Question

If the line y = $\sqrt{3}$x cuts the curve x³ + y³ + 3xy + 5x² + 3y² + 4x + 5y – 1 = 0 at the points A, B, C, then OA · OB · OC is equal to–

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

Answer 1

Answer: (A)

$\frac{4}{13}$ (3$\sqrt{3}$ – 1)

Answer 2

The abscissa of the intersection points of the given line and the given curve is given by the equation

(3$\sqrt{3}$ + 1)x³ + (3$\sqrt{3}$ + 14)x² + (5$\sqrt{3}$ + 4) x – 1 = 0

If x₁, x₂, x₃ be the roots of the above equation, then

A ŗ (x₁,$\sqrt{3}$x₁), B ŗ (x₂, $\sqrt{3}$x₂) and C ŗ (x₃, $\sqrt{3}$x₃)

Hence, we have

OA · OB · OC = 2x₁ · 2x₂ · 2x₃ = 8x₁x₂x₃

= 8 × $\frac{1}{3\sqrt{3} + 1}$

= 8 × $\frac{3\sqrt{3} - 1}{26}$ = $\frac{4}{13}$ (3$\sqrt{3}$ – 1).