Answer
$y=2 x+1$.
Work Step by Step
Given
$$x=1+\ln t,\ \ \ y=t^2+2 \ \ \ ;(1,3)$$
(a) Since at $(1,3)$, we have
\begin{aligned} x&=1+\ln t\\
1&=1+\ln (t)\\
0&=\ln (t)\\
t&=1 \end{aligned}
and
\begin{aligned}
\frac{d y}{d t}=2 t, \ \ \ \ \frac{d x}{d t}=\frac{1}{t}\end{aligned}
Then
\begin{aligned}
\frac{d y}{d x}&=\frac{d y / d t}{d x / d t}\\
&=\frac{2 t}{1 / t}\\
&=2 t^2
\end{aligned}
Hence the slope at $(1,3)$
\begin{aligned}
m&= 2(1)^2=2
\end{aligned}
The tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-3}{x-1}&=2\\
y-3&= 2x-2\\
y&=2x+1
\end{aligned}
(b) Since
$$x=1+\ln t \Rightarrow \ln t=x-1 \Rightarrow t=e^{x-1}$$
Then
\begin{aligned}
y&= t^2+2\\
&=\left(e^{x-1}\right)^2+2\\
&=e^{2 x-2}+2
\end{aligned}
At $(1,3)$, we have
$$y^{\prime}=e^{2(1)-2} \cdot 2=2$$
Then the tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-3}{x-1}&=2\\
y-3&= 2x-2\\
y&=2x+1
\end{aligned}
