Answer
(a) See the graph.
(b) $x^2=y+1,~~x\geq1,~~y\geq0$
Work Step by Step
(a)
$t=0$:
$x=sec~0=1$
$y=tan^20=0$
Point: $(1,0)$
$t=\frac{\pi}{4}$:
$x=sec~\frac{\pi}{4}=\sqrt 2$
$y=tan^2\frac{\pi}{4}=1$
Point: $(\sqrt 2,1)$
When $t→\frac{\pi}{2}$, $x→+∞$ and $y→+∞$. So, $x\geq1$ and $y\geq0$
(b)
$x=sec~t$
$x^2=sec^2t$
$y=tan^2t$
$sec^2t=tan^2t+1$
$x^2=y+1$