Answer
$y =4ex-7e$
Work Step by Step
Given
$$x=1+\sqrt{ t},\ \ \ y=e^{t^2} \ \ \ ;(2,e)$$
(a) Since at $(2,e)$, we have
\begin{aligned} x&=1+\sqrt{ t}\\
2&=1+\sqrt{ t}\\
1&=1+\sqrt{ t}\\
t&=1 \end{aligned}
and
\begin{aligned}
\frac{d y}{d t}=2 te^{t^2}, \ \ \ \ \frac{d x}{d t}=\frac{1}{2\sqrt{t}}\end{aligned}
Then
\begin{aligned}
\frac{d y}{d x}&=\frac{d y / d t}{d x / d t}\\
&=\frac{2 te^{t^2}}{1 /2 \sqrt{t}}\\
&=4 t\sqrt{t}e^{t^2}
\end{aligned}
Hence the slope at $(2,e)$
\begin{aligned}
m&= 4e
\end{aligned}
The tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-e}{x-2}&=4e\\
y-e&= 4ex-8e\\
y&=4ex-7e
\end{aligned}
(b) Since
$$x=1+\sqrt{ t} \Rightarrow \sqrt{ t}=x-1 \Rightarrow t=(x-1)^2$$
Then
\begin{aligned}
y&= e^{t^2}\\
&=\left(e^{(x-1)^4}\right)
\end{aligned}
At $(2,e)$, we have
$$y^{\prime}= 4(x-1)^3e^{(x-1)^4}\ \ \ \to \ \ m= 4e$$
Then the tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-e}{x-2}&=4e\\
y-e&= 4ex-8e\\
y&=4ex-7e
\end{aligned}
