Question

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 $P$ is $\sqrt{3},0$)

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

Answer 1

Answer: (A)

$y-\sqrt{3}x+3=0$ can be written as
$\frac{y-0}{\frac{\sqrt{3}}{2}}=\frac{x-\sqrt{3}}{\frac{1}{2}}=r...\left(1\right)$
Solving $\left(1\right)$ with the parabola $y^2=x+2$
we get $\frac{3r^2}{4}=\frac{r}{2}+\sqrt{3}+2$
$\Rightarrow 3r^2-2r-\left(4\sqrt{3}+8\right)=0$
$\therefore PA.PB=\left|r_1,r_2\right|=\frac{4\sqrt{3}+8}{3}$
$=\frac{4\left(\sqrt{3}+2\right)}{3}$