Tangent line of parametric curve I have not seen a problem like this so I have no idea what to do.
Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.
$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$
I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$
But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.

2nd method: eliminating parameter:
$x=1+ \ln t , y=t^2+2 \Leftrightarrow  t=\exp(x-1), y=2+\exp(2x-2)$
Consider the function:  $f(x)=2+ \exp(2x-2)$, then: $f'(x)=2 \exp(2x-2)$
The  tangent at $x=1$ has equation: $Y=f'(1)(X-1)+ f(1)$, thus: $Y=2(X-1)+3$, thus:$$Y=2X+1$$
