Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 604: 63


(55, 0)

Work Step by Step

Slope of the tangent line= $c'(t)=\frac{(t^{2}-8t)'}{(t^{2}-9)'}=\frac{2t-8}{2t}=\frac{1}{2}$ $\implies 2t-8=t$ $\implies t=8$ When $t=8$, the point is $ ((8)^{2}-9, (8)^{2}-8(8))=(55,0)$
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