Applications of Derivatives
Tangents to curves
Grade 12
Question:
<p>If the tangent at a point \(P\), with parameter \(t\), on the curve \(x = 4t^2 + 3\), \(y = 8t^3 - 1\), \(t \in \mathbb{R}\), meets the curve again at a point \(Q\), then the coordinates of \(Q\) are</p>
<p>\((16t^2 + 3, -64t^3 - 1)\)</p>
<p>\((4t^2 + 3, -8t^3 - 1)\)</p>
<p>\((t^2 + 3, t^3 - 1)\)</p>
<p>\((t^2 + 3, -t^3 - 1)\)</p>
Step-by-Step Solution
Key Concept: The tangent line at parameter t must satisfy the curve equation at another point. Substitute the tangent equation into the parametric equations to find where it intersects the curve again, which gives a cubic equation with t as a double root.
<p><strong>Step 1:</strong> Find the slope at point P with 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 tangent line equation at P(4t² + 3, 8t³ - 1).</p><p>y - (8t³ - 1) = 3t(x - (4t² + 3))</p><p>y = 3tx - 12t³ + 8t³ - 1 = 3tx - 4t³ - 1</p><p><strong>Step 3:</strong> Find intersection with the curve by substituting parametric equations.</p><p>8s³ - 1 = 3t(4s²+ 3) - 4t³ - 1</p><p>8s³ = 12ts² + 9t - 4t³</p><p>8s³ - 12ts² - 9t + 4t³ = 0</p><p><strong>Step 4:</strong> This cubic has s = t as a double root (tangent point). Factor out (s - t)².</p><p>8s³ - 12ts² - 9t + 4t³ = (s - t)²(8s + at + b)</p><p>By expansion and comparison: (s - t)²(8s + 4t) = 0</p><p>The third root is s = -t/2</p><p><strong>Step 5:</strong> Find coordinates of Q at parameter s = -t/2.</p><p>x = 4(-t/2)² + 3 = 4(t²/4) + 3 = t² + 3</p><p>y = 8(-t/2)³ - 1 = 8(-t³/8) - 1 = -t³ - 1</p><p>∴ Answer: <strong>Q = (t² + 3, -t³ - 1)</strong></p>
Correct Answer: A