Question:
If the line \( y - \sqrt{3}x + 3 = 0 \) cuts the parabola \( y^2 = x + 2 \) at \( A \) and \( B \), then \( PA \cdot PB \) where \( P = (\sqrt{3},0) \) is
If the line \( y - \sqrt{3}x + 3 = 0 \) cuts the parabola \( y^2 = x + 2 \) at \( A \) and \( B \), then \( PA \cdot PB \) where \( P = (\sqrt{3},0) \) is
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Avoid direct distance formulas; use roots for product problems.
Updated On: Apr 23, 2026
- $\frac{4(2-\sqrt{3})}{3}$
- $\frac{4(\sqrt{3}+2)}{3}$
- $\frac{4\sqrt{3}}{3}$
- $\frac{2(\sqrt{3}+2)}{3}$
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The Correct Option is B
Solution and Explanation
Concept:
Product of distances using roots
Step 1: Substitute line in parabola: \[ (\sqrt{3}x - 3)^2 = x + 2 \]
Step 2: \[ 3x^2 - (6\sqrt{3}+1)x + 7 = 0 \]
Step 3: Roots = $x_1, x_2$.
Step 4: Use distance relation with point $P$.
Step 5: Simplify product: \[ PA \cdot PB = \frac{4(\sqrt{3}+2)}{3} \] Conclusion:
Answer = $\frac{4(\sqrt{3}+2)}{3}$
Step 1: Substitute line in parabola: \[ (\sqrt{3}x - 3)^2 = x + 2 \]
Step 2: \[ 3x^2 - (6\sqrt{3}+1)x + 7 = 0 \]
Step 3: Roots = $x_1, x_2$.
Step 4: Use distance relation with point $P$.
Step 5: Simplify product: \[ PA \cdot PB = \frac{4(\sqrt{3}+2)}{3} \] Conclusion:
Answer = $\frac{4(\sqrt{3}+2)}{3}$
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