First slide

Equation of line in Straight lines

Question

If the line $y = \sqrt{3} x$ cuts the curve $x^{3} + y^{3} + 3 x y + 5 x^{2} + 3 y^{2} + 4 x + 5 y - 1 = 0$ at the points A,B,C, then $O A . O B . O C$ is

Moderate

A

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

$3 \sqrt{3} + 1$
C

$\frac{2}{\sqrt{3}} + 7$
D

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

Solution

$T a n θ = \sqrt{3} \Rightarrow θ = 60^{0}$
Any point on the line is $(x_{1} + r \cos θ, y_{1} + r \sin θ) = (\frac{r}{2}, \frac{\sqrt{3} r}{2})$
where r is distance from origin substituting in the curve $r^{3} (\frac{1 + 3 \sqrt{3}}{8}) + r^{2} (....) + r (.......) - 1 = 0$
this is cubic equation in ‘r’
$∴ O A . O B . O C = r_{1} r_{2} r_{3} = \frac{1}{\frac{1 + 3 \sqrt{3}}{8}} = \frac{4}{13} (3 \sqrt{3} - 1)$

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