left(3\sqrt{3}+14\right)r^2+4\left(4+5\sqrt{3}\right)r-8=0$
If $OA=r_1,OB=r_2$ and $OC=r_3$
Then $OA.OBOC=r_1{r}_2{r}_3=\left|\frac{8}{3\sqrt{3}+1}\right|$
$=\frac{4}{13}\left(3\sqrt{3}-1\right)$