Multivariable Calculus, 7th Edition

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 711: 53

Answer

$\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$

Work Step by Step

We need to re-arrange the given equations as: $x^2=4(\dfrac{-1}{4})(y-100)$ Vertex: $(0,100)$ and Focus:F $(0,100-\dfrac{1}{4})=F(0,\dfrac{399}{4})$ with Centre: $(0,\dfrac{399}{8})$ The equation of the ellipse: $\dfrac{x^2}{b^2}+\dfrac{(y-\dfrac{399}{8})^2}{a^2}=1$ Eccentricity $e$ of the ellipse: $ae=\dfrac{399}{8}$ $\implies a=100-\dfrac{399}{8}=\dfrac{401}{8}$; Also, we have $b^2=a^2(1-e^2)=a^2-(ae)^2$ This gives: $b^2=(\dfrac{401}{8})^2-(\dfrac{399}{8})^2=25$ Now, the equation of the ellipse is: $\dfrac{x^2}{25}+\dfrac{(y-399/8)^2}{(401/8)^2}=1$ Hence, we have $\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
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