If the line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^{2} = x + 2$ at $A$ and $B$, and if $P \equ

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If the line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^{2} = x + 2$ at $A$ and $B$, and if $P \equiv (\sqrt{3},0)$, then $PA \cdot PB$ is equal to:

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Sol. [Key Concept: Put $(x,y) = (x_{0} + r\cos	heta,\ r\sin	heta)$ into the parabola; then $PA \cdot \ PB = {|r}_{1}\ r_{2}|$ from the quadratic in $r$.]$$PA = r_{1},PB = - r_{2}$$Put ( $\sqrt{3} + rcos	heta,rsin	heta$ ) to $y^{2} = x + 2$$$\Rightarrow r^{2}\sin^{2}	heta - rcos	heta - (\sqrt{3} + 2) = 0^{r_{1}} $$$${(PA)(PB) = - r_{1}r_{2} = \frac{\sqrt{3} + 2}{\sin^{2}	heta} }{= (\sqrt{3} + 2)\left( 1 + \cot^{2}	heta ight) }{= (\sqrt{3} + 2)\left( 1 + \frac{1}{3} ight)\ \lbrack\because tan	heta = \sqrt{3}brack }{(PA)(PB) = \frac{4}{3}(2 + \sqrt{3}) }$$

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