Find equations of the tangents to the curve x = 9t^2 + 6, y = 6t^3 + 6 that pass through the point (15, 12).

Answer 1

dy/dx = (dy/dt)/(dx/dt) = 18t²/(18t) = t; 6t³ + 6 = 12; t³ = 1; t = 1;

dy/dx = 1. Equation of tangent line: y = 12 + 1(x - 15); y = x - 3