If the line $y - \sqrt{3} x + 3 = 0 c u t s t h e c u r v e y^{2} = x + 2$ at A and B and point on the line P is $(\sqrt{3}, 0) t h e n P A . P B =$

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$\frac{2 (\sqrt{3} + 2)}{3}$

$\frac{4 (\sqrt{3} + 2)}{3}$

$\frac{4 \sqrt{3}}{2}$

$\frac{4 (2 - \sqrt{3})}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} S l o p e o f y = \sqrt{3} x - 3 i s \sqrt{3} \\ ∴ Tan θ = \sqrt{3} \Rightarrow θ = 60^{0} \\ T h e p o int s A = (\sqrt{3} \pm r \cos 60^{0}, 0 \pm r \sin 60^{0}) \\ \Rightarrow (\sqrt{3} \pm \frac{r}{2}, \pm \frac{\sqrt{3} r}{2}) \\ A l i e s o n y^{2} = x + 2 \Rightarrow \frac{3 r^{2}}{4} = \sqrt{3} \pm \frac{r}{2} + 2 \\ \Rightarrow 3 r^{2} = 4 (\frac{2 \sqrt{3} \pm r + 4}{2}) \end{array}$

$\begin{array}{l} \Rightarrow 3 r^{2} = 4 \sqrt{3} \pm 2 r + 8 \\ \Rightarrow 3 r^{2} \pm 2 r - (8 + 4 \sqrt{3}) = 0 \\ r_{1} = P A a n d r_{2} = P B a r e r o o t s \\ ∴ r_{1} . r_{2} = P A . P B = \frac{8 + 4 \sqrt{3}}{3} = \frac{4 (\sqrt{3} + 2)}{3} \end{array}$