(a) Eliminate the parameter to find a Cartesian equation of the curve.
(b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.
$$x = \sqrt {t + 1}$$
$$y = \sqrt {t - 1}$$
a) $y = \sqrt {t - 1} \implies t = y^2 + 1 \implies x = \sqrt {y^2 + 2} \implies x^2 - y^2 = 2$
b) Now I'm not sure how to sketch this since its undefined at certain points, also I feel that I'll have a hard time finding the direction the parameter increases
Your answer for (a) is correct. Putting that into standard form, we get
$$\frac {x^2}2 - \frac {y^2}2 = 1$$
We therefore see that the graph is a hyperbola centered at the origin with branches to the left and the right. (You do know enough analytical geometry to see that, I hope!) At least, your graph is part of that hyperbola. You need to find which part.
Since both $x$ and $y$ result from square roots, we see that $x \ge 0$ and $y \ge 0$. This means the graph is limited to the first quadrant, cutting out the other three-fourths of the hyperbola. We next want to find the domain for the parameter $t$. From the square roots in the formulas for $x$ and $y$ we see that
$$t+1 \ge 0 \text { and } t-1 \ge 0$$
The intersection for that is
$$t \ge 1$$
Setting $t=1$ we get $x = \sqrt 2$, $y=0$. As $t$ increases both $x$ and $y$ increase, up to infinity. We therefore get confirmation that our graph is the portion of the hyperbola in the first quadrant.
You can plot a few more values of $x$ and $y$ for $t$-values greater than 1, and combine that with your knowledge of analytical geometry that the asymptote is $y=x$ to sketch the graph. Plotting these points in order will confirm the direction of our arrow, which will point up and to the right, toward the corner of your graph area.
Of course, if you have a graphing calculator or program you can easily check all this, but we saw that was not actually necessary. Here is the graph on my calculator, which agrees with our analysis.