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

Question

The problem asks for the smaller and larger slope, I'm a bit confused on what that means and how to find them. The way we did them in class the problems weren't phrased this way.

Answer 1

Slope of tangent line m =dy/dx = (dy/dt)/(dx/dt) = 18t²/18t = t. To get t corresponding to point (15, 12).

t is common solution of equations 9t² + 6 = 15 and 6t³ + 6 = 12

For first equation t = ±1 and from second t = 1. Common t = 1

m = t at t = 1, so m = 1. Equation of tangent line y = 12 + 1(x - 15) or y = x - 3