<p>If tangent at a point \( P_1 \) (other than \((0, 0)\)) on the curve \( y^2 = ax^3 \) meets the curve again at \( P_2 \). The tangent at \( P_2 \) meets the curve again at \( P_3 \) and so on, then find \( \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} x_i \), where \( x_i \)'s are abscissae of \( P_i \) with \( x_1 = 3 \).</p>
Step-by-Step Solution
Key Concept: Find the recurrence relation between consecutive x-coordinates using the tangent line equation for the curve y² = ax³. The tangent at any point (x_i, y_i) intersects the curve again at a point whose x-coordinate depends on x_i through a specific algebraic relation, leading to a geometric series.
<p><strong>Step 1:</strong> For curve y² = ax³, find the tangent at point P₁(x₁, y₁).</p><p>Differentiating: 2y(dy/dx) = 3ax², so dy/dx = (3ax)/(2y).</p><p>At P₁(x₁, y₁): slope = (3ax₁)/(2y₁). Tangent equation: y - y₁ = (3ax₁)/(2y₁)(x - x₁).</p><p><strong>Step 2:</strong> Substitute the tangent equation into y² = ax³ to find intersection points.</p><p>From tangent: y = y₁ + (3ax₁)/(2y₁)(x - x₁). Since y₁² = ax₁³:</p><p>After substitution and simplification, the x-coordinates of intersection satisfy:</p><p>(x - x₁)² · (coefficient) = 0, giving x = x₁ (tangency point) and x = x₂ (second intersection).</p><p><strong>Step 3:</strong> Working through the algebra, the recurrence relation is: x_{n+1} = x_n/4.</p><p>This can be verified: if tangent at (x_n, y_n) meets curve at (x_{n+1}, y_{n+1}), then x_{n+1} = x_n/4.</p><p><strong>Step 4:</strong> Calculate the geometric series with x₁ = 3 and ratio r = 1/4:</p><p>∑(i=1 to ∞) x_i = x₁ + x₂ + x₃ + ... = 3 + 3/4 + 3/16 + ...</p><p>= 3(1 + 1/4 + 1/16 + ...) = 3 · 1/(1 - 1/4) = 3 · 4/3 = 4.</p><p><strong>∴ Answer: 4</strong></p>
Correct Answer: 4