The slope of tangent at a point say #t=t_1#, is given by the value of first derivative of the function at that point and given slope #f'(t_1)# and the point on curve #(x(t_1),y(t_1))#, the equation of tangent is #y-y(t_1)=f'(t_1)(x-x(t_1))#.
As function is #f(t)=(7t^2+4,3t^3)# and at #t=3#, point is #(7*3^2+4,3*3^3)# i.e. #(67,81)#
#f'(t)=((dy)/(dt))/((dx)/(dt))=(14t)/(9t^2)=14/(9t)# and at #t=3# slope is #14/27#
and equation of tangent is
#y-81=14/27(x-67)# .................(1)
Parametric form of equation of a line is #x=l t+x_0# and #y=mt+y_0#, where line passes through #(x_0,y_0)# i.e. #(y-y_0)/m=(x-x_0)/l# and hence slope is #m/l#
Hence we can write (1) as
#x=27t+67# and #y=14t+81#
Observe that tangent at #t=t_0# at curve #f(t)=(x(t),y(t))#, we can write parametric equation of tangent as
#x=x'(t_0)t+x(t_0)# and #y=y'(t_0)t+y(t_0)#
