I am having difficulty finding parametric equations of the curve of intersection of $z = x^2 - y^2$ and cylinder $x^2 + y^2 = 1$
I am aware that the first equation represents a hyperboloid, and the second equation represents a cylinder. I think for the second one I can just do $x = cos(t)$ and $y = sin(t)$. Not sure about the other one.
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$\begingroup$We have the two surfaces
$$ x^2-y^2 = z\\ x^2+y^2 = 1 $$
Solving for $x^2, y^2$ we have
$$ x^2 = \frac 12(1+z)\\ y^2 = \frac 12(1-z) $$
so a parametric representation can be read as
$$ (x(s),y(s),z(s)) = \left(\pm\sqrt{\frac 12(1+s)},\pm\sqrt{\frac 12(1-s)},s\right) $$
for $-1\le s \le 1$
Attached a plot with the four leafs
As you point out, the second equation implies that $x = \cos(t)$ and $y = \sin(t)$. Substituting these into the first equation yields $$ z = x^2 - y^2 = \cos^2(t) - \sin^2(t) \, . $$ Thus the curve is parametrized by $$ (x(t), y(t), z(t)) = (\cos(t), \sin(t), \cos^2(t) - \sin^2(t)) = (\cos(t), \sin(t), \cos(2t)) \, . $$
Here's a SageMathCell illustrating the surfaces and curve of intersection, which produces the following plot.
$\hspace 4cm$
For the first one use that $$\cosh^2(t)-\sinh^2(t)=1$$ And since$$x^2=z+y^2$$ we get
$$z+2y^2=1$$ so.... $$z=1-2y^2$$
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