Question

If the line y = $\sqrt{3}x$ intersects 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 (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. $\frac{1}{26}(3\sqrt{3}+1)$

Answer 1

The correct option is B $\frac{4}{13}(3\sqrt{3}-1)$

The lines $y=\sqrt{3x}$ intersects the curve at three points $A$, $B$ and $C$.

The coordinates of these points can be written as ,

$A(x_1,\sqrt{3}{x}_1)$

$B(x_2,\sqrt{3}{x}_2)$

$C(x_3,\sqrt{3}{x}_3)$

If $O(0,0)$ is the origin then $OA=\sqrt{(x_1{)}^2+(\sqrt{3x_1}{)}^2}$

$\Rightarrow OA=2x_1$

Similarly $OB=2x_2$

and $OC=2x_3$

Hence $OA.OB.OC=8(x_1.x_2.x_3)$

Now putiing value of $y=\sqrt{3}$ into equation of given curve, we get,

$\Rightarrow x^3+(\sqrt{3}x{)}^3+3.x.\sqrt{3}x+5x^2+3(\sqrt{3}x{)}^2+4x-\sqrt{3}x-1=0$

$\Rightarrow (1+3\sqrt{3})x^3+(14+3\sqrt{3})x^2+(4-\sqrt{3})x-1=0$ ...$\left(1\right)$

The equation $\left(1\right)$ contains the abscissa of the intersection points of the given line and curve, which are $x_1$ , $x_2$ and $x_3$

From equation $\left(1\right)$ we can see that the product of roots is $x_1.x_2.x_3=-\left(\frac{-1}{1+3\sqrt{3}}\right)=\frac{1}{1+3\sqrt{3}}=\frac{1-3\sqrt{3}}{-26}$

Hence $OA.OB.OC=8(x_1.x_2.x_3)=8\times \frac{1-3\sqrt{3}}{-26}$

$\Rightarrow OA.OB.OC=\frac{4}{13}(3\sqrt{3}-1)$

So correct option is $B$.