so the sun is one of the foci, and we have an eccentricity of 0.97. So the ellipse will look more or less like the picture below.
[tex] \bf \textit{ellipse, horizontal major axis}\\\\\cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1\qquad \begin{cases}center\ ( h, k)\\vertices\ ( h\pm a, k)\\c=\textit{distance from}\\\qquad \textit{center to foci}\\\qquad \sqrt{ a ^2- b ^2}\\eccentricity\quad e=\cfrac{c}{a}\end{cases}\\\\-------------------------------\\\\\begin{cases}h=0\\k=0\\a=35.88\\e=0.97\end{cases}\implies \cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{b^2}=1\\\\------------------------------- [/tex]
[tex] \bf e=\cfrac{c}{a}\implies 0.97=\cfrac{c}{35.88}\implies \boxed{34.8036=c}
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c=\sqrt{a^2-b^2}\implies b=\sqrt{a^2-c^2}\implies b=\sqrt{35.88^2-34.8036^2}
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\boxed{b\approx 8.72}\\\\
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\cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{8.72^2}=1\implies \cfrac{x^2}{1287.3744}+\cfrac{y^2}{76.0384}=1 [/tex]
