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 $y-\sqrt{3}x+3=0$   is  $P$  $\left(\sqrt{3},\text{ 0}\right)$ then  PA . PB =

 

 

$The given line is y-\sqrt{3}x+3=0$

$The slope of the above line is \sqrt{3}$

$Inclination is \theta ={60}^0$and$P(\sqrt{3},0)$

$General point on the line y-\sqrt{3}x+3=0 at a dis\tan ce of r units from the point P is A\left(\sqrt{3}+\frac{r}{2},\frac{\sqrt{3}}{2}r\right)$

 $Since this point lies on the curve y^2=x+2$

                 ${\left(\frac{\sqrt{3}r}{2}\right)}^2=\sqrt{3}+\frac{r}{2}+2$

            $3r^2-2r-\left(8+4\sqrt{3}\right)=0$

product of roots=$r_1{r}_2=PA·PB=\frac{c}{a}$

            $PA.PB=\left|\frac{8+4\sqrt{3}}{3}\right|$

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