If the line $y-\sqrt{3}x+3=0$ cuts the curve $y^2=x+2$ at $A$ and $B$ and point on the line $P$ is $(\sqrt{3},\ 0)$ then $|PA\cdot PB|=?$

equation of line $A B: y-\sqrt{3} x+3=0$

$\begin{array}{l} \Rightarrow y=\sqrt{3} x-3 \\ \Rightarrow y=\sqrt{3}(x-\sqrt{3}) \\ \Rightarrow y=\tan 60^{\circ}(x-\sqrt{3}) \\ \Rightarrow \frac{y}{\sin 60^{\circ}}=\frac{x-\sqrt{3}}{\cos 60^{\circ}}=\gamma \text { (say) } \\ \Rightarrow x=\cos 60^{\circ} \cdot r+\sqrt{3} \& y=\sin 60^{\circ} \cdot r \\ \Rightarrow x=\frac{r}{2}+\sqrt{3} \& y=\frac{\sqrt{3}}{2} r \end{array}$

$\therefore$ this point, will lie on $y^{2}=x+2$.

$\begin{array}{l} \Rightarrow \frac{3 r^{2}}{4}=\sqrt{3}+\frac{r}{2}+2 \\ \Rightarrow \frac{3 r^{2}}{4}-\frac{r}{2}-(2+\sqrt{3})=0 \end{array}$

let the roots be $\gamma_{1}$ and $\gamma_{2}$, then the product,

$\begin{aligned} r_{1} \cdot r_{2}=P A \cdot P B & =\frac{c}{a} \quad\left[\begin{array}{c} c=-(2+\sqrt{3}) \\ a=3 / 4 \end{array}\right] \\ |P A \cdot P B| & =\left|\frac{-(2+\sqrt{3})}{3 / 4}\right| \\ |P A \cdot P B| & =\frac{4(2+\sqrt{3})}{3} \end{aligned}$