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Q.
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)
Straight Lines
Solution:
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 $OA. OB OC =r_{1} r_{2}r_{3} = \left|\frac{8}{3\sqrt{3} + 1 } \right|$
$ = \frac{4}{13} \left(3\sqrt{3} - 1\right)$