Answer
$x=y^2-8y+12, 1\leq y\leq 6$
Work Step by Step
$x=t^2+4t=(2-y)^2+4(2-y)$
$x=y^2-8y+12$
Therefore,
$x=y^2-8y+12, 1\leq y\leq 6$
Also see attached graph.
Multivariable Calculus, 7th Edition
by Stewart, James
Published by
Brooks Cole
ISBN 10:
0-53849-787-4
ISBN 13:
978-0-53849-787-9
Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 710: 1
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