Question

If the line $y-\sqrt{3}x+3=0$ cuts the parabola $y^2=x+2$ at $A$ and $B$, then $PA\cdot PB$ 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 $AB$ 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 $y^2=x+2$
$\therefore \frac{3r^2}{4}=\sqrt{3}+\frac{r}{2}+2$
$\Rightarrow \frac{3r^2}{4}-\frac{r}{2}-(2+\sqrt{3})=0$
Let the roots be $r_1$ and $r_2$, then the product
$r_1\times r_2=PA\cdot PB=\left|\frac{-(2+\sqrt{3})}{\frac{3}{4}}\right|$
$=\frac{4(2+\sqrt{3})}{3}$