The line y - sqrt{3}x + 3 = 0 cuts the parabo | vedclass

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(A) The line equation is $y = \sqrt{3}x - 3$. The slope $m = \sqrt{3}$, so $	an 	heta = \sqrt{3} \Rightarrow 	heta = 60^\circ$. Any point on the li

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$\frac{4(2+\sqrt{3})}{3}$

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(A) The line equation is $y = \sqrt{3}x - 3$. The slope $m = \sqrt{3}$, so $	an 	heta = \sqrt{3} \Rightarrow 	heta = 60^\circ$. Any point on the line at a distance $r$ from $X(\sqrt{3}, 0)$ is given by $x = \sqrt{3} + r \cos 60^\circ = \sqrt{3} + \frac{r}{2}$ and $y = 0 + r \sin 60^\circ = \frac{r\sqrt{3}}{2}$. Substituting these into the parabola equation $y^2 = x + 2$: $(\frac{r\sqrt{3}}{2})^2 = (\sqrt{3} + \frac{r}{2}) + 2$ $\frac{3r^2}{4} = \sqrt{3} + 2 + \frac{r}{2}$ $3r^2 - 2r - 4(\sqrt{3} + 2) = 0$. Let $r_1$ and $r_2$ be the roots of this quadratic equation, which represent the directed distances $XP$ and $XQ$. The product of the roots $r_1 r_2 = \frac{-4(\sqrt{3} + 2)}{3}$. Since $XP \cdot XQ = |r_1 r_2|$, we have $XP \cdot XQ = |\frac{-4(\sqrt{3} + 2)}{3}| = \frac{4(2 + \sqrt{3})}{3}$.

The line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = x + 2$ at the points $P$ and $Q$. If the coordinates of the point $X$ are $(\sqrt{3}, 0)$, then the value of $XP \cdot XQ$ is

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