Let the line y - sqrt{3}x + 3 = 0 cut the par | vedclass

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(C) The line equation is $y = \sqrt{3}x - 3$,which can be written as $\frac{y - 0}{x - \sqrt{3}} = \sqrt{3} = 	an(60^{\circ})$.Using the parametric f

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$\frac{\sqrt{76 + 48\sqrt{3}}}{3}$

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(C) The line equation is $y = \sqrt{3}x - 3$,which can be written as $\frac{y - 0}{x - \sqrt{3}} = \sqrt{3} = 	an(60^{\circ})$.Using the parametric form of the line passing through $P(\sqrt{3}, 0)$ with angle $	heta = 60^{\circ}$:$x = \sqrt{3} + r \cos(60^{\circ}) = \sqrt{3} + \frac{r}{2}$$y = 0 + r \sin(60^{\circ}) = \frac{r\sqrt{3}}{2}$Substitute these into the parabola equation $2y^2 = 2x + 3$:$2(\frac{r\sqrt{3}}{2})^2 = 2(\sqrt{3} + \frac{r}{2}) + 3$$\frac{3r^2}{2} = 2\sqrt{3} + r + 3$$3r^2 - 2r - (6 + 4\sqrt{3}) = 0$Let $r_1 = PA$ and $r_2 = -PB$ be the roots of this quadratic equation (since $P$ lies between $A$ and $B$ or on one side,we consider the directed distances).The sum of roots $r_1 + r_2 = \frac{2}{3}$ and the product $r_1 r_2 = -\frac{6 + 4\sqrt{3}}{3}$.We need $|PA - PB| = |r_1 - r_2|$.Using $|r_1 - r_2| = \sqrt{(r_1 + r_2)^2 - 4r_1 r_2}$:$|r_1 - r_2| = \sqrt{(\frac{2}{3})^2 - 4(-\frac{6 + 4\sqrt{3}}{3})} = \sqrt{\frac{4}{9} + \frac{24 + 16\sqrt{3}}{3}} = \sqrt{\frac{4 + 72 + 48\sqrt{3}}{9}} = \frac{\sqrt{76 + 48\sqrt{3}}}{3}$.

Let the line $y - \sqrt{3}x + 3 = 0$ cut the parabola $2y^2 = 2x + 3$ at $A$ and $B$. If $P(\sqrt{3}, 0)$,then the value of $|PA - PB|$ is [where $PA$ denotes the distance between points $P$ and $A$].

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