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 $C,D$ be the point on $x$-axis such that $AC\perp CP,BD\perp PD$ respectively.
Then, $\triangle{PAC},\triangle{PBD}$ are triangles with $30^\circ,60^\circ,90^\circ$ from which we have $$PA=\frac{2}{\sqrt 3}AC=\frac{2}{\sqrt{3}}|A_y|,\qquad PB=\frac{2}{\sqrt 3}|B_y|$$
Eliminating $x$ from the system gives $$y^2=\frac{y+3}{\sqrt 3}+2\implies \sqrt 3\ y^2-y-3-2\sqrt 3=0$$ from which we have $$A_y+B_y=\frac{1}{\sqrt 3},\qquad A_yB_y=\frac{-3-2\sqrt 3}{\sqrt 3}$$
(a) is not correct since $$PA+PB\gt \frac{2(A_y+B_y)}{\sqrt 3}=\frac{2}{\sqrt 3}\cdot \frac{1}{\sqrt 3}=\frac 23$$
(b) is correct since $$|PA-PB|=\left|\frac{2(A_y+B_y)}{\sqrt 3}\right|=\frac 23$$
(c) is correct since $$PA\times PB=\frac 43\left|\frac{-3-2\sqrt 3}{\sqrt 3}\right|=\frac{4(2+\sqrt 3)}{3}$$
(d) is not correct since $$\frac{1}{PA}+\frac{1}{PB}=\frac{PA+PB}{PA\cdot PB}\gt \frac{2(A_y+B_y)}{\sqrt 3}\div \frac{4(2+\sqrt 3)}{3}=\frac{2-\sqrt 3}{2}$$