Applications of Derivatives
Tangent to a curve
Grade 12
Question:
<p>Let \(P_1 = (t_1^2,\, \sqrt{a}\, t_1^3)\) and \(P_2 = (t_2^2,\, \sqrt{a}\, t_2^3)\) be two points on the curve \(y^2 = ax^3\). The tangent at \(P_1\) passes through \(P_2\). Which of the following relations holds?</p>
<p>\(t_1 + 2t_2 = 0\)</p>
<p>\(t_1^2 - t_2^2 + t_1 t_2 = 0\)</p>
<p>\((t_1 + 2t_2)(t_1 - t_2) = 0\)</p>
<p>\(3(t_1)(t_1 + t_2) = 2(t_1^2 + t_2^2 + 2t_1 t_2)\)</p>
Step-by-Step Solution
Key Concept: Find the tangent line at P₁ using implicit differentiation of y² = ax³, then use the condition that this tangent passes through P₂ to establish a relationship between t₁ and t₂.
<p><strong>Step 1: Find dy/dx at P₁</strong></p><p>From y² = ax³, differentiating implicitly: 2y(dy/dx) = 3ax²</p><p>∴ dy/dx = 3ax²/(2y)</p><p>At P₁(t₁², √a·t₁³): dy/dx = 3a(t₁²)²/(2√a·t₁³) = 3at₁⁴/(2√a·t₁³) = (3√a·t₁)/2</p><p><strong>Step 2: Write equation of tangent at P₁</strong></p><p>Tangent line: y - √a·t₁³ = (3√a·t₁/2)(x - t₁²)</p><p><strong>Step 3: Use condition that P₂ lies on this tangent</strong></p><p>Substituting P₂(t₂², √a·t₂³):</p><p>√a·t₂³ - √a·t₁³ = (3√a·t₁/2)(t₂² - t₁²)</p><p>Dividing by √a: t₂³ - t₁³ = (3t₁/2)(t₂² - t₁²)</p><p><strong>Step 4: Factor and simplify</strong></p><p>(t₂ - t₁)(t₂² + t₁t₂ + t₁²) = (3t₁/2)(t₂ - t₁)(t₂ + t₁)</p><p>Since t₂ ≠ t₁: t₂² + t₁t₂ + t₁² = (3t₁/2)(t₂ + t₁)</p><p>Multiply by 2: 2t₂² + 2t₁t₂ + 2t₁² = 3t₁t₂ + 3t₁²</p><p>∴ 2t₂² - t₁t₂ - t₁² = 0</p><p><strong>Step 5: Factor</strong></p><p>(2t₂ + t₁)(t₂ - t₁) = 0</p><p>Since t₂ ≠ t₁: 2t₂ + t₁ = 0</p><p>∴ <strong>t₁ + 2t₂ = 0</strong> or equivalently <strong>t₁ = -2t₂</strong></p>
Correct Answer: C