A line through $P=(\sqrt3,0)$ and at an angle of $\pi/3$ with the positive direction of x axis cuts the parabola $y^2=x+2$ at A and B 
A line passing through $P=(\sqrt3,0)$ and making an angle of $\pi/3$ with the positive direction of x axis cuts the parabola $y^2=x+2$ at A and B, then:
  (a)$PA+PB=2/3$
  (b)$|PA-PB|=2/3$
  (c)$(PA)(PB)=\frac{4(2+\sqrt3)}{3}$
  (d)$\frac{1}{PA}+\frac{1}{PB}=\frac{2-\sqrt3}{2}$    

Equation of line:
$y=\sqrt3(x-\sqrt3)$
$y=\sqrt3x-3$    
substituting $y=\sqrt3x-3$ in $y^2=x+2$     
$(\sqrt3x-3)^2=x+2$
$3x^2-6\sqrt3x+9=x+2$
$3x^2-(6\sqrt3+1)x+7=0$     
$x_A=\frac{(6\sqrt3+1)+\sqrt{(6\sqrt3+1)^2-84}}{6}$
$x_B=\frac{(6\sqrt3+1)-\sqrt{(6\sqrt3+1)^2-84}}{6}$    
But this gives a very complicated value of $x$ and that makes me feel that there ought to be a shorter way to solve this question, just can't figure out what it is.

Let the feet of A and B on the x-axis be A’ and B’ respectively.

Compute $A’P$. It is … $\dfrac {1}{6} + \dfrac {\sqrt(\delta)}{6}$.
Similarly, $PB’ = -\dfrac {1}{6} + \dfrac {\sqrt(\delta)}{6}$.
Then, $|PA – PB| = 2 \times |A’P – PB’| = \dfrac {2}{3}$.
