Find an equation of the tangent line to the curve for the given value of t.

Question

x=t^2-2t, y=t^2+2t when t=1

Answer 1

dx/dt = 2t-2
dy/dt = 2t+2

when t=1
slope = dy/dx = (dy/dt) / (dx/dt) = 4/0 , ahhh, a vertical line

when t=1
x = 1-2 = -1
y = 1 + 2 = 3

so a vertical line through (-1,2) is

x = -1