Let x(t) = 3 + 3/t and y(t) = 3t - 3/t
dx/dt = -3/t^2 and dy/dt = 3 + 3/t^2
slope = dy/dt / dx/dt = dy/dx = (3 + 3/t^2) / (-3/t^2) = -(t^2 + 1)
At t = 2, the slope of the tangent is -5 and the cartesian points x(2) = 9/2 and y(2) = 9/2 are calculated
The point-slope form of the normal to the curve at t = 2 is then given by
y - y1 = -1/slope(x - x1) or y - 9/2 = 1/5(x - 9/2) or y = 1/5(x - 9/2) + 9/2