Parabola
Intersection with Lines
Grade 11
Question:
<p>If the line \(y - \sqrt{3}x + 3 = 0\) cuts the parabola \(y^2 = x + 2\) at \(A\) and \(B\), then \(PA \times PB\) is equal to (where \(P = (\sqrt{3}, 0)\))</p>
<p>(a) \(\frac{4(\sqrt{3} + 2)}{3}\)</p>
<p>(b) \(\frac{4(2 - \sqrt{3})}{3}\)</p>
<p>(c) \(4\sqrt{3}\)</p>
<p>(d) \(2(\sqrt{3} + 2)\)</p>
Step-by-Step Solution
Key Concept: Parametrize the line through point P and substitute into the parabola equation to find the product of distances using Vieta's formulas.
<p><strong>Step 1:</strong> Given line: \(y - \sqrt{3}x + 3 = 0\) or \(y = \sqrt{3}x - 3\), and parabola: \(y^2 = x + 2\) with point \(P = (\sqrt{3}, 0)\).</p><p><strong>Step 2:</strong> The line passes through \(P\) at \(60°\) inclination. We can parametrize the line as \(\frac{x - \sqrt{3}}{\cos 60°} = \frac{y - 0}{\sin 60°} = r\), giving \(x = \sqrt{3} + \frac{r}{2}\) and \(y = \frac{r\sqrt{3}}{2}\).</p><p><strong>Step 3:</strong> Substitute into the parabola equation \(y^2 = x + 2\):</p><p>\[\left(\frac{r\sqrt{3}}{2}\right)^2 = \sqrt{3} + \frac{r}{2} + 2\]</p><p>\[\frac{3r^2}{4} = \sqrt{3} + 2 + \frac{r}{2}\]</p><p>\[3r^2 - 2r = 4(\sqrt{3} + 2)\]</p><p><strong>Step 4:</strong> For the two intersection points \(A\) and \(B\) with parameters \(r_1\) and \(r_2\), we have \(PA \times PB = |r_1 \times r_2| = \frac{4(\sqrt{3} + 2)}{3}\).</p><p>∴ Answer is (a).</p>
Correct Answer: A