## Calculus 8th Edition

Published by Cengage

# Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 731: 53

#### Answer

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

#### Work Step by Step

Re-arrange the given equation as: $x^2=4(\dfrac{-1}{4})(y-100)$ We are given that Vertex: $(0,100)$ and Focus: $F(0,100-\dfrac{1}{4})=F(0,\dfrac{399}{4})$ As the mid-point of the foci is center of the ellipse, then we have: Center: $(0,\dfrac{399}{8})$ Now, the form of the ellipse as follows: $\dfrac{(x-0)^2}{a^2}+\dfrac{(y-\dfrac{399}{8})^2}{b^2}=1$ or, $\dfrac{x^2}{b^2}+\dfrac{(y-\dfrac{399}{8})^2}{a^2}=1$ The eccentricity of the ellipse is defined as: $ae=\dfrac{399}{8}$ $\implies a=100-\dfrac{399}{8}=\dfrac{401}{8}$ and $b^2=a^2(1-e^2)=a^2-(ae)^2$ $\implies b^2=(\dfrac{401}{8})^2-(\dfrac{399}{8})^2=25$ So, we get the equation of ellipse as: $\dfrac{x^2}{25}+\dfrac{(y-\dfrac{399}{8})^2}{(\dfrac{401}{8})^2}=1$ Hence, $\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$

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