15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T606,

The line $y-\sqrt{3} x+3=0$ passes through $X(\sqrt{3}, 0)$ because $0-\sqrt{3}(\sqrt{3})+3=0$

The distance from $X(\sqrt{3}, 0)$ to any point $(x, y)$ on the line is

$$ d=\sqrt{(x-\sqrt{3})^2+y^2} $$

Since, $$ y=\sqrt{3}(x-\sqrt{3}) $$

we have$$ y^2=3(x-\sqrt{3})^2 $$

$$ \begin{aligned}So,\,\, d & =\sqrt{(x-\sqrt{3})^2+3(x-\sqrt{3})^2} \\ & =\sqrt{4(x-\sqrt{3})^2} \\ & =2|x-\sqrt{3}| \end{aligned} $$

The line intersects the parabola $y^2=x+2$

$$ \begin{aligned} & \text { Substitute } \quad y=\sqrt{3} x-3 \\ & \qquad \begin{aligned} &(\sqrt{3} x-3)^2=x+2 \\ & 3 x^2-6 \sqrt{3} x+9=x+2 \\ & \Rightarrow 3 x^2-(6 \sqrt{3}+1) x+7=0 \end{aligned} \end{aligned} $$

Let the roots of this quadratic be $x_P$ and $x_Q$

$$ \begin{array}{r} x_P \cdot x_Q=\left(2\left|x_P-\sqrt{3}\right|\right)\left(2\left|x_Q-\sqrt{3}\right|\right)=4 \\ \left|\left(x_P-\sqrt{3}\right)\left(x_Q-\sqrt{3}\right)\right|=4\left|x_P x_Q-\sqrt{3}\left(x_P+x_Q\right)+3\right| \end{array} $$

From vieta's formulas, $x_P+x_Q=\frac{6 \sqrt{3}+1}{3}$ and

$$ \begin{aligned} &x_P x_Q=\frac{7}{3} .\\ &\text { substitute these values }\\ &\begin{aligned} x_P \cdot x_Q & =4\left|\frac{7}{3}-\sqrt{3}\left(\frac{6 \sqrt{3}+1}{3}\right)+3\right| \\ & =4\left|\frac{7-(18+\sqrt{3})+9}{3}\right| \\ & =4\left|\frac{-2-\sqrt{3}}{3}\right|=4\left(\frac{2+\sqrt{3}}{3}\right) \end{aligned} \end{aligned} $$

Parabola

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