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If the line $y - \sqrt{3} x + 3 = 0$ cuts the parabola $y^{2} = x + 2$ at $A$ and $B$ , then $P A . P B$ is equal to [where $P \equiv (\sqrt{3}, 0)$ ]

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

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

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

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

answer is A.

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Detailed Solution

$y = \sqrt{3} x - 3 \Rightarrow θ = 60^{0},$ point on the line
$A = (\sqrt{3} + \frac{r}{2}, \frac{\sqrt{3} r}{2})$ lies on parabola
$\Rightarrow 3 r^{2} - 2 r + 4 (2 + \sqrt{3}) = 0$
$r_{1} r_{2} = \frac{4 (2 + \sqrt{3})}{3}$

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