Parabola
Tangent to Parabola
Grade 11
Question:
<p>For the curve \(x = 4t^2 + 3\) and \(y = 8t^3 - 1\), the tangent at parameter \(t\) passes through another point \(Q\) on the curve at parameter \(t_1\). Find the coordinates of \(Q\).</p>
<p>\(Q\left(4t^2 + 3,\, 8t^3 - 1\right)\)</p>
<p>\(Q\left(t^2 + 3,\, -t^3 - 1\right)\)</p>
<p>\(Q\left(4\cdot\dfrac{t^2}{4} + 3,\, -8\cdot\dfrac{t^3}{8} - 1\right)\)</p>
<p>\(Q\left(4t^2 + 3,\, -t^3 - 1\right)\)</p>
Step-by-Step Solution
Key Concept: Use parametric differentiation to find the tangent line equation at parameter t, then substitute another point (4t₁² + 3, 8t₁³ - 1) on the curve into this tangent equation to find the relationship between t and t₁.
<p><strong>Step 1:</strong> Find the slope of tangent at parameter t.</p><p>dx/dt = 8t, dy/dt = 24t²</p><p>dy/dx = 24t²/(8t) = 3t</p><p><strong>Step 2:</strong> Write the equation of tangent at point (4t² + 3, 8t³ - 1).</p><p>y - (8t³ - 1) = 3t(x - (4t² + 3))</p><p>y = 3tx - 12t³ + 8t³ - 1</p><p>y = 3tx - 4t³ - 1</p><p><strong>Step 3:</strong> The tangent passes through Q at parameter t₁: (4t₁² + 3, 8t₁³ - 1).</p><p>Substitute into tangent equation:</p><p>8t₁³ - 1 = 3t(4t₁² + 3) - 4t³ - 1</p><p>8t₁³ = 12tt₁² + 9t - 4t³</p><p><strong>Step 4:</strong> Rearrange to find relationship:</p><p>8t₁³ - 12tt₁² + 4t³ - 9t = 0</p><p>Factor: (2t₁ - t)(4t₁² - 4tt₁ - 4t²) - 9t = 0</p><p>This yields: t₁ = -t/2</p><p><strong>Step 5:</strong> Find coordinates of Q:</p><p>x = 4(-t/2)² + 3 = t² + 3</p><p>y = 8(-t/2)³ - 1 = -t³ - 1</p><p>∴ Answer: Q = (t² + 3, -t³ - 1)</p>
Correct Answer: D