Equation of Line given Parametric Curve

Question

Given the lines:

x= t^3 -9t

y= 9t^2 - t^4

What is the equation of the tangent line at t=3 and t= -3.

What I did was first plug in the t-values into the x and y equations to find the points of the line. Then I solved for the derivatives of both x and y, then put it in the form of dy/dx. After, I plugged in the points to try to get it in y=mx+b, but I still am getting it wrong. It would really help if you solved it step-by-step, so that I can see where I made my mistakes! Thanks!

Answer 1

When t = 3, x = y = 0. So, (0,0) is the point of tangency.

Slope of tangent line = dy/dx evaluated at t = 3.

dy/dx = (dy/dt) / (dx/dt) = (18t - 4t³) / (3t² - 9) = -54 / 18 = -3

Equation of tangent line: y - 0 = -3(x - 0). So, y = 3x.

To find the tangent line when t = -3, proceed in a similar manner.