Question

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. $\frac{4(2-\sqrt{3})}{3}$
• B. $\frac{4(\sqrt{3}+2)}{3}$
• C. $\frac{4 \sqrt{3}}{3}$
• D. $\frac{2(\sqrt{3}+2)}{3}$

Answer 1

Answer: (B)

Given $P=(\sqrt{3}, 0)$
$\therefore $ Equation of line $A B$ is
$\frac{x-\sqrt{3}}{\cos 60^{\circ}}=\frac{y-0}{\sin 60^{\circ}}=r$ (say)

$\Rightarrow x=\sqrt{3}+\frac{r}{2}, y=\frac{r \sqrt{3}}{2}$
$\therefore $ point $\left(\sqrt{3}+\frac{r}{2}, \frac{r \sqrt{3}}{2}\right)$ lies on $y^{2}=x+2$
$\therefore \frac{3 r^{2}}{4}=\sqrt{3}+\frac{r}{2}+2$
$\Rightarrow \frac{3 r^{2}}{4}-\frac{r}{2}-(2+\sqrt{3})=0$
Let the roots be $r_{1}$ and $r_{2}$, then the product
$r_{1} \times r_{2}=P A \cdot P B=\left|\frac{-(2+\sqrt{3})}{\frac{3}{4}}\right|$
$=\frac{4(2+\sqrt{3})}{3}$