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

The correct option is A
$y-\sqrt{3}x+3=0$ can be rewritten as
$\frac{y-0}{\frac{\sqrt{3}}{2}}$ $=\frac{x-\sqrt{3}}{\frac{1}{2}}=r\left(1\right)$
Solving (1) with the parabola $y^2=x+2\frac{3r^2}{4}-\frac{r}{2}-\sqrt{3}-2=0$ $\Rightarrow PAPB=r_1{r}_2=\frac{4(\sqrt{3}+2)}{3}$
 

Mathematics