Question

If the line $y=\sqrt{3}x$ cuts the curve $x^3+y^3+3xy+5x^2+3y^3+4x+5y-1=0$ at the points $A,B,C$. Then $OA.OB.OC=$

• A. $\sqrt{336-350\sqrt{3}}$
• B. $3\sqrt{3}+1$
• C. $\frac{2}{\sqrt{3}}+7$
• D. $\frac{4}{13}(3\sqrt{3}-1)$

Answer 1

The correct option is A $\sqrt{336-350\sqrt{3}}$

$x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0$

And $y=\sqrt{3}$

$\therefore x^3+3\sqrt{3}+3\sqrt{3}x+5x^2+9+4x+5\sqrt{3}-1=0$

$x^3+5x^2(4+3\sqrt{3})x+8\sqrt{3}+8=0$

$x_1{x}_2{x}_3=-8(\sqrt{3}+1)$

$x_1{x}_2+x_2{x}_3+x_1{x}_3=(4+3\sqrt{3})$

$x_1+x_2+x_3=-5$

$\therefore OA\cdot OB\cdot OC$

$=\sqrt{(x_1^2+3)\left(x_2^3\right)(x_3^2+3)}$

$=\sqrt{(x_1^2{x}_2^2+3(x_1^2+x_2^2)+9)(x_3^2+3)}$

$=\sqrt{(x_1{x}_2{x}_3{)}^2+3(x_1^2+x_3^2+x_2^2{x}_3^2)+9x_3^2+3x_1^2{x}_2^2+9x_1^2+9x_2^2+27}$

$=\sqrt{64(4-2\sqrt{3})+3(x_1^2{x}_2^2+x_2^2{x}_3^2+x_1^2{x}_3^2)+9(x_1^2+x_2^2+x_3^2)+27}$

$x_1^2+x_2^2+x_3^2=(-5{)}^2-2(4+3\sqrt{3})$

$=25-86\sqrt{3}$

$17-6\sqrt{3}$

&
$x_1^2{x}_2^2+x_2^2{x}_3^2+x_3^2{x}_1^2=(4+3\sqrt{3}{)}^2-2x_1{x}_2{x}_3(x_1+x_2+x_3)$

$=\left(43\right)+24\sqrt{3}-2(-5)(-8(\sqrt{3}+1\left)\right)$

$=43+24\sqrt{3}-80\sqrt{3}-80$

$=-37-56\sqrt{3}$

$\therefore OA\cdot OB\cdot OC\cdot =\sqrt{256-128\sqrt{3}-101-168\sqrt{3}+153-56\sqrt{3}+27}$

$=\sqrt{336-350\sqrt{3}}$