Answer
$$ y=\pm \sqrt{1+x^2}.$$
Work Step by Step
We have $x=\tan t$, $y=\sec t$.
Use the trig identity:
$$1+\tan^2 t=\sec^2t$$
Combine with $x$ and $y$:
$$y^2=1+x^2.$$
Solve for $y$:
$$ y=\pm \sqrt{1+x^2}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 638: 12
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