Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.5 - Parametric Equations - Exercise Set - Page 1019: 42


$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$; Ellipse equation

Work Step by Step

Re-arrange the parametric equations as shown below: $\cos t=\dfrac{x-h}{a}$ and $\sin t=\dfrac{y-k}{b}$ Since, $\sin^2 t+\cos^2 t=1$ $\implies (\dfrac{y-k}{b})^2 + (\dfrac{x-h}{a})^2=1$. or, $\dfrac{(y-k)^2}{b^2}+\dfrac{(x-h)^2}{a^2}=1`$ Therefore, $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ This is the ellipse equation
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