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.

One way to do this is by considering the parametric form of the curve:
$(x,y)(t) = (1 + \log t, t^2 + 2)$, so $(x,y)'(t) = (\frac{1}{t}, 2t)$
We need to find the value of $t$ when $(x,y)(t) = (1 + \log t, t^2 + 2) = (1,3)$, from where we deduce $t=1$. The tangent line at $(1,3)$ has direction vector $(x,y)'(1) = (1,2)$, and since it passes by the point $(1,3)$ its parametric equation is given by: $s \mapsto (1,2)t + (1,3)$.
Another way (I suppose this is eliminating the parameter) would be to express $y$ in terms of $x$ (this can't be done for any curve, but in this case it is possible). We solve for $t$: $x = 1 + \log x \Rightarrow x = e^{x-1}$, so $y = t^2 + 2 = (e^{x-1})^2 + 2 = e^{2x-2} + 2$. The tangent line has slope $\frac{dy}{dx}=y_x$ evaluated at $1$: we have $y_x=2e^{2x-2}$ and $y_x(1)=2e^0 = 2$, so it the line has equation $y=2x +b$. Also, it passes by the point $(1,3)$, so we can solve for $b$: $3 = 2 \cdot 1 + b \Rightarrow b = 1$. Then, the equation of the tangent line is $y = 2x + 1$.
Note that $s \mapsto (1,2)t + (1,3)$ and $y = 2x + 1$ are the same line, expressed in different forms.
