Answer
$$y=4ex-7e$$
Work Step by Step
$a)$ To find the slope $m$ of the tangent line, use the following formula:
$$m = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$ individually, then take the quotient to obtain $m$.
$$\frac{dy}{dt} = \frac{d}{dt}(e^{t^2}) = 2te^{t^2}$$$$\frac{dx}{dt} = \frac{d}{dt}(1+\sqrt{t}) = \frac{1}{2\sqrt{t}}$$$$m = \frac{2te^{t^2}}{(\frac{1}{2\sqrt{t}})}=4\sqrt{t^3}e^{t^2}$$
Since we want the slope at the point $(2,e),$ find the value of $t$ at this point. The $x$-coordinate is $2,$ so set the equation for $x$ equal to $2$ and solve for $t$.
$$x = 1 + \sqrt t = 2$$$$t = 1$$
Evaluate $m$ at $t=1$.
$$m=4\sqrt{(1)^3}e^{(1)^2}=4e$$
Now use the point-slope formula to find the equation of the tangent line at $(2,e)$:
$$y-y_{0}=m(x-x_{0})$$$$y-e=4e(x-2)$$$$y=4ex-7e$$
$b)$ To eliminate the parameter, first solve for $t$ in terms of $x$:
$$x = 1 + \sqrt t$$$$t = (x-1)^2$$
Now plug this expression for $t$ into the equation for $y$:
$$y=e^{t^2}$$$$y=e^{(x-1)^4}$$
Take the derivative with respect to $x$ to find the slope:
$$\frac{dy}{dx}=4(x-1)^3e^{(x-1)^4}$$
Evaluate $\frac{dy}{dx}$ at the point $(2,e)$ to find $m$ at that point:
$$m=4((2)-1)^3e^{((2)-1)^4}=4e$$
Again, use the point-slope formula to find the equation of the tangent line at $(2,e)$ to get:
$$y=4ex-7e$$
