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

• A

  $\frac{4(2-\sqrt{3})}{3}$
• Correct answer
  B

  $\frac{4(\sqrt{3}+2)}{3}$
• C

  $\frac{4 \sqrt{3}}{3}$
• D

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

Solution

Given $P=(\sqrt{3}, 0)$ $\therefore$ Equation of line $A B$ is $\frac{x-\sqrt{3}}{\cos 60^{\circ}}=\frac{y-0}{\sin 60^{\circ}}=r$
$\begin{aligned} & \Rightarrow \quad x=\sqrt{3}+\frac{r}{2}, y=\frac{r \sqrt{3}}{2} \\ & \therefore \text { point }\left(\sqrt{3}+\frac{r}{2}, \frac{r \sqrt{3}}{2}\right) \text { lies on } y^2=x+2 \\ & \therefore \\ & \Rightarrow \\ & \Rightarrow \quad \frac{3 r^2}{4}=\sqrt{3}+\frac{r}{2}+2 \\ & \Rightarrow-\frac{r}{2}-(2+\sqrt{3})=0 \end{aligned}$
Let the roots be $r_1$ and $r_2$, then the product $\begin{aligned} r_1 \times r_2 & =P A \cdot P B=\left|\frac{-(2+\sqrt{3})}{\frac{3}{4}}\right| \\ & =\frac{4(2+\sqrt{3})}{3} \end{aligned}$

Hence, the correct answer is option: 'B'.