Applications of Derivatives
Tangent to Curves
Grade 12
Question:
<p>\(P_r(x_r, y_r);\, r = 1, 2, 3, \ldots\) is a sequence of points on the curve \(y^3 = x\). The tangent at \(P_n\) cuts the curve again at \(P_{n+1}\) for all \(n \in \mathbb{N}\). Then \(y_1, y_2, y_3, \ldots\) are in</p>
<p>A. AP</p>
<p>B. GP</p>
<p>C. HP</p>
<p>D. none of these</p>
Step-by-Step Solution
Key Concept: Find the tangent line at point P_n on y³ = x, then determine where it intersects the curve again to get P_{n+1}, and identify the relationship between consecutive y-coordinates.
<p><strong>Step 1:</strong> For point P_n(x_n, y_n) on curve y³ = x, we have x_n = y_n³.</p><p><strong>Step 2:</strong> Find the tangent at P_n. From y³ = x, differentiate: 3y² dy/dx = 1, so dy/dx = 1/(3y²). At P_n, slope = 1/(3y_n²).</p><p><strong>Step 3:</strong> Tangent equation: y - y_n = (1/(3y_n²))(x - y_n³), which gives 3y_n²y - 3y_n³ = x - y_n³, so x = 3y_n²y - 2y_n³.</p><p><strong>Step 4:</strong> Substitute into curve y³ = x: y³ = 3y_n²y - 2y_n³.</p><p><strong>Step 5:</strong> Rearrange: y³ - 3y_n²y + 2y_n³ = 0. This cubic factors as (y - y_n)²(y + 2y_n) = 0.</p><p><strong>Step 6:</strong> The roots are y = y_n (double root, tangency point) and y = -2y_n (the other intersection point P_{n+1}).</p><p><strong>Step 7:</strong> Therefore y_{n+1} = -2y_n, giving the ratio y_{n+1}/y_n = -2 (constant).</p><p>∴ The sequence y₁, y₂, y₃, ... is in <strong>Geometric Progression with common ratio -2</strong> (Answer: B)</p>
Correct Answer: B