## Calculus: Early Transcendentals 8th Edition

$\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
Given: $x^2+y=100$ Re-arrange as: $x^2=4(\dfrac{-1}{4})(y-100)$ Here, we have the vertex: $(0,100)$ and the focus is: $(0,100-\dfrac{1}{4})=(0,\dfrac{399}{4})$ Since, the mid-point of the foci is the center of the ellipse, we get Center: $(0,\dfrac{399}{8})$ Thus, the equation of the ellipse is: $\dfrac{(x-0)^2}{a^2}+\dfrac{(y-\dfrac{399}{8})^2}{b^2}=1$ or, $\dfrac{x^2}{b^2}+\dfrac{(y-\dfrac{399}{8})^2}{a^2}=1$ ...(1) The eccentricity of the ellipse is: $ae=\dfrac{399}{8}$ Thus, we have $a=100-\dfrac{399}{8}=\dfrac{401}{8}$ and $b^2=a^2(1-e^2)=a^2-(ae)^2$ This gives: $b^2=(\dfrac{401}{8})^2-(\dfrac{399}{8})^2=25$ Therefore, equation (1) of the ellipse becomes: $\dfrac{x^2}{25}+\dfrac{(y-\dfrac{399}{8})^2}{(\dfrac{401}{8})^2}=1$ Hence, $\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$