Question

If the line $y=2\sqrt{2}x$ cuts the curve $x^3+y^3+6xy+6x^2+2y^2+5x+5y+1=0$ at the point $A,B,C$ then $OA.OB.OC$ is equal to $\frac{a(1-16\sqrt{b})}{c}$. Find $a+b+c$.

(where $O$ is the origin)

• A. $511$
• B. $540$
• C. $420$
• D. $-340$

Answer 1

The correct option is B $540$

Given Equation of curve:

$x^3+y^3+6xy+6x^2+2y^2+5x+5y+1=0$

Substituting $y=2\sqrt{2}x$ in the above curve

$\Rightarrow (16\sqrt{2}+1)x^3+(22+12\sqrt{2})x^2+(5+10\sqrt{2})x+1=0$

$\Rightarrow$ Product of roots $=-\frac{d}{a}=-\frac{1}{16\sqrt{2}+1}=\frac{1-16\sqrt{2}}{511}$

$OA=\sqrt{x_1^2+(2\sqrt{2}{x}_1{)}^2}=\sqrt{9x_1^2}=3x_1$

$OA.OB.OC=27x_1{x}_2{x}_3$

$=\frac{27(1-16\sqrt{2})}{511}=\frac{a(1-16\sqrt{b})}{c}$

So,

$a+b+c=27+2+511=540$