Question

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

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

Answer 1

The correct option is A $\frac{4(2+\sqrt{3})}{3}$

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

$\Rightarrow x=\sqrt{3}+\frac{r}{2},y=\frac{r\sqrt{3}}{2}$
Therefore, point $(\sqrt{3}+\frac{r}{2},\frac{r\sqrt{3}}{2})$ lies on $y^2=x+2$
$\Rightarrow \frac{3r^2}{4}=\sqrt{3}+\frac{r}{2}+2$
$\Rightarrow \frac{3r^2}{4}-\frac{r}{2}-(2+\sqrt{3})=0$
Let the roots be $r_1$ and $r_2$, then the product $r_1\times r_2=PA\cdot PB=\left|\frac{-(2+\sqrt{3})}{\frac{3}{4}}\right|$
$=\frac{4(2+\sqrt{3})}{3}$