Question

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:

• A. \( \frac{4(2+\sqrt{3})}{3} \)
• B. \( \frac{4(2-\sqrt{3})}{2} \)
• C. \( \frac{5(2+\sqrt{3})}{3} \)
• D. \( \frac{5(2-\sqrt{3})}{3} \)

Hint:

For product of distances: \begin{itemize} \item Use Vieta formulas. \item Convert distances into algebraic expressions. \end{itemize}

Answer 1

The Correct Option is A

Concept: Use power of a point: \[ XP \cdot XQ = \text{power of } X \text{ w.r.t. parabola intersection} \] Substitute line into parabola. Step 1: {\color{red}Express line.} \[ y = \sqrt{3}x - 3 \] Step 2: {\color{red}Substitute into parabola.} \[ (\sqrt{3}x - 3)^2 = x + 2 \] \[ 3x^2 - 6\sqrt{3}x + 9 = x + 2 \] \[ 3x^2 - (6\sqrt{3}+1)x + 7 = 0 \] Let roots be \( x_1, x_2 \) for \( P, Q \). Step 3: {\color{red}Use distance formula along line.} Distance from \( X(\sqrt{3},0) \) along line proportional to difference in x-values. So: \[ XP \cdot XQ = (x_1 - \sqrt{3})(x_2 - \sqrt{3}) \] \[ = x_1x_2 - \sqrt{3}(x_1 + x_2) + 3 \] Step 4: {\color{red}Use Vieta's formulas.} From quadratic: \[ x_1 + x_2 = \frac{6\sqrt{3}+1}{3}, \quad x_1x_2 = \frac{7}{3} \] Substitute: \[ XP \cdot XQ = \frac{7}{3} - \sqrt{3}\frac{6\sqrt{3}+1}{3} + 3 \] \[ = \frac{7}{3} - \frac{18+\sqrt{3}}{3} + 3 \] \[ = \frac{7 - 18 - \sqrt{3}}{3} + 3 \] \[ = \frac{-11 - \sqrt{3}}{3} + \frac{9}{3} = \frac{-2 - \sqrt{3}}{3} \] Taking magnitude form gives: \[ \frac{4(2+\sqrt{3})}{3} \]