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 $P\equiv (\sqrt{3},0))$

• A. $\frac{4(\sqrt{3}+2)}{3}$
• B. $\frac{4(2-\sqrt{3})}{3}$
• C. $\frac{4\sqrt{3}}{3}$
• D. none of these

Answer 1

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

$y-\sqrt{3}x+3=0$ can be rewritten as $\frac{y-0}{\frac{\sqrt{3}}{2}}=\frac{x-\sqrt{3}}{\frac{1}{2}}=r$ ...(1)
solving (1) with the parabola $y^2=x+2$
$\Rightarrow \frac{3r^2}{4}=\frac{r}{2}+\sqrt{3}+2\Rightarrow 3r^2-2r-(4\sqrt{3}+8)=0$
$\Rightarrow PA.PB=\left|r_1{r}_2\right|=\frac{4(\sqrt{3}+2)}{3}$