Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 11 - Section 11.4 - Shifted Conics - 11.4 Exercises - Page 815: 60

Answer

The graph of $4x^2+9y^2-36y=0$ Parametric equations: $x=3~cos~t$ $y=2+2~sin~t$

Work Step by Step

If you can not insert the function into your graphics device as shown, do the following: $4x^2+9y^2-36y=0$ $4x^2+9y^2-36y+36=36$ $4x^2+9(y^2-4y+4)=36$ $4x^2+9(y-2)^2=36$ $\frac{x^2}{9}+\frac{(y-2)^2}{4}=1$ $(\frac{x}{3})^2+(\frac{y-2}{2})^2=1$ Now, use parametric equations: Make: $\frac{x}{3}=cos~t$ $x=3~cos~t$ $\frac{y-2}{2}=sin~t$ $y-2=2~sin~t$ $y=2+2~sin~t$ Remember: $cos^2t+sin^2t=1$
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