Question

If $O$ is the origin and the line $y=\sqrt{3}x$ 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 :

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

Answer 1

Answer: (A)

$\frac{4}{13}(3\sqrt{3}-1)$

$x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0$
Substituting $y=\sqrt{3}x$,
$x^3+3\sqrt{3}{x}^3+3\sqrt{3}{x}^2+5x^2+9x^2+4x+5\sqrt{3}x=1$
$\therefore x_1+x_2+x_3=\frac{-(14+3\sqrt{3})}{(1+3\sqrt{3})}$
$x_1{x}_2+x_1{x}_3+x_2{x}_3=\frac{4+5\sqrt{3}}{1+3\sqrt{3}}$
$x_1{x}_2{x}_3=\frac{1}{1+3\sqrt{3}}$
$OA\cdot OB\cdot OC=\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}\sqrt{x_3^2+y_3^2}$
$=\sqrt{4x_1^2\times 4x_2^2\times 4x_3^2}$
$=8|x_1{x}_2{x}_3|$
$=8\times \frac{1}{1+3\sqrt{3}}$
$=\frac{8}{1+3\sqrt{3}}$
$=\frac{8}{26}(3\sqrt{3}-1)$
$=\frac{4}{13}(3\sqrt{3}-1)$