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.

Method 1 Eliminating. I think they want to write everything in terms of x first. 
$x = 1 + ln(t) \iff e^{x - 1} = t$
$y = t^2 + 2 = e^{2x -2} + 2 \implies y' = 2e^{2x -2}$
At (1,3) $y' = 2$. So the tangent line is $y = 2(x- 1) + 3$ or parametrically let $x - 1 = t 
\iff x = 1 + t$ and $y = 2t + 3$
Method 2 No eliminating. 
Let $r = (1 + ln(t),2 + t^2) \implies r' = (1/t, 2t)$. At (1,3), $t = 1$, therefore $r'(1) = (1,2)$
$r = (1,3) + s(1,2)$ which gives you $x = 1 + s$ and $y = 3 + 2s$
