Parametric equation: x=3cos(t), y=2sin(t)... How do I find an equation of the tangent line to the graph at the point where T=(pi)/6 ??

x = 3 cos(t) --- y = 2 sin(t)
cos^2(t) = x^2 /9 ---- sin^2(t) = y^2 /4 ---- sin^2(t) + cos^2(t) = 1
1 =(x^2 /9) + (y^2 /4) <<< ellipse
---- point --- t = pi/6 --- x = 3(sq rt 3)/2 --- y = 1
0 = (2x /9) dx + (2y / 4) dy
dy/dx = -(2x /9)/(y / 2)
dy/dx = -(2(3(sq rt 3)/2) /9)/(1 / 2)
dy/dx = -(2sqrt3)/3 <<< slope

y - y1 = m(x-x1)
y = -(2sqrt3)/3 (x - 3(sq rt 3)/2) + 1 <<< simplify from there