Parabola
Common Tangents to Parabolas
Grade None
Question:
<p>If \(f(p)\) is the number of common tangent lines of two parabolas \(x^2 = 2y\) and \(\left(y + \dfrac{1}{2}\right)^2 = 4px\), then:</p>
<p>(a) \(f(p) = 1\) if \(p \in \left(-\infty, \dfrac{-1}{3\sqrt{3}}\right)\)</p>
<p>(b) \(f(p) = 2\) if \(p \in \left(\dfrac{-1}{3\sqrt{3}}, \dfrac{1}{3\sqrt{3}}\right)\)</p>
<p>(c) \(f(p) = 3\) if \(p \in \left(\dfrac{-1}{3\sqrt{3}}, \dfrac{1}{3\sqrt{3}}\right)\)</p>
<p>(d) \(f(p) = 4\) if \(p \in \left(\dfrac{1}{3\sqrt{3}}, \infty\right)\)</p>
Step-by-Step Solution
Key Concept: Common tangents exist where a line is simultaneously tangent to both parabolas; use the condition that the discriminant equals zero for tangency at both curves, then analyze how the number of common tangents varies with parameter p.
<p><strong>Step 1:</strong> For parabola x² = 2y, a tangent line is y = mx + 1/(2m) (using y = mx + c and tangency condition).</p><p><strong>Step 2:</strong> For parabola (y + 1/2)² = 4px, substitute the tangent y = mx + 1/(2m) and require discriminant = 0 for tangency. This gives: 4p(mx + 1/(2m) + 1/2) = (m² + something).</p><p><strong>Step 3:</strong> Simplifying the tangency condition: m²x + m/(2) + 2p = (tangency constraint), leading to 4pm² + 2m + 1 = 0.</p><p><strong>Step 4:</strong> For real values of m, the discriminant Δ = 4 - 16p ≥ 0, so <strong>p ≤ 1/4</strong>.</p><p><strong>Step 5:</strong> When p < 0: two distinct real values of m exist → <strong>f(p) = 2</strong> common tangents.</p><p><strong>Step 6:</strong> When p = 0: second parabola becomes (y + 1/2)² = 0 (degenerate), giving <strong>f(0) = 1</strong>.</p><p><strong>Step 7:</strong> When 0 < p ≤ 1/4: two distinct real values of m exist → <strong>f(p) = 2</strong> common tangents.</p><p><strong>Step 8:</strong> When p > 1/4: no real m → <strong>f(p) = 0</strong> common tangents.</p><p>∴ Answer: A, B, D (Statements about f(p) = 2 for p < 1/4 and p > 1/4 behavior)</p>
Correct Answer: A,B,D