• # question_answer If the line $y=\sqrt{3}\,x$ intersects the curve ${{x}^{3}}+{{y}^{3}}+3xy+5{{x}^{2}}+3{{y}^{2}}+4x+5y-1=0$ at the points A, B, C then OA. OB. OC is (Here 'O' is origin) A) $\frac{4}{13}(3\sqrt{3}+1)$ B) $\frac{4}{13}(3\sqrt{3}-1)$ C) $\frac{1}{26}(3\sqrt{3}-1)$        D) $\frac{1}{26}(3\sqrt{3}+1)$

 Line $y=\sqrt{3}\,x$ and curve ${{x}^{3}}+{{y}^{3}}+3xy+5{{x}^{2}}+3{{y}^{2}}+4x+5y-1=0$ Solving (1) & (2) then$\Rightarrow$${{x}^{3}}+3\sqrt{3}\,{{x}^{3}}+3\sqrt{3}\,{{x}^{2}}+5{{x}^{2}}+9{{x}^{2}}+4x+5\sqrt{3}\,x-1=0$ Let roots ${{x}_{1}},$${{x}_{2}},$${{x}_{3}}$ Then     ${{x}_{1}}\,{{x}_{2}}\,{{x}_{3}}=$ Co-ordinates of A, B, C are $({{x}_{1}},\,\,\sqrt{3}\,{{x}_{1}}),$$({{x}_{2}},\,\,\sqrt{3}\,{{x}_{2}})$ and $({{x}_{3}},\,\,\sqrt{3}\,{{x}_{3}})$ respectively. then   $OA.\,\,OB.\,\,OC=8\,{{x}_{1}}\,{{x}_{2}}\,{{x}_{3}}$ $=\frac{8}{3\sqrt{3}+1}=\frac{8\,(3\sqrt{3}-1)}{26}=\frac{4}{13}\,\,(3\sqrt{3}-1)$